Class 11 - Maths - Sequences Series

Exercise 9.1

Question 1:

Write the first five terms of the sequences whose nth term is an= n(n + 2).

Given, an= n(n + 2)

Substituting n = 1, 2, 3, 4, and 5, we get

a1 = 1(1 + 2) = 1 * 3 = 3

a2 = 2(2 + 2) = 2 * 4 = 8

a3 = 3(3 + 2) = 3 * 5 = 15

a2 = 4(4 + 2) = 4 * 6 = 24

a2 = 5(5 + 2) = 5 * 7 = 36

Therefore, the required terms are 3, 8, 15, 24, and 35.

Question 2:

Write the first five terms of the sequences whose nth term is an = n/(n + 1)

Given, an= n/(n + 1)

Substituting n = 1, 2, 3, 4, 5, we obtain

a1 = 1/(1 + 1) = 1/2

a2 = 2/(2 + 1) = 2/3

a1 = 3/(3 + 1) = 3/4

a1 = 4/(4 + 1) = 4/5

a1 = 5/(5 + 1) = 5/6

Therefore, the required terms are 1/2, 2/3, 3/4, 4/5 and 5/6

Question 3:

Write the first five terms of the sequences whose nth term is an = 2n

Given, an = 2n

Substituting n = 1, 2, 3, 4, 5, we obtain

a1 = 21  = 2

a2 = 22  = 4

a3 = 23  = 8

a4 = 24  = 16

a5 = 25  = 32

Therefore, the required terms are 2, 4, 8, 16, and 32.

Question 4:

Write the first five terms of the sequences whose nth term is an = (2n - 3)/6

Substituting n = 1, 2, 3, 4, 5, we obtain

a1 = (2 * 1 - 3)/6 = (2 - 3)/6 = -1/6

a2 = (2 * 2 - 3)/6 = (4 - 3)/6 = 1/6

a3 = (2 * 3 - 3)/6 = (6 - 3)/6 = 3/6 = 1/2

a4 = (2 * 4 - 3)/6 = (8 - 3)/6 = 5/6

a5 = (2 * 5 - 3)/6 = (10 - 3)/6 = 7/6

Therefore, the required terms are -1/6, 1/6, 1/2, 5/6 and 7/6

Question 5:

Write the first five terms of the sequences whose nth term is an = (-1)n-1 5n+1

Given, an = (-1)n-1 5n+1

Substituting n = 1, 2, 3, 4, 5, we obtain

a1 = (-1)1-1 51+1 = (-1)0 52 = 1 * 25 = 25

a2 = (-1)2-1 52+1 = (-1)1 53 = -1 * 125 = -125

a3 = (-1)3-1 53+1 = (-1)2 54 = 1 * 625 = 625

a4 = (-1)4-1 54+1 = (-1)3 55 = -1 * 3125 = -3125

a5 = (-1)5-1 55+1 = (-1)0 56 = 1 * 15625 = 15625

Therefore, the required terms are 25, –125, 625, –3125, and 15625.

Question 6:

Write the first five terms of the sequences whose nth term is an = n * (n2 + 5)/4

Given, an = n * (n2 + 5)/4

Substituting n = 1, 2, 3, 4, 5, we get

a1 = 1 * (12 + 5)/4 = 1 * 6/4 = 6/4 = 3/2

a2 = 2 * (22 + 5)/4 = 2 * 9/4 = 18/4 = 9/2

a3 = 3 * (32 + 5)/4 = 3 * 14/4 = 42/4 = 21/2

a4 = 4 * (42 + 5)/4 = 4 * 21/4 = 21

a5 = 5 * (52 + 5)/4 = 5 * 30/4 = 150/4 = 72/2

Therefore, the required terms are 3/2, 9/2, 21/2, 21 and 75/2

Question 7:

Find the 17th term in the following sequence whose nth term is an = 4n – 3, a17; a24

Given, an = 4n – 3

Substituting n = 17, we get

a17 = 4 * 17 – 3 = 68 – 3 = 65

Substituting n = 24, we obtain

a24 = 4 * 24 – 3 = 96 – 3 = 93

Question 8:

Find the 7th term in the following sequence whose nth term is an = n2/2n, a7

Given, an = n2/2n

Substituting n = 7, we obtain

a7 = 72/27 = 49/128

Question 9:

Find the 9th term in the following sequence whose nth term is an = (-1)n-1 n3, a9

Given, an = (-1)n-1 n3

Substituting n = 9, we obtain

a9 = (-1)9-1 93 = (-1)8 93 = 1 * 729 = 729

Question 10:

Find the 20th term in the following sequence whose nth term is an = n(n - 2)/(n + 3); a20

Given, an = n(n - 2)/(n + 3)

Substituting n = 20, we get

a20 = 20(20 - 2)/(20 + 3) = (20 * 18)/23 = 360/23

Question 11:

Write the first five terms of the following sequence and obtain the corresponding series:

a1 = 3, an = 3an-1 + 2 for all n > 1

Given, a1 = 3, an = 3an-1 + 2 for all n > 1

Now, a2 = 3a2-1 + 2 = 3a1 + 2 = 3 * 3 + 2 = 9 + 2 = 11

a3 = 3a3-1 + 2 = 3a2 + 2 = 3 * 11 + 2 = 33 + 2 = 35

a4 = 3a4-1 + 2 = 3a3 + 2 = 3 * 35 + 2 = 105 + 2 = 107

a5 = 3a5-1 + 2 = 3a4 + 2 = 3 * 107 + 2 = 321 + 2 = 323

Hence, the first five terms of the sequence are 3, 11, 35, 107, and 323.

The corresponding series is 3 + 11 + 35 + 107 + 323 + ……….

Question 12:

Write the first five terms of the following sequence and obtain the corresponding series:

a1 = -1, an = an-1/n, n ≥ 2

Given, a1 = -1, an = an-1/n, n ≥ 2

Now, a2 = a2-1/2 = a1/2 = -1/2

a3 = a3-1/3 = a2/3 = (-1/2)/3 = -1/6

a4 = a4-1/4 = a3/4 = (-1/6)/4 = -1/24

a5 = a5-1/5 = a4/5 = (-1/24)/5 = -1/120

Hence, the first five terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120

The corresponding series is (-1) + (-1/2) + (-1/6) + (-1/24) + (-1/120) + ………..

Question 13:

Write the first five terms of the following sequence and obtain the corresponding series:

a1 = a2 = 2, an = an-1 - 1, n > 2

Given, a1 = a2 = 2, an = an-1 - 1, n > 2

Now, a3 = a2 – 1 = 2 – 1 = 1

a4 = a3 – 1 = 1 – 1 = 0

a5 = a4 – 1 = 0 – 1 = -1

Hence, the first five terms of the sequence are 2, 2, 1, 0, and –1.

The corresponding series is 2 + 2 + 1 + 0 + (–1) + ………….

Question 14:

The Fibonacci sequence is defined by 1 = a1 = a2 and an = an-1 + an-2; n > 2

Find an+1/an for n = 1, 2, 3, 4, 5

Given, 1 = a1 = a2 and an = an-1 + an-2 , n > 2

Now, a3 = a3-1 + a3-2 = a2 + a1 = 1 + 1 = 2

a4 = a4-1 + a4-2 = a3 + a2 = 2 + 1 = 3

a5 = a5-1 + a5-2 = a4 + a3 = 3 + 2 = 5

a6 = a6-1 + a6-2 = a5 + a4 = 5 + 3 = 8

Again for n = 1, an+1/an = a1+1/a1 = a2/a1 = 1/1 = 2

for n = 2, an+1/an = a2+1/a2 = a3/a2 = 2/1 = 2

for n = 3, an+1/an = a3+1/a2 = a4/a3 = 3/2

for n = 4, an+1/an = a4+1/a4 = a5/a4 = 5/3

for n = 5, an+1/an = a5+1/a5 = a6/a5 = 8/5

Exercise 9.2

Question 1:

Find the sum of odd integers from 1 to 2001.

The odd integers from 1 to 2001 are 1, 3, 5 ...1999, 2001.

This sequence forms an A.P.

Here, first term, a = 1

Common difference, d = 2

Last term l = 2001

Now, l = a + (n - 1)d

=> 2001 = 1 + 2(n - 1)

=> 2001 - 1 = 2(n - 1)

=> 2000 = 2(n - 1)

=> n – 1 = 2000/2

=> n – 1 = 1000

=> n = 1000 + 1

=> n = 1001

Sum Sn = (n/2)[2a + (n - 1)d]

= (1001/2)[2 * 1 + (1001 - 1)2]

= 1001[1 + 1001 - 1]

= 1001 * 1001

= 1002001

Thus, the sum of odd numbers from 1 to 2001 is 1002001.

Question 2:

Find the sum of all natural numbers lying between 100 and 1000, which are multiples of 5.

The natural numbers lying between 100 and 1000, which are multiples of 5 are

105, 110, 115, .........., 995.

Here, first term, a = 105

Common difference, d = 5

Last term l = 995

Now, l = a + (n - 1)d

=> 995 = 105 + 5(n - 1)

=> 995 - 105 = 5(n - 1)

=> 890 = 5(n - 1)

=> n – 1 = 890/5

=> n – 1 = 178

=> n = 178 + 1

=> n = 179

Sum Sn = (n/2)[2a + (n - 1)d]

= (179/2)[2 * 105 + (179 - 1)5]

= (179/2)[2 * 105 + 178 * 5]

= 179[1 * 105 + 89 * 5]

= 179[105 + 445]

= 179 * 550

= 98450

Thus, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is

98450.

Question 3:

In an A.P, the first term is 2 and the sum of the first five terms is one-fourth of the next five terms. Show that 20th term is –112.

First term a = 2

Let d be the common difference of the A.P.

Therefore, the A.P. is 2, 2 + d, 2 + 2d, 2 + 3d, ...........

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

According to the given condition,

10 + 10d = (10 + 35d)/4

=> 4(10 + 10d) = 10 + 35d

=> 40 + 40d = 10 + 35d

=> 40d – 35d = 10 – 40

=> 5d = -30

=> d = -30/5

=> d = -6

Now, a20 = a + (20 - 1)d

=> a20 = 2 + (20 - 1)(-6)

=> a20 = 2 – 19 * 6

=> a20 = 2 – 114

=> a20 = -112

Thus, the 20th term of the A.P. is –112.

Question 4:

How many terms of the A.P.  -6, -11/2, -5, ……… are needed to give the sum –25?

Let the sum of n terms of the given A.P. be –25.

It is known that,

Sn = (n/2)[2a + (n - 1)d]

Where n = number of terms, a = first term, and d = common difference

Here, a = –6

Common difference d = -11/2 + 6 = (-11 + 12)/2 = 1/2

Therefore, we obtain

-25 = (n/2)[2 * (-6) + (n - 1)(1/2)]

=> -25 * 2 = n[-12 + n/2 – 1/2]

=> -50 = n[-25/2 + n/2]

=> -100 = n(-25 + n)

=> -100 = -25n + n2

=> n2- 25n + 100 = 0

=> n2- 5n – 20 n + 100 = 0

=> n(n - 5) – 20(n - 5) = 0

=> (n - 5)(n - 20) = 0

=> n = 5, 20

Question 5:

In an A.P., if pth term is 1/q and qth term is 1/p, prove that the sum of first pq terms is   (pq + 1)/2, where p ≠ q.

It is known that the general term of an A.P. is an = a + (n – 1)d

According to the given information,

pth term = ap = a + (p - 1)d

=> a + (p - 1)d = 1/q  ………1

qth term = aq = a + (q - 1)d

=> a + (q - 1)d = 1/p  ………2

Subtracting equation 2 from 1, we get

(p - 1)d – (q - 1)d = 1/q – 1/p

=> (p – 1 – q + 1)d = (p - q)/pq

=> (p – q)d = (p - q)/pq

=> d = 1/pq

Put value of d in equation 1, we get

a + (p - 1)/pq = 1/q

=> a + p/pq - 1/pq = 1/q

=> a + 1/q - 1/pq = 1/q

=> a = 1/q – 1/q + 1/1pq

=> a = 1/pq

Spq = (pq/2)[2a + (pq - 1)d]

= (pq/2)[2/pq + (pq - 1)/pq]

= (pq/2)[2/pq + 1 - 1/pq]

= (pq/2)[1/pq + 1]

= 1 + (pq - 1)/2

= 1 + pq/2 – 1/2

= 1/2 + pq/2

= (1 + pq)/2

Thus, the sum of first pq terms of the A.P. is (pq + 1)/2

Question 6:

If the sum of a certain number of terms of the A.P. 25, 22, 19, ... is 116. Find the last term.

Let the sum of n terms of the given A.P. be 116.

Sn = (n/2)[2a + (n - 1)d]

Here, a = 25 and d = 22 – 25 = – 3

Sn = (n/2)[2 * 25 + (n - 1)(-3)]

=> 116 = (n/2)[50 - 3n + 3]

=> 232 = n(53 - 3n)

=> 232 = 53n - 3n2

=> 3n2 – 53n + 232 = 0

=> 3n2 – 24n – 29n + 232 = 0

=> 3n(n - 8) – 29(n - 8) = 0

=> (n - 8)(3n - 29) = 0

=> n = 8, 29/3

However, n cannot be equal to 29/3. Therefore, n = 8

So, a8 = Last term = a + (n – 1)d

= 25 + (8 – 1) (– 3)

= 25 + (7) (– 3)

= 25 – 21

= 4

Thus, the last term of the A.P. is 4.

Question 7:

Find the sum to n terms of the A.P., whose kth term is 5k + 1.

It is given that the kth term of the A.P. is 5k + 1.

kth term = ak = a + (k – 1)d

=> a + (k – 1)d = 5k + 1

=> a + kd – d = 5k + 1

Comparing the coefficient of k, we get d = 5

=> a – 5 = 1

=> a = 6

Now, Sn = (n/2)[2a + (n - 1)d]

= (n/2)[2 * 6 + 5(n - 1)]

= (n/2)[12 + 5n - 5]

= (n/2)[5n + 7]

Question 8:

If the sum of n terms of an A.P. is (pn + qn2), where p and q are constants, find the common difference.

It is known that:

Sn = (n/2)[2a + (n - 1)d]

According to the given condition,

(n/2)[2a + (n - 1)d] = pn + qn2

=> (n/2)[2a + dn - d] = pn + qn2

=> na + dn2/2 – dn/2 = pn + qn2

Comparing the coefficients of n2 on both sides, we get

d/2 = q

=> d = 2q

Thus, the common difference of the A.P. is 2q.

Question 9:

The sums of n terms of two arithmetic progressions are in the ratio 5n + 4: 9n + 6. Find the ratio of their 18th terms.

Let a1, a2, and d1, d2 be the first terms and the common difference of the first and second

arithmetic progression respectively.

According to the given condition,

Sum of n terms of first AP/Sum of n terms of second AP = (5n + 4)/(9n + 6)

=> (n/2)[2a1 + (n - 1)d1]/ (n/2)[2a2 + (n - 1)d2] = (5n + 4)/(9n + 6)

=> [2a1 + (n - 1)d1]/ [2a2 + (n - 1)d2] = (5n + 4)/(9n + 6)   ……………1

Substituting n = 35 in equation 1, we get

[2a1 + (35 - 1)d1]/ [2a2 + (35 - 1)d2] = (5 * 35 + 4)/(9 * 35 + 6)

=> [2a1 + 34d1]/ [2a2 + 34d2] = 179/321

=> [a1 + 17d1]/ [a2 + 17d2] = 179/321    ……………..2

=> 18th term of first AP/18th term of second AP = 179/321

Thus, the ratio of 18th term of both the A.P.s is 179 : 321.

Question 10:

If the sum of first p terms of an A.P. is equal to the sum of the first q terms, then find the sum of the first (p + q) terms.

Let a is the first term and d is the common difference of the AP

So, the sum of p terms Sp = (p/2){2a + (p - 1)d}

and Sum of q terms Sq = (q/2){2a + (q - 1)d}

Now since sum of first p terms is equal to the sum of first q terms, then

Sp = Sq

=> (p/2){2a + (p - 1)d} = (q/2){2a + (q - 1)d}

=> p{2a + (p - 1)d} = q{2a + (q - 1)d}

=> 2ap + p2d - pd = 2aq + q2d - qd

=> 2ap + p2d - pd - 2aq – q2d + qd = 0

=> 2a(p - q) + d(p2 - d2) - d(p-q) = 0

=>(p - q)*{2a + d(p + q) - d} = 0

=> (p - q)*{2a + d(p + q - 1)} = 0

Now p - q = 0  ...1

and 2a + d(p + q - 1) = 0 ........2

Multiply (p + q)/2 in equation 2, we get

{(p + q)/2}*{2a + d(p + q - 1)} = 0

=> Sp+q = 0

So, the sum of p + q terms is 0.

Question 11:

Sum of the first p, q and r terms of an A.P. are a, b and c, respectively. Prove that

a(q - r)/p + b(r - p)/q + c(p - q)/r = 0

Let a1 and d be the first term and the common difference of the A.P. respectively.

According to the given information,

Sp = (p/2)[2a + (p - 1)d] = a

=> 2a1 + (p - 1)d = 2a/p   …………1

Sq = (q/2)[2a + (q - 1)d] = b

=> 2a1 + (q - 1)d = 2b/q   …………2

Sr = (r/2)[2a + (r - 1)d] = c

=> 2a1 + (r - 1)d = 2c/r   …………3

Subtract equation 2 from equation 1, we get

(p - 1)d – (q - 1)d = 2a/p – 2b/q

=> (p - q)d = (2aq – 2bp)/pq

=> d = (2aq – 2bp)/{pq(p - q)}   ……….4

Subtract equation 3 from equation 2, we get

(q - 1)d – (r - 1)d = 2b/q – 2c/r

=> (q - r)d = (2br – 2qc)/qr

=> d = (2br – 2qc)/{qr(q - r)}   ……….5

From equation 4 and 5, we get

(2aq – 2bp)/{pq(p - q)} = (2br – 2qc)/{qr(q - r)}

=> (aq – bp)/{pq(p - q)} = (br – qc)/{qr(q - r)}

=> qr(q - r)(aq – bp) = pq(p - q)(br – qc)

=> r(q - r)(aq – bp) = p(p - q)(br – qc)

=> (q - r)(aqr – bpr) = (p - q)(bpr – pqc)

Divide both side by pqr, we get

(a/p – b/q)(q - r) = (p - q)(b/q – c/r)

=> (a/p)(q - r) – (b/q)(q - r) = (p - q)(b/q) – (p - q)(c/r)

=> (a/p)(q - r) – (b/q)(q – r + p - q) + (p - q)(c/r) = 0

=> (a/p)(q - r) + (b/q)(r - p) + (p - q)(c/r) = 0

Question 12:

The ratio of the sums of m and n terms of an A.P. is m2 : n2. Show that the ratio of mth and nth term is (2m – 1): (2n – 1).

Let a and b be the first term and the common difference of the A.P. respectively.

According to the given condition,

Sum of m terms of AP/Sum of n terms of AP = m2/n2

=> (m/2)[2a + (m - 1)d]/ (n/2)[2a + (n - 1)d] = m2/n2

=> [2a + (m - 1)d]/ [2a + (n - 1)d] = m/n   ……………1

Substituting m = 2m – 1 and n = 2n - 1 in equation 1, we get

[2a + (2m - 1)d]/ [2a + (2n - 1)d] = (2m - 1)/(2n - 1)

=> [a + (m - 1)d]/ [a + (n - 1)d] = (2m - 1)/(2n - 1)

=> mth term of AP/nth term of AP = (2m - 1)/(2n - 1)

=> mth term of AP : nth term of AP = (2m - 1) : (2n - 1)

This is the required ratio.

Question 13:

If the sum of n terms of an A.P. is 3n2 + 5n and its mth term is 164, find the value of m.

Let a and b be the first term and the common difference of the A.P. respectively.

am = a + (m – 1)d = 164      …m………1

Sum of n terms: Sn = (n/2)[2a + (n - 1)d]

Given, sum of n terms of an A.P. is n2 + 5n

=> (n/2)[2a + (n - 1)d] = 3n2 + 5n

=> na + n2d/2 – nd/2 = 3n2 + 5n

Comparing the coefficient of n2 on both sides, we obtain

d/2 = 3

=> d = 6

Comparing the coefficient of n on both sides, we obtain

a - d/2 = 5

=> a – 3 = 5

=> a = 8

Therefore, from equation 1, we obtain

8 + (m – 1) 6 = 164

⇒ (m – 1) 6 = 164 – 8 = 156

⇒ m – 1 = 26

⇒ m = 27

Thus, the value of m is 27.

Question 14:

Insert five numbers between 8 and 26 such that the resulting sequence is an A.P.

Let A1, A2, A3, A4, and A5 be five numbers between 8 and 26 such that 8, A1, A2, A3, A4, A5, 26 is

an A.P.

Here, a = 8, b = 26, n = 7

Therefore, 26 = 8 + (7 – 1) d

⇒ 6d = 26 – 8 = 18

⇒ d = 3

A1 = a + d = 8 + 3 = 11

A2 = a + 2d = 8 + 2 * 3 = 8 + 6 = 14

A3 = a + 3d = 8 + 3 * 3 = 8 + 9 = 17

A4 = a + 4d = 8 + 4 * 3 = 8 + 12 = 20

A5 = a + 5d = 8 + 5 * 3 = 8 + 15 = 23

Thus, the required five numbers between 8 and 26 are 11, 14, 17, 20, and 23.

Question 15:

If (an + bn)/(an-1 + bn-1) is the A.M. between a and b, then find the value of n.

A.M. of a and b = (a + b)/2

According to the given condition,

(a + b)/2 = (an + bn)/(an-1 + bn-1)

=> (a + b)(an-1 + bn-1) = 2(an + bn)

=> an + abn-1 + ban-1 + bn = 2an + 2bn

=> abn-1 + ban-1 = an + bn

=> abn-1 - bn = an - an-1b

=> bn-1(a - b) = an-1(a - b)

=> bn-1 = an-1

=> (b/a)n-1 = 0

=> (b/a)n-1 =(b/a)0

=> n – 1 = 0

=> n = 1

Hence, the value of n is 1

Question 16:

Between 1 and 31, m numbers have been inserted in such a way that the resulting sequence is an A.P. and the ratio of 7th and (m – 1)th numbers is 5 : 9.

Find the value of m.

Let A1, A2, ... Am be m numbers such that 1, A1, A2, ... Am, 31 is an A.P.

Here, a = 1, b = 31, n = m + 2

So, 31 = 1 + (m + 2 – 1)d

⇒ 30 = (m + 1)d

⇒ d = 30/(m + 1)   ………..1

A1 = a + d

A2 = a + 2d

A3 = a + 3d

………………

A7 = a + 7d

Am–1 = a + (m – 1) d

According to the given condition,

(a + 7d)/{a + (m - 1)d} = 5/9

=> [1 + 7{30/(m + 1)}]/{1 + (m - 1) * 30/(m + 1)} = 5/9            [From equation 1]

=> [m + 1 + 7 * 30]/[m + 1 + 30(m - 1)] = 5/9

=> [m + 1 + 210]/[m + 1 + 30m - 30] = 5/9

=> [m + 211]/[31m - 29] = 5/9

=> 9[m + 211] = 5[31m - 29]

=> 9m + 1899 = 155m - 145

=> 155m – 9m = 1899 + 145

=> 146m = 2044

=> m = 2044/146

=> m = 14

Thus, the value of m is 14.

Question 17:

A man starts repaying a loan as first installment of Rs. 100.

If he increases the installment by Rs 5 every month, what amount he will pay in the 30th installment?

The first installment of the loan is Rs 100.

The second installment of the loan is Rs 105 and so on.

The amount that the man repays every month forms an A.P.

The A.P. is 100, 105, 110, ...........

First term, a = 100

Common difference, d = 5

A30 = a + (30 – 1)d

= 100 + 29 * 5

= 100 + 145

= 245

Thus, the amount to be paid in the 30th installment is Rs 245.

Question 18:

The difference between any two consecutive interior angles of a polygon is 5°. If the smallest angle is 120°,

find the number of the sides of the polygon.

Let there are n side of the polygon.

The angles of the polygon will form an A.P. with common difference d as 5° and first term a as

120°.

Now sum of all interior angles of the polygon = (2n - 4)*90

=> (n/2)*[2a + (n - 1)*d] = (2n - 4) * 90

=> (n/2)*[2*120 + (n - 1)*5] = (2n - 4) * 90

=> (n/2)*[240 + (n - 1) * 5] = (2n - 4) * 90

=> (n/2)*[240 + 5n - 5] = (2n - 4) * 90

=> n*[235 + 5n] = (2n - 4) * 90 * 2

=> 5*[47n + n2] = (2n - 4)*180

=> 47n + n2 = (2n - 4)*180/5

=> 47n + n2 = (2n - 4)*36

=> 47n + n2 = 2n * 36 – 4 * 36

=> 47n + n2 = 72n - 144

=> n2 + 47n - 72n + 144 = 0

=> n2 + 47n - 72n + 144 = 0

=> n2 - 25n + 144 = 0

=>(n - 9)*(n - 16) = 0

=> n = 9, 16

When n =16 then largest interior angle of the polygon = a + (n - 1)*d

= 120 + (16 - 1)*5

= 120 + 15*5

= 120 + 75

= 195

Which is not possible.

So, the number of side n =9

Exercise 9.3

Question 1:

Find the 20th and nth terms of the G.P. 5/2, 5/4, 5/8, ………..

Given, G.P. is: 5/2, 5/4, 5/8, ………..

First term a = 5/2

Common ratio r = (5/4)/(5/2) = 1/2

Now, a20 = ar20-1 = (5/2)(1/2)19 = 5/(2 * 219) = 5/220

an = arn-1 = (5/2)(1/2)n-1 = 5/(2 * 2n-1)  = 5/2n

Question 2:

Find the 12th term of a G.P. whose 8th term is 192 and the common ratio is 2.

Given, common ratio r = 2

Let a be the first term of the G.P.

So, a8 = ar8–1 = ar7

⇒ ar7 = 192

⇒ a(2)7 = 192

⇒ a(2)7 = 26 * 3

⇒ a = (26 * 3)/(2)7

⇒ a = 3/2

Now, a12 = ar12–1 = ar11

= (3/2) * 211

= 3 * 210 = 3 * 1024 = 3072

Question 3:

The 5th, 8th and 11th terms of a G.P. are p, q and s, respectively. Show that q2 = ps.

Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

a5 = ar5–1 = ar4 = p ..............1

a8 = ar8–1 = ar7 = q ..............2

a11 = ar11–1 = ar10 = s ...........3

Dividing equation 2 by 1, we get

ar7/ar4 = q/p

=> r3 = q/p  …………4

Dividing equation 3 by 2, we get

ar10/ar7 = s/q

=> r3 = s/q  ………….5

Equating the values of r3 obtained in 4 and 5, we get

q/p = s/q

=> q2 = ps

Question 4:

The 4th term of a G.P. is square of its second term, and the first term is –3. Determine its 7th term.

Let a be the first term and r be the common ratio of the G.P.

So, a = -3

It is known that, an = arn–1

So, a4 = ar3 = (-3)r3

a2 = ar1 = (-3)r

According to the given condition,

(-3)r3  = [(-3)r]2

⇒ -3r3 = 9r2

⇒ r = -3

Now, a7 = ar7–1

= ar6

= (-3) * (-3)6

= -(3)7

= -2187

Thus, the seventh term of the G.P. is -2187

Question 5:

Which term of the following sequences:

(a) 2, 2√2, 4, ... is 128?             (b) √3, 3, 3√3, ... is 729?    (c) 1/3, 1/9,1/27, ... is 1/19683 ?

(a) The given sequence is 2, 2√2, 4, ...

Here, a = 2 and r = (2√2)/2 = √2

Let nth term of the given sequence be 128

=> an = arn-1

=> 2 * (√2)n-1  = 128

=> 2 * 2(n-1)/2  = 128

=> 2(n-1)/2 + 1  = 128

=> 2(n-1)/2 + 1  = 27

=> (n - 1)/2 + 1 = 7

=> (n - 1)/2 = 6

=> n – 1 = 12

=> n = 13

Thus, the 13th term of the given sequence is 128.

(b) Given sequence is : √3, 3, 3√3, ...

Here, a = √3 and r = 3/√3 = √3

Let nth term of the given sequence be 729

=> an = arn-1

=> √3 * (√3)n-1  = 729

=> 31/2 * 3(n-1)/2  = 729

=> 3(n-1)/2 + 1/2  = 729

=> 3(n-1)/2 + 1/2  = 36

=> (n - 1)/2 + 1/2 = 6

=> (n - 1)/2 = 6 – 1/2

=> (n – 1)/2 = 11/2

=> n – 1 = 11

=> n = 12

Thus, the 12th term of the given sequence is 729.

(c) Given sequence is: 1/3, 1/9,1/27, ...

Here, a = 1/3 and r = (1/9)/(1/3) = 1/3

Let nth term of the given sequence be 1/19683

=> an = arn-1

=> (1/3) * (1/3)n-1  = 1/19683

=> (1/3)n  = (1/3)9

=> n = 9

Thus, the 9th term of the given sequence is 1/19683

Question 6:

For what values of x, the numbers 2/7, x, -7/2 are in G.P?

The given numbers are 2/7, x, -7/2

Common ratio = x/(-2/7) = -7x/2   …………1

Also, common ratio = (-7/2)/x = -7/2x  ………..2

From equation 1 and 2, we get

-7x/2 = -7/2x

=> x2 = (-2 * 7)/ (-2 * 7)

=> x2 = 1

=> x = ±1

Thus, for x = ± 1, the given numbers will be in G.P.

Question 7:

Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 ...

The given G.P. is 0.15, 0.015, 0.00015 ...

Here, a = 0.15 and r = 0.015/0.15 = 0.1

Now, Sn = a(1 - rn)/(1 - r)

=> S20 = 0.15{1 – (0.1)20}/(1 – 0.1)

= 0.15{1 – (0.1)20}/0.9

= 15{1 – (0.1)20}/90

= {1 – (0.1)20}/6

Question 8:

Find the sum to n terms in the geometric progression √7 , √21, 3√7, ...

The given G.P. is √7, √21, 3√7, ...

Here, a = √7 and r = √21/√7 = √3

Now, Sn = a(1 − rn)/(1 – r)

⟹ Sn = √7 [1 − (√3)n]/(1 − √3)

⟹ Sn = √7 [1 − (√3)n]/(1 − √3) * (1 + √3)/(1 + √3)

⟹ Sn = √7(√3 + 1) [1 − (√3)n]/(1 – 3)

⟹ Sn = −√7(√3 + 1) [1 − (√3)n]/2

⟹ Sn = √7(√3 + 1) [(√3)n – 1]/2

Question 9:

Find the sum to n terms in the geometric progression 1, -a, a2, -a3, ……. (if a ≠ -1)

The given G.P. is 1, -a, a2, -a3, …….

Here, first term = a = 1

Common ratio = r = – a

Now, Sn = a(1 − rn)/(1 – r)

= {1 – (-a)n}/{1 – (-a)}

= {1 – (-a)n}/(1 + a)

Question 10:

Find the sum to n terms in the geometric progression x3, x5, x7, ……. (if x ≠ ±1)

The given G.P. is x3, x5, x7, …….

Here, first term = a = x3

Common ratio = r = x5/x3 = x2

Now, Sn = a(1 − rn)/(1 – r)

= x3{1 – (x2)n}/(1 – x2)

= x3(1 – x2n)/{1 – x2)

Question 11:

Evaluate Σk=111 (2 + 3k)

Given, Σk=111 (2 + 3k) = Σk=111 2 + Σk=111  3k

= 2 * 11 + Σk=111  3k

= 22 + Σk=111  3k    ………….1

Now, Σk=111  3k = 31 + 32 + 33 + ….+ 311

The terms of this sequence 3, 32, 33, ……forms a GP

Now, Sn = a(rn – 1)/(r - 1)

So, S11 = 3(311 – 1)/(3 – 1)

= 3(311 – 1)/2

=> Σk=111  3k = 3(311 – 1)/2

Put this value in equation 1, we get

Given, Σk=111 (2 + 3k) = 22 + Σk=111 3k = 22 + 3(311 – 1)/2

Question 12:

The sum of first three terms of a G.P. is 39/10 and their product is 1. Find the common ratio and the terms.

Let a/r, a, ar be the first three terms of the GP

Now, a/r + a + ar = 39/10   ……….1

And (a/r) * a * ar = 1

=> a3 = 1

=> a = 1

Substituting a = 1 in equation 1, we obtain

1/r + 1 + r = 39/10

=> 1 + r + r2 = 39r/10

=> 10(1 + r + r2) = 39r

=> 10 + 10r + 10r2 = 39r

=> 10 + 10r + 10r2 - 39r = 0

=> 10r2 - 29r + 10 = 0

=> 10r2 - 25r – 4r + 10 = 0

=> 5r(2r - 5) – 2(2r - 5) = 0

=> (2r - 5)(5r - 2) = 0

=> r = 2/5 or 5/2

Thus, the first three terms of the G.P. are 5/2, 1 and 2/5

Question 13:

How many terms of G.P. 3, 32, 33 ... are needed to give the sum 120?

The given G.P. is 3, 32, 33, ........

Let n terms of this G.P. be required to obtain the sum as 120.

Sn = a(1 − rn)/(1 – r)

Here, a = 3 and r = 3

Sn = 120 = 3(3n - 1)/(3 – 1)

=> 120/3 = (3n - 1)/2

=> 40 = (3n - 1)/2

=> 40 * 2 = 3n – 1

=> 3n = 80 + 1

=> 3n = 81

=> 3n = 34

=> n = 4

Thus, four terms of the given G.P. are required to obtain the sum as 120.

Question 14:

The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128.

Determine the first term, the common ratio and the sum to n terms of the G.P.

Let the G.P. be a, ar, ar2, ar3, .........

According to the given condition,

a + ar + ar2 = 16 and ar3 + ar4 + ar5 = 128

⇒ a (1 + r + r2) = 16   ................ 1

ar3(1 + r + r2) = 128   ................2

Dividing equation 2 by 1, we get

ar3(1 + r + r2)/a(1 + r + r2) = 128/16

=> r3 = 8

=> r = 2

Substituting r = 2 in equation 1, we get

a(1 + 2 + 4) = 16

⇒ a * 7 = 16

⇒ a = 16/7

Now, Sn = a(rn - 1)/(r – 1)

= {(16/7)(2n - 1)}/(2 – 1)

= 16(2n - 1)/7

Question 15:

Given a G.P. with a = 729 and 7th term 64, determine S7.

a = 729, a7 = 64

Let r be the common ratio of the G.P.

It is known that, an = arn–1

=> a7 = ar7–1 = (729)r6

=> 64 = 729 r6

=> r6 = 64/729

=> r6 = (2/3)6

=> r = 2/3

Also, it is known that,

Sn = a(1 − rn)/(1 – r)

S7 = 729(1 – (2/3)7)/(1 – 2/3)

= 3 * 729[1 – (2/3)7]

= 37[(37 – 27)/37]

= 37 – 27

= 2187 - 128

= 2059

Question 16:

Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term.

Let a be the first term and r be the common ratio of the G.P.

According to the given conditions,

S2 = -4 = a(1 − rn)/(1 – r)   ………..1

a5 = 4 * a3

⇒ ar4 = 4ar2

⇒ r2 = 4

⇒ r = ± 2

From equation 1, we obtain

-4 = a[1 – 22]/(1 – 2) for r = 2

=> -4 = a(1 - 4)/(-1)

=> -4 = 3a

=> a = -4/3

Again,

-4 = a[1 – (-2)2]/(1 + 2) for r = -2

=> -4 = a(1 - 4)/3

=> -4 = -3a/3

=> a = 4

Thus, the required G.P. is -4/3, -8/3, -16/3, ….. or 4, –8, 16, –32 ...

Question 17:

If the 4th, 10th and 16th terms of a G.P. are x, y and z, respectively. Prove that x, y, z are in G.P.

Let a be the first term and r be the common ratio of the G.P.

According to the given condition,

a4 = ar3 = x ......................... (1)

a10 = ar9 = y ........................ (2)

a16 = ar15 = z ....................... (3)

Dividing (2) by (1), we obtain

y/x = ar9/ar3 = r6

Dividing (3) by (2), we obtain

z/y = ar15/ar9 = r6

Now, y/x = z/y

=> y2 = xz

Thus, x, y, z are in G. P.

Question 18:

Find the sum to n terms of the sequence, 8, 88, 888, 8888...

The given sequence is 8, 88, 888, 8888...

Now, Sn = 8 + 88 + 888 + ……… to n terms

=> Sn = 8[1 + 11 + 111 + ……… to n terms]

=> Sn = (8/9)[9 + 99 + 999 + ……… to n terms]

=> Sn = (8/9)[(10 - 1) + (102 - 1) + (103 - 1) + ……… to n terms]

=> Sn = (8/9)[(10 + 102 + 103 + ……… to n terms) – (1 + 1 + 1 + …. to n terms)]

=> Sn = (8/9)[(10(10n - 1)/(10 - 1) – n]

=> Sn = (8/9)[(10(10n - 1)/9 – n]

=> Sn = 80(10n - 1)/81 – 8n/9

Question 19:

Find the sum of the products of the corresponding terms of the sequences  2, 4, 8, 16, 32 and 128, 32, 8, 2, 1/2 .

Required sum = 2 * 128 + 4 * 32 + 8 * 8 + 16 * 2 + 32 * 1/2

= 256 + 128 + 64 + 32 + 16

= 496

Question 20:

Show that the products of the corresponding terms of the sequences form a, ar, ar2, …., arn-1 and A, AR, AR2, …., ARn-1 form a G.P,

and find the common ratio.

We have to be prove that the sequence: aA, arAR, ar2AR2, …., arn-1 ARn-1  forms a G.P.

Now, second term/first term = arAR/aA = rR

and third term/second term = ar2AR2/arAR = rR

Thus, the above sequence forms a G.P. and the common ratio is rR.

Question 21:

Find four numbers forming a geometric progression in which third term is greater than the first term by 9,

and the second term is greater than the 4th by 18.

Let a be the first term and r be the common ratio of the G.P.

a1 = a, a2 = ar, a3 = ar2, a4 = ar3

By the given condition,

a3 = a1 + 9

⇒ ar2 = a + 9 ................... (1)

a2 = a4 + 18

⇒ ar = ar3 + 18 ............ (2)

From equation (1) and (2), we obtain

a(r2 – 1) = 9 .......................... (3)

ar (1– r2) = 18 ...................... (4)

Dividing (4) by (3), we obtain

ar (1– r2)/a(r2 – 1) = 18/9

=> -r = 2

=> r = -2

Substituting the value of r in equation (1), we obtain

4a = a + 9

⇒ 3a = 9

⇒ a = 3

Thus, the first four numbers of the G.P. are 3, 3(– 2), 3(–2)2, and 3(–2)3 i.e., 3 ̧–6, 12, and –24.

Question 22:

If pth, qth and rth terms of a G.P. are a, b and c respectively. Prove that aq – r * br – p * cp – q = 1

Let A be the first term and R be the common ratio of the G.P.

According to the given information,

ARp–1 = a

ARq–1 = b

ARr–1 = c

Now, aq – r * br – p * cp – q

= Aq–r * R(p–1) (q–r) * Ar–p × R(q–1) (r-p) * Ap–q × R(r –1)(p–q)

= Aq – r + r – p + p – q * R(pr – pr – q + r) + (rq – r + p – pq) + (pr – p – qr + q)

= A0 * R0

= 1

Thus, the given result is proved.

Question 23:

If the first and the nth term of a G.P. are a ad b, respectively, and if P is the product of n terms, prove that P2 = (ab)n

The first term of the G.P is a and the last term is b.

Therefore, the G.P. is a, ar, ar2, ar3, ........, arn–1, where r is the common ratio.

b = arn–1 ................. (1)

P = Product of n terms

= (a) (ar) (ar2) ... (arn–1)

= (a * a *...a) (r * r2 * ...rn–1)

= an r1 + 2 +...(n–1)

= an rn(n-1)/2

Now, P2 = (an rn(n-1)/2)2

= (a2n rn(n-1))

= (a2 r(n-1))n

= [a * ar(n-1)]n

= (ab)n                     [From equation 1]

Thus, the given result is proved.

Question 24:

Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th term is 1/rn.

Let a be the first term and r be the common ratio of the G.P.

Sum of first n terms = a(1 - rn)/(1 - r)

Since there are n terms from (n +1)th to (2n)th term,

Sum of terms from (n + 1)th to (2n)th term

Sn = an+1(1 - rn)/(1 - r)

Now, an+1 = arn+1-1 = arn

Thus, required ratio = [a(1 - rn)/(1 - r)] * [an+1(1 - rn)/(1 - r)]

= [a(1 - rn)/(1 - r)] * [arn (1 - rn)/(1 - r)]

= 1/rn

Thus, the ratio of the sum of first n terms of a G.P. to the sum of terms from (n + 1)th to (2n)th

term is 1/rn.

Question 25:

If a, b, c and d are in G.P. show that: (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2

Given, a, b, c, d are in G.P. Therefore,

b2 = ac ........................ (2)

c2 = bd ....................... (3)

It has to be proved that,

(a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2

R.H.S.

= (ab + bc + cd)2

= (ab + ad + cd)2              [Using equation (1)]

= [ab + d (a + c)]2

= a2b2 + 2abd (a + c) + d2(a + c)2

= a2b2 +2a2bd + 2acbd + d2 (a2 + 2ac + c2)

= a2b2 + 2a2c2+ 2b2c2+ d2a2 + 2d2b2 + d2c2         [Using equation (1) and (2)]

= a2b2 + a2c2 + a2c2 + b2c2 + b2c2 + d2a2 + d2b2 + d2b2 + d2c2

= a2b2 + a2c2 + a2d2 + b2 * b2 + b2c2 + b2d2 + c2b2 + c2 * c2 + c2d2

[Using (2) and (3) and rearranging terms]

= a2 (b2 + c2 + d2) + b2(b2 + c2 + d2)+ c2(b2 + c2 + d2)

= (a2 + b2 + c2) (b2 + c2 + d2) = LHS

So, LHS = RHS

Hence, (a2 + b2 + c2) (b2 + c2 + d2) = (ab + bc + cd)2

Question 26:

Insert two numbers between 3 and 81 so that the resulting sequence is G.P.

Let G1 and G2 be two numbers between 3 and 81 such that the series, 3, G1, G2, 81, forms a

G.P.

Let a be the first term and r be the common ratio of the G.P.

So, 81 = 3 * r3

⇒ r3 = 81/3

⇒ r3 = 27

⇒ r = 3              (Taking real roots only)

For r = 3,

G1 = ar = 3 * 3 = 9

G2 = ar2 = 3 * 32 = 3 * 9 = 27

Thus, the required two numbers are 9 and 27.

Question 27:

Find the value of n so that (an+1 + bn+1)/(an + bn) may be the geometric mean between a and b.

Geometric Mean of a and b is √ab

By the given condition:

(an+1 + bn+1)/(an + bn) = √ab

Squaring both sides, we obtain

[(an+1 + bn+1)/(an + bn)]2 = ab

=> (an+1 + bn+1)2/(an + bn)2 = ab

=> (an+1 + bn+1)2 = ab(an + bn)2

=> a2n+2 + b2n+2 + 2an+1 bn+1 = ab(a2n + b2n + 2an bn)

=> a2n+2 + b2n+2 + 2an+1 bn+1 = a2n+1b + ab2n+1 + 2an+1 bn+1

=> a2n+2 + b2n+2 = a2n+1b + ab2n+1

=> a2n+2 - a2n+1 b = ab2n+1 - ab2n+2

=> a2n+1(a – b) = b2n+1(a – b)

=> (a/b)2n+1 = 1

=> (a/b)2n+1 = (a/b)0

=> 2n + 1 = 0

=> n = -1/2

Question 28:

The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio (3 + 2√2) : (3 - 2√2)

Let the two numbers be a and b.

G.M. = √ab

According to the given condition,

a + b = 6√ab    …………..1

=> (a + b)2 = 36ab

Also, (a - b)2 = (a + b)2 – 4ab

=> (a - b)2 =36ab – 4ab

=> (a - b)2 = 32ab

=> (a - b)2 = 4√2√ab    ………………2

Adding equation 1 and 2, we get

2a = (6 + 4√2)√ab

=> a = (3 + 2√2)√ab

Put the value of a in equation 1, we get

(3 + 2√2)√ab + b = 6√ab

=> b = 6√ab - (3 + 2√2)√ab

=> b = (3 - 2√2)√ab

Now, a/b = (3 + 2√2)√ab/(3 - 2√2)√ab = (3 + 2√2)/(3 - 2√2)

This the required ratio = (3 + 2√2) : (3 - 2√2)

Question 29:

If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are A ± √{(A + G)(A - G)}

It is given that A and G are A.M. and G.M. between two positive numbers.

Let these two positive numbers be a and b.

Now, AM = A = (a + b)/2   …………..1

and GM = G = √ab   ………….2

From equation 1 and 2, we obtain

a + b = 2A ................... 3

ab = G2 .......................4

Substituting the value of a and b from 3 and 4 in the identity (a – b)2 = (a + b)2 – 4ab, we get

(a – b)2 = 4A2 – 4G2

=> (a – b)2 = 4(A2 – G2)

=> (a – b)2 = 4(A – G)(A + G)

=> (a – b) = 2√{(A + G) (A – G)}  ………..5

From equation 3 and 5, we get

2a = 2A + 2√{(A + G)(A – G)}

=> a = A + √{(A + G)(A – G)}

Substituting the value of a in 3, we obtain

b = 2A - A - √{(A + G)(A – G)}

=> b = A - √{(A + G)(A – G)}

Thus, the two numbers are A ± √{(A + G)(A - G)}

Question 30:

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally,

how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?

It is given that the number of bacteria doubles every hour. Therefore, the number of bacteria

after every hour will form a G.P.

Here, a = 30 and r = 2

So, a3 = ar2 = 30 * 22 = 30 * 4 = 120

Therefore, the number of bacteria at the end of 2nd hour will be 120.

a5 = ar4 = 30 * 24 = 30 * 16 = 480

The number of bacteria at the end of 4th hour will be 480.

an +1 = arn = 30 * 2n

Thus, number of bacteria at the end of nth hour will be 30 * 2n

Question 31:

What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?

The amount deposited in the bank is Rs 500.

At the end of 1st year, amount = Rs 500(1 + 1/10) = Rs 500 * 1.1

At the end of 2nd year, amount = Rs 500 * 1.1 * 1.1

At the end of 3rd year, amount = Rs 500 * 1.1 * 1.1 *1.1 and so on

Amount at the end of 10 years = Rs 500 * 1.1 * 1.1 * ... (10 times) = Rs 500 * (1.1)10

Question 32:

If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.

Let the root of the quadratic equation be a and b.

According to the given condition,

AM = (a + b)/2 =8

=> a + b = 16   ………..1

GM = √ab = 5

=> ab = 25   …………..2

The quadratic equation is given by,

x2 – x (Sum of roots) + (Product of roots) = 0

=> x2 – x(a + b) + ab = 0

=> x2 – 16x + 25 = 0                      [Using equation 1 and 2]

Thus, the required quadratic equation is x2 – 16x + 25 = 0

Exercise 9.4

Question 1:

Find the sum to n terms of the series 1 * 2 + 2 * 3 + 3 * 4 + 4 * 5 + ......nth term.

The given series is 1 * 2 + 2 * 3 + 3 * 4 + 4 * 5 + ... nth term

an = n(n + 1)

So, Sn = ∑n k=1 ak

=∑n k=1   k (k+1)

=∑n k=1   k2 + ∑n k=1   k

= {n(n + 1)(2n + 1)}/6 + n(n + 1)/2

= {n(n + 1)/2}{(2n + 1)/3 + 1}

= {n(n + 1)/2}(2n + 4)/3

= {n(n + 1)(n + 2)}/3

Question 2:

Find the sum to n terms of the series 1 * 2 * 3 + 2 * 3 * 4 + 3 * 4 * 5 + ...

The given series is 1 * 2 * 3 + 2 * 3 * 4 + 3 * 4 * 5 + ... nth term,

an = n(n + 1)(n + 2)

= (n2 + n) (n + 2)

= n3 + 3n2 + 2n

Now, Sn = ∑n k=1   ak

= ∑n k=1   k3 + 3∑n k=1   k2  + 2 ∑n k=1   k

= [n(n + 1)/2]2 + {3n(n + 1)(2n + 1)}/6 + {2n(n + 1)}/2

= [n(n + 1)/2]2 + {n(n + 1)(2n + 1)}/2 + {2n(n + 1)}/2

= [n(n + 1)/2][{n(n + 1)}/2 + 2n + 1 + 2}

= [n(n + 1)/2][n2 + n + 4n + 6]/2

= [n(n + 1)/2][n2 + 5n + 6]/2

= [n(n + 1)/2][(n + 2)(n + 3)]/2

= [n(n + 1)(n + 2)(n + 3)]/4

Question 3:

Find the sum to n terms of the series 3 * 12 + 5 * 22 + 7 * 32 + ...........

The given series is 3 * 12 + 5 * 22 + 7 * 32 + ......... nth term,

an = (2n + 1)n2 = 2n3 + n2

Now, Sn = ∑n k=1   ak

= 2 ∑n k=1   k3 + ∑n k=1   k2

= 2[n(n + 1)/2]2 + {n(n + 1)(2n + 1)}/6

= [n2(n + 1)2]/2 + {n(n + 1)(2n + 1)}/6

= [n(n + 1)/2][n(n + 1) + (2n + 1)/3]

= [n(n + 1)/2][(3n2 + 3n + 2n + 1)/3]

= [n(n + 1)/2][(3n2 + 5n + 1)/3]

= [n(n + 1)(3n2 + 5n + 1)]/6

Question 4:

Find the sum to n terms of the series 1/(1 * 2) + 1/(2 * 3) +1/(3 * 4) + ……………

The given series is 1/(1 * 2) + 1/(2 * 3) +1/(3 * 4) + …………… nth term,

an = 1/{n(n + 1)} = 1/n – 1/(n + 1)                     [By partial fraction]

a1 = 1/1 – 1/(1 + 1) = 1/1 – 1/2

a2 = 1/2 – 1/(2 + 1) = 1/2 – 1/3

a3 = 1/3 – 1/(3 + 1) = 1/3 – 1/4

………………………………………………

an = 1/n – 1/(n + 1)

Adding the above terms column wise, we get

a1 + a2 + a3 + ……+ an = 1/1 – 1/2 + 1/2 - 1/3 + 1/3 – 1/4 + ………..+ 1/n – 1/(n + 1)

= 1 – 1/(n + 1)

= (n + 1 - 1)/(n + 1)

= n/(n + 1)

Question 5:

Find the sum to n terms of the series 52 + 62 + 72 + ... + 202

The given series is 52 + 62 + 72 + ... + 202 nth term,

an = (n + 4)2 = n2 + 8n + 16

Now, Sn = ∑n k=1 ak

= ∑n k=1   k2 + 8 ∑n k=1   k +  ∑n k=1   16

= {n(n + 1)(2n + 1)}/6 + 8n(n + 1)/2 + 16n

16th term = (16 + 4)2 = 202

So, S16 = {16(16 + 1)(2 * 16 + 1)}/6 + {8 * 16(16 + 1)}/2 + 16 * 16

= (16 * 17 * 33)/6 + (8 * 16 * 17)/2 + 265

= 1496 + 1088 + 256

= 2840

So, 52 + 62 + 72 + ... + 202 = 2840

Question 6:

Find the sum to n terms of the series 3 * 8 + 6 * 11 + 9 * 14 +........

The given series is 3 * 8 + 6 * 11 + 9 * 14 + ...

an = (nth term of 3, 6, 9 ...) * (nth term of 8, 11, 14 ...)

= (3n) (3n + 5)

= 9n2 + 15n

Now, Sn = ∑n k=1 ak

= 9 ∑n k=1   k+ 15 ∑n k=1 k

= 9{n(n + 1)(2n + 1)}/6 + 15n(n + 1)/2

= 3{n(n + 1)(2n + 1)}/2 + 15n(n + 1)/2

= [3{n(n + 1)}/2][2n + 1 + 5]

= [3{n(n + 1)}/2][2n + 6]

= 3n(n + 1)(n + 3)

Question 7:

Find the sum to n terms of the series 12 + (12 + 22) + (12 + 22 + 32) + ........

The given series is 12 + (12 + 22) + (12 + 22 + 32) + ........+ an

an = (12 + 22 + 33 +.......+ n2)

= {n(n + 1)(2n + 1)}/6

= {n(2n2 + 3n + 1)}/6

= (2n3 + 3n2 + n)}/6

= n3/3 + n2/2 + n/6

Now, Sn =  ∑n k=1 ak

= (1/3) ∑n k=1   k3   + (1/2) ∑n k=1   k2 + (1/6) ∑n k=1   k

= (1/3)[n2(n + 1)2/2] + (1/2){n(n + 1)(2n + 1)}/6 + (1/6){n(n + 1)/2}

= {n(n + 1)/6}[n(n + 1)/2 + (2n + 1)/2 + 1/2]

= {n(n + 1)/6}[{n2 + n + 2n + 1 + 1}/2]

= {n(n + 1)/6}[{n2 + 3n + 2}/2]

= {n(n + 1)/6}[{(n + 1)(n + 2)}/2]

= {n(n + 1)2(n + 2)}/12

Question 8:

Find the sum to n terms of the series whose nth term is given by n(n + 1)(n + 4).

Given, an = n(n + 1)(n + 4)

= n(n2 + 5n + 4)

= n3 + 5n2 + 4n

Now, Sn = ∑n k=1 ak

= ∑n k=1   k3   + 5 ∑n k=1   k2 + 4 ∑n k=1   k

= n2(n + 1)2/4 + 5{n(n + 1)(2n + 1)}/6 + 4n(n + 1)/2

= n(n + 1)/2[n(n + 1)/2 + 5(2n + 1)}/3 + 4]

= n(n + 1)/2[(3n2 + 3n + 20n + 10 + 24)/6]

= n(n + 1)/2[(3n2 + 23n + 34)/6]

= [n(n + 1)(3n2 + 23n + 34)]/12

Question 9:

Find the sum to n terms of the series whose nth terms is given by n2 + 2n

Given, an = n(n + 1)(n + 4)

= n(n2 + 5n + 4)

= n3 + 5n2 + 4n

Now, Sn = ∑n k=1 ak

= ∑n k=1   k3   + 5 ∑n k=1   k2 + 4 ∑n k=1   k

= n2(n + 1)2/4 + 5{n(n + 1)(2n + 1)}/6 + 4n(n + 1)/2

= n(n + 1)/2[n(n + 1)/2 + 5(2n + 1)}/3 + 4]

= n(n + 1)/2[(3n2 + 3n + 20n + 10 + 24)/6]

= n(n + 1)/2[(3n2 + 23n + 34)/6]

= [n(n + 1)(3n2 + 23n + 34)]/12

Question 10:

Find the sum to n terms of the series whose nth terms is given by (2n – 1)2

Given, an = (2n – 1)2 = 4n2 – 4n + 1

Now, Sn = ∑n k=1 ak

= 4 ∑n k=1  k2 - 4  + ∑n k=1  k + ∑n k=1  1

= {4n(n + 1)(2n + 1)}/6 – 4n(n + 1)/2 + n

= {2n(n + 1)(2n + 1)}/3 – 2n(n + 1) + n

= n[{2(n2 + 3n + 1)}/3 – 2(n + 1) + 1]

= n[4n2 + 6n + 2 – 6n - 6 + 3]/3

= n[4n2 - 1]/3

= {n(2n + 1)(2n - 1)}/3

Miscellaneous Exercise on chapter 9

Question 1:

Show that the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Let a and d be the first term and the common difference of the A.P. respectively.

It is known that the kth term of an A. P. is given by

ak = a + (k –1) d

So, am + n = a + (m + n –1) d

and am – n = a + (m – n –1) d

am = a + (m –1) d

Now, am + n + am – n = a + (m + n –1) d + a + (m – n –1) d

= 2a + (m + n –1 + m – n –1) d

= 2a + (2m – 2) d

= 2a + 2 (m – 1) d

= 2 [a + (m – 1) d]

= 2am

Thus, the sum of (m + n)th and (m – n)th terms of an A.P. is equal to twice the mth term.

Question 2:

If the sum of three numbers in A.P., is 24 and their product is 440, find the numbers.

Let the three numbers in A.P. be a – d, a, and a + d.

According to the given information,

(a – d) + a + (a + d) = 24    …………. (1)

⇒ 3a = 24

⇒ a = 8

(a – d) * a * (a + d) = 440     ……………… (2)

⇒ (8 – d) * 8 * (8 + d) = 440

⇒ (8 – d)(8 + d) = 55

⇒ 64 – d2 = 55

⇒ d2 = 64 – 55

⇒ d2 = 9

⇒ d = ± 3

Therefore, when d = 3, the numbers are 5, 8, and 11 and

When d = -3, the numbers are 11, 8, and 5.

Thus, the three numbers are 5, 8, and 11.

Question 3:

Let the sum of n, 2n, 3n terms of an A.P. be S1, S2 and S3, respectively, show that   S3 = 3 (S2– S1)

Let a and b be the first term and the common difference of the A.P. respectively.

Therefore,

S1 = (n/2)[2a + (n - 1)d]   ………..1

S2 = (2n/2)[2a + (2n - 1)d] = n[2a + (2n - 1)d]  ………..2

S3 = (3n/2)[2a + (3n - 1)d]   ………….3

From equation 1 and 2, we get

S2 – S1 = n[2a + (2n - 1)d] - (n/2)[2a + (n - 1)d]

= n[4a + 4nd – 2d – 2a – nd + d]/2

= n[2a + 3nd – d]/2

= (n/2)[2a + (3n – 1)d]/2

=> 3(S2 – S1) = (3n/2)[2a + (3n – 1)d]/2 = S3

Hence, the given result is proved.

Question 4:

Find the sum of all numbers between 200 and 400 which are divisible by 7.

The numbers lying between 200 and 400, which are divisible by 7, are

203, 210, 217 … 399

Now, First term, a = 203

Last term, l = 399

Common difference, d = 7

Let the number of terms of the A.P. be n.

So, an = 399 = a + (n –1) d

⇒ 399 = 203 + (n –1) 7

⇒ 7(n –1) = 196

⇒ n –1 = 28

⇒ n = 29

Now, S29 = (n/2)(a + l)

= (29/2)(203 + 399)

= (29/2) * 602

= 29 * 301

= 8729

Thus, the required sum is 8729.

Question 5:

Find the sum of integers from 1 to 100 that are divisible by 2 or 5.

The integers from 1 to 100, which are divisible by 2, are 2, 4, 6… 100.

This forms an A.P. with both the first term and common difference equal to 2.

⇒ 100 = 2 + (n –1)2

⇒ 50 = 1 + n –1

⇒ n = 50

So, 2 + 4 + 6 + ……100 = (50/2)[2 * 2 + (50 - 1)2]

= 25[4 + 98]

= 25 * 102

= 2550

The integers from 1 to 100, which are divisible by 5, are 5, 10… 100.

This forms an A.P. with both the first term and common difference equal to 5.

So, 100 = 5 + (n –1) 5

⇒ 5n = 100

⇒ n = 20

So, 5 + 10 + 15 + ……100 = (20/2)[2 * 5 + (20 - 1)5]

= 10[10 + 95]

= 10 * 105

= 1050

The integers which are divisible by both 2 and 5 are 10, 20, … 100.

This also forms an A.P. with both the first term and common difference equal to 10.

So, 100 = 10 + (n –1) (10)

⇒ 100 = 10n

⇒ n = 10

So, 10 + 20 + 30 + ……100 = (10/2)[2 * 10 + (10 - 1)10]

= 5[20 + 90]

= 5 * 110

= 550

Hence, required sum = 2550 + 1050 – 550 = 3050

Thus, the sum of the integers from 1 to 100, which are divisible by 2 or 5 is 3050.

Question 6:

Find the sum of all two digit numbers which when divided by 4, yields 1 as remainder.

The two-digit numbers, which when divided by 4, yield 1 as remainder, are 13, 17, … 97.

This series forms an A.P. with first term 13 and common difference 4.

Let n be the number of terms of the A.P.

It is known that the nth term of an A.P. is given by,

an = a + (n –1)d

97 = 13 + (n –1)4

⇒ 4(n –1) = 84

⇒ n – 1 = 21

⇒ n = 22

Sum of n terms of an A.P. is given by

Sn = (n/2)[2a + (n - 1)d]

So, S22 = (22/2)[2 * 13 + (22 - 1)4]

= 11[26 + 84]

= 11 * 110

= 1210

Thus, the required sum is 1210.

Question 7:

If f is a function satisfying f(x + y) = f(x).f(y) for all x, y ∈ N, such that f(1) = 3 and ∑1n 𝑓(𝑥) = 120, find the value of n.

It is given that,

f (x + y) = f (x) × f (y) for all x, y ∈ N ………………. (1)

f(1) = 3

Taking x = y = 1 in equation (1), we obtain

f(1 + 1) = f (2) = f (1) * f (1) = 3 * 3 = 9

Similarly,

f (1 + 1 + 1) = f (3) = f (1 + 2) = f (1) * f (2) = 3 * 9 = 27

f (4) = f (1 + 3) = f (1) * f (3) = 3 * 27 = 81

So, f (1), f (2), f (3), …, that is 3, 9, 27, …, forms a G.P. with both the first term and common

ratio equal to 3.

It is known that, Sn = a(rn - 1)/(r - 1)

It is given that ∑1n 𝑓(𝑥) = 120

=> 120 = 3(3n - 1)/(3 - 1)

=> 40 = (3n - 1)/2

=> 80 = 3n – 1

=> 3n = 81

=> 3n = 34

=> n = 4

Thus, the value of n is 4.

Question 8:

The sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively.

Find the last term and the number of terms.

Let the sum of n terms of the G.P. be 315.

It is known that Sn = a(rn - 1)/(r - 1)

It is given that the first term a is 5 and common ratio r is 2.

Now, 315 = 5(2n - 1)/(2 - 1)

=> 63 = 2n – 1

=> 2n = 64

=> 2n =26

=> n = 6

So, last term of the G.P = 6th term = ar6 – 1 = 5 * 25 = 5 * 32 = 160

Thus, the last term of the G.P. is 160.

Question 9:

The first term of a G.P. is 1. The sum of the third term and fifth term is 90. Find the common ratio of G.P.

Let a and r be the first term and the common ratio of the G.P. respectively.

Now a = 1, a3 = ar2 = r2, a5 = ar4 = r4

r2 + r4 = 90

⇒ r4 + r2 – 90 = 0

⇒ r2 = [-1 ± √(1 + 360)]/2

= [-1 ± √(361)]/2

= [-1 ± 19]/2

= -10, 9

⇒ r = ±3             [Taking real roots]

Thus, the common ratio of the G.P. is ±3.

Question 10:

The sum of three numbers in G.P. is 56. If we subtract 1, 7, 21 from these numbers in that order, we obtain an arithmetic progression.

Find the numbers.

Let the three numbers in G.P. be a, ar, and ar2.

From the given condition,

a + ar + ar2 = 56

⇒ a(1 + r + r2) = 56 ……………………….… (1)

a – 1, ar – 7, ar2 – 21 forms an A.P.

So, (ar – 7) – (a – 1) = (ar2 – 21) – (ar – 7)

⇒ ar – a – 6 = ar2 – ar – 14

⇒ ar2 – 2ar + a = 8

⇒ ar2 – ar – ar + a = 8

⇒ a(r2 + 1 – 2r) = 8

⇒ a(r – 1)2 = 8 ……………………………… (2)

From equation (1) and (2), we get

⇒7(r2 – 2r + 1) = 1 + r + r2

⇒7r2 – 14 r + 7 – 1 – r – r2 = 0

⇒ 6r2 – 15r + 6 = 0

⇒ 6r2 – 12r – 3r + 6 = 0

⇒ 6r(r – 2) – 3(r – 2) = 0

⇒ (6r – 3)(r – 2) = 0

⇒ r = 2, 1/2

When r = 2, a = 8

When r = 1/2, a = 32

Therefore, when r = 2, the three numbers in G.P. are 8, 16, and 32.

When, r = 1/2, the three numbers in G.P. are 32, 16, and 8.

Thus, in either case, the three required numbers are 8, 16, and 32.

Question 11:

A G.P. consists of an even number of terms. If the sum of all the terms is 5 times the sum of terms occupying odd places,

then find its common ratio.

Let the G.P. be T1, T2, T3, T4, …….., T2n.

Number of terms = 2n

According to the given condition,

T1 + T2 + T3 + …+ T2n = 5[T1 + T3 + … +T2n–1]

⇒ T1 + T2 + T3 + …+ T2n - 5[T1 + T3 + … +T2n–1] = 0

⇒ T2 + T4 + T6 + …+ T2n = 4[T1 + T3 + … +T2n–1]

Let the G.P. be a, ar, ar2, ar3 ……..

ar(rn - 1)/(r - 1) = 4 * a(rn - 1)/(r - 1)

=> ar = 4a

=> r = 4

Thus, the common ratio of the G.P. is 4.

Question 12:

The sum of the first four terms of an A.P. is 56. The sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Let the A.P. be a, a + d, a + 2d, a + 3d ... a + (n – 2) d, a + (n – 1)d.

Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d

Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d]

= 4a + (4n – 10) d

According to the given condition,

4a + 6d = 56

⇒ 4 * 11 + 6d = 56                  [Since a = 11 (given)]

⇒ 6d = 12

⇒ d = 2

So, 4a + (4n –10) d = 112

⇒ 4 * 11 + (4n – 10)2 = 112

⇒ (4n – 10)2 = 68

⇒ 4n – 10 = 34

⇒ 4n = 44

⇒ n = 11

Thus, the number of terms of the A.P. is 11.

Question 13:

If (a + bx)/(a - bx) = (b + cx)/(b - cx) = (c + dx)/(c - dx) (x ≠ 0) then show that a, b, c and d are in G.P.

It is given that,

(a + bx)/(a - bx) = (b + cx)/(b - cx)

=> (a + bx)(b - cx) = (b + cx)(a - bx)

=> ab – acx + b2 x – bcx2 = ab –b2 x + acx – bcx2

=> 2b2 x = 2acx

=> b2 = ac

=> b/a = c/b  ………..1

Also, (b + cx)/(b - cx) = (c + dx)/(c - dx)

=> (b + cx)/(c - dx) = (b - cx)/(c + dx)

=> bc – bdx + c2 x – cdx2 = bc + bdx - c2 x –cdx2

=> 2c2 x = 2bdx

=> c2 = bd

=> c/d = d/c   …………2

From equation 1 and 2, we obtain

b/a = c/b = d/c

Thus, a, b, c, and d are in G.P.

Question 14:

Let S be the sum, P the product and R the sum of reciprocals of n terms in a G.P. Prove that P2Rn = Sn

Let the G.P. be a, ar, ar2, ar3, … arn – 1

According to the given information,

S = a(rn - 1)/(r - 1)

P = an * r1 + 2 + 3 + …..+ n-1 = an rn(n-1)/2

R = 1/a + 1/ar + ………+ 1/arn-1

= (rn-1 + rn-2 +………+ r + 1)/arn-1

= {1(rn - 1)/(r - 1)} * (1/ arn-1)

= (rn - 1)/{(r - 1) * arn-1}

Now, P2Rn = a2n * rn(n - 1) * [(rn - 1)/{(r - 1) * arn-1}]n

= a2n * rn(n - 1) * [(rn - 1)n/{(r - 1)n * an rn(n -1)}]

= an [(rn - 1)n/(r - 1)n]

= [a(rn - 1)/(r - 1)]n

= Sn

Hence, P2Rn = Sn

Question 15:

The pth, qth and rth terms of an A.P. are a, b, c respectively. Show that (q - r)a + (r - p)b + (p - q)c = 0

Let t and d be the first term and the common difference of the A.P. respectively.

The nth term of an A.P. is given by, an = t + (n – 1) d

Therefore,

ap = t + (p – 1) d = a ......................... (1)

aq = t + (q – 1)d = b .......................... (2)

ar = t + (r – 1) d = c .......................... (3)

Subtracting equation (2) from (1), we obtain

(p – 1 – q + 1) d = a – b

⇒ (p – q) d = a – b

⇒ d = (a – b)/(p - q)

Subtracting equation (3) from (2), we obtain

(q – 1 – r + 1) d = b – c

⇒ (q – r) d = b – c

⇒ d = (b – c)/(q - r)

Equating both the values of d obtained in (4) and (5), we obtain

(a – b)/(p - q) = (b – c)/(q - r)

⇒ (a – b)(q - r) = (b – c)(p - q)

⇒ aq – bq – ar + br = bp – bq – cp + cq

⇒ bp – cp + cq – aq + ar – br = 0

⇒ (-aq + ar) + (bp - br) + (-cp + cq) = 0

⇒ -a(q – r) – b(r - p) – c(p - q) = 0

⇒ a(q – r) + b(r - p) + c(p - q) = 0

Thus, the given result is proved.

Question 16:

If a(1/b + 1/c), b(1/c + 1/a), c(1/a +1/b) are in A.P., prove that a, b, c are in A.P.

It is given that a(1/b + 1/c), b(1/c + 1/a), c(1/a +1/b) are in A.P.

=> b(1/c + 1/a) - a(1/b + 1/c) = c(1/a +1/b) - b(1/c + 1/a)

=> b(a + c)/ac - a(b + c)/bc = c(a + b)/ab - b(a + c)/ac

=> (b2 a + b2 c – a2 b – a2 c)/abc = (c2 a + c2 b – b2 a – b2 c)/abc

=> b2 a + b2 c – a2 b – a2 c = c2 a + c2 b – b2 a – b2 c

=> ab(b - a) + c(b2 – a2) = a(c2 – b2) + bc(c - b)

=> ab(b - a) + c(b – a)(b + a) = a(c – b)(c + b) + bc(c - b)

=> (b - a)(ab + bc + ca) = (c - b)(ab + bc + ca)

=> b – a = c - b

Thus, a, b, and c are in A.P.

Question 17:

If a, b, c, d are in G.P, prove that (an + bn), (bn + cn), (cn + dn) are in G.P.

It is given that a, b, c, and d are in G.P.

b2 = ac ................................. (1)

c2 = bd ..................................(2)

It has to be proved that (an + bn), (bn + cn), (cn + dn) are in G.P. i.e.,

(bn + cn)2 = (an + bn)(cn + dn)

L.H.S.

(bn + cn)2 = b2n + 2bncn + c2n

= (b2)n + 2bncn + (c2)n

= (ac)n + 2bncn + (bd)n                  [Using equation (1) and (2)]

= an cn + bncn + bncn + bndn

= an cn + bncn + andn + bndn          [Using equation (3)]

= cn (an + bn) + dn (an + bn)

= (an + bn)(cn + dn)

= R.H.S.

Thus, (an + bn), (bn + cn), (cn + dn) are in G.P.

Question 18:

If a and b are the roots of x2 – 3x + p = 0 and c, d are roots of x2 – 12x + q = 0, where a, b, c, d, form a G.P. Prove that (q + p) : (q – p) = 17 : 15.

It is given that a and b are the roots of x2 – 3x + p = 0

So, a + b = 3 and ab = p ………………………………… (1)

Also, c and d are the roots of x2 – 12x + q = 0

So, c + d = 12 and cd = q ………………………………… (2)

It is given that a, b, c, d are in G.P.

Let a = x, b = xr, c = xr2, d = xr3

From equations (1) and (2), we get

x + xr = 3

⇒ x(1 + r) = 3

xr2 + xr3 =12

⇒ xr2(1 + r) = 12

On dividing, we obtain

xr2(1 + r)/ x(1 + r) = 12/3

=> r2 = 4

=> r = ±2

When r = 2, x = 3/(1 + 2) = 3/3 = 1

When r = -2, x = 3/(1 - 2) = 3/(-1) = -3

Case 1:

When r = 2 and x =1, ab = x2 r = 2, cd = x2r5 = 32

So, (q + p)/(q – p) = (32 + 2)/(32 – 2) = 34/30 = 17/15

Hence, (q + p) : (q – p) = 17 : 15

Case 2:

When r = -2 and x = -3, ab = x2 r = -18, cd = x2r5 = -288

So, (q + p)/(q – p) = (-288 - 18)/(-288 + 18) = -306/(-270) = 17/15

Hence, (q + p) : (q – p) = 17 : 15

Thus, in both the cases, we obtain (q + p) : (q – p) = 17 : 15

Question 19:

The ratio of the A.M and G.M. of two positive numbers a and b, is m : n. Show that

a : b = [m + √(m2 – n2)] : [m + √(m2 – n2)]

Let the two numbers be a and b.

A.M = (a + b)/2 and GM = √ab

According to the given condition,

(a + b)/2√ab = m/n

=> (a + b)2/4ab = m2/n2

=> (a + b)2 = 4abm2/n2

=> (a + b) = 2m√(ab)/n   …………….1

Using this in the identity (a – b)2 = (a + b)2 – 4ab, we obtain

(a – b)2 = 4abm2/n2 – 4ab

= 4ab(m2 - n2)/n2

=> a – b = 2√(ab) √(m2 - n2)/n   …………2

Adding equation (1) and (2), we obtain

2a = [2√(ab){m + √(m2 - n2)}]/n

=> a = [√(ab){m + √(m2 - n2)}]/n

Substitute the value of a in equation (1), we get

b = 2m√(ab)/n - [√(ab){m + √(m2 - n2)}]/n

=> b = m√(ab)/n – [√(ab) *  √(m2 - n2)}]/n

=> b = {√(ab)/n}[m - √(m2 - n2)]

Now, a/b = [[√(ab){m + √(m2 - n2)}]/n]/[{√(ab)/n}[m - √(m2 - n2)]]

=> a/b = [m + √(m2 - n2)]/[m - √(m2 - n2)]

=> a : b = [m + √(m2 - n2)] : [m - √(m2 - n2)]

Question 20:

If a, b, c are in A.P; b, c, d are in G.P and 1/c, 1/d, 1/e are in A.P. prove that a, c, e are in G.P.

It is given that a, b, c are in A.P.

b – a = c – b ………………………….. (1)

It is given that b, c, d, are in G.P.

c2 = bd …………………………………… (2)

Also, 1/c, 1/d, 1/e are in HP

=> 1/d – 1/c = 1/e – 1/d

=> 2/d = 1/c + 1/e   ………………..(3)

It has to be proved that a, c, e are in G.P. i.e., c2 = ae

From equation (1), we obtain

2b = a + c

=> b = (a + c)/2

From equation (2), we obtain

d = c2/b

Substituting these values in equation (3), we obtain

2b/c2 = 1/c + 1/e

=> 2(a + c)/2c2 = (c + e)/ce

=> (a + c)/c2 = (c + e)/ce

=> (a + c)/c = (c + e)/e

=> (a + c)e = c(c + e)

=> ae + ce = c2 + ce

=> c2 = ae

Thus, a, c, and e are in G.P.

Question 21:

Find the sum of the following series up to n terms:

(i) 5 + 55 + 555 + …….                        (ii) .6 +.66 +. 666 +………….

(i) The given series is 5 + 55 + 555 + ……..

Now, Sn = 5 + 55 + 555 + ……… to n terms

=> Sn = 5[1 + 11 + 111 + ……… to n terms]

=> Sn = (5/9)[9 + 99 + 999 + ……… to n terms]

=> Sn = (5/9)[(10 - 1) + (102 - 1) + (103 - 1) + ……… to n terms]

=> Sn = (5/9)[(10 + 102 + 103 + ……… to n terms) – (1 + 1 + 1 + …. to n terms)]

=> Sn = (5/9)[(10(10n - 1)/(10 - 1) – n]

=> Sn = (5/9)[(10(10n - 1)/9 – n]

=> Sn = 50(10n - 1)/81 – 5n/9

(ii) The given series is .6 +.66 +. 666 +………….

Now, Sn = .6 +.66 +. 666 +………….… to n terms

=> Sn = 6[0.1 + 0.11 + 0.111 + ……… to n terms]

=> Sn = (6/9)[0.9 + 0.99 + 0.999 + ……… to n terms]

=> Sn = (6/9)[(1 – 1/10) + (1 – 1/102) + (1 – 1/103) + ……… to n terms]

=> Sn = (6/9)[ (1 + 1 + 1 + …. to n terms) - (1/10 + 1/102 + 1/103 + ……… to n terms)]

=> Sn = (2/3)[n - (1/10){(1 – (1/10)n)/(1 – 1/10 )]

=> Sn = 2n/3 – (2/30) * (10/9) * (1 – 10-n )

=> Sn = 2n/3 – (2/27) * (1 – 10-n )

Question 22:

Find the 20th term of the series 2 * 4 + 4 * 6 + 6 * 8 + … + n terms.

The given series is 2 * 4 + 4 * 6 + 6 * 8 + … n terms

So, nth term = an = 2n * (2n + 2) = 4n2 + 4n

a20 = 4 * 202 + 4 * 20

= 4 * 400 + 80

= 1600 + 80

= 1680

Thus, the 20th term of the series is 1680.

Question 23:

Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 + …

The given series is 3 + 7 + 13 + 21 + 31 + …

S = 3 + 7 + 13 + 21 + 31 + …+ an–1 + an

S = 3 + 7 + 13 + 21 + …. + an – 2 + an–1 + an

On subtracting both the equations, we obtain

S – S = [3 + (7 + 13 + 21 + 31 + …+ an–1 + an)] – [(3 + 7 + 13 + 21 +31 + …+ an–1) + an]

=> S – S = 3 + [(7 – 3) + (13 – 7) + (21 – 13) + … + (an – an–1)] – an

=> 0 = 3 + [4 + 6 + 8 + … (n –1) terms] – an

=> an = 3 + [4 + 6 + 8 + … (n –1) terms]

=> an = 3 + {(n - 1)/2}[2 * 4 + (n – 1 - 1)2]

=> an = 3 + {(n - 1)/2}[8 + (n – 2)2]

=> an = 3 + {(n - 1)/2}[2n + 4]

=> an = 3 + (n - 1)(n + 2)

=> an = 3 + (n2 + n - 2)

=> an = n2 + n + 1

Now, Sn = ak

=    +    +

= {n(n + 1)(2n + 1)}/6 + n(n + 1)/2 + n

= n[{(n + 1)(2n + 1)}/6 + (n + 1)/2 + 1]

= n[2n2 + 3n + 1 + 3n + 3 + 6]/6

= n[2n2 + 6n + 10]/6

= n[n2 + 3n + 5]/3

Question 24:

If S1, S2, S3 are the sum of first n natural numbers, their squares and their cubes, respectively, show that 9S22 = S3(1 + 8S1)

Given, S1 is the sum of first n natural numbers

=> S1 = n(n + 1)/2

Again, S3 is the sum of square first n natural numbers

=> S3 = n2(n + 1)2/4

Now, S3(1 + 8S1) = [n2(n + 1)2/4][1 + 8n(n + 1)/2]

= [n2(n + 1)2/4][1 + 4n(n + 1)]

= [n2(n + 1)2/4][1 + 4n2 + 4n]

= [n2(n + 1)2/4](2n + 1)2   ……………1

Also, 9S22 = 9[{n(n + 1)(2n + 1)}/6]2

= 9{n(n + 1)(2n + 1)}2/36

= {n(n + 1)(2n + 1)}2/4      …………….2

From equation 1 and 2, we get

9S22 = S3(1 + 8S1)

Question 25:

Find the sum of the following series up to n terms:

13/1 + (13 + 23)/(1 + 3) + (13 + 23 + 33)/(1 + 3 + 5) + ………..

The nth term of the given series is an = (13 + 23 + 33 + …….+ n3)/(1 + 3 + 5 + ………..(2n - 1))

= [n(n + 1)/2]2/[(n/2){2 * 1 + (n - 1)2}]

= [n(n + 1)/2]2/n2

= (n + 1)2/4

=> an = n2/4 + n/2 + 1/4

Now, Sn = ak

= (1/4)  + (1/2)  + /4

= (1/4){n(n + 1)(2n + 1)}/6 + (1/4) * n(n + 1)/2 + n/4

= n[{(n + 1)(2n + 1)} + 6(n + 1) + 6]/24

= n[2n2 + 3n + 1 + 6n + 6 + 6]/24

= n[2n2 + 9n + 13]/24

Question 26:

Show that {1 * 22 + 2 * 23 + …..n * (n + 1)2}/ {12 * 2 + 22 * 2 + …..n2 * (n + 1)} = (3n + 5)/(3n + 1)

nth term of numerator = n(n + 1)2 = n3 + 2n2 + n

nth term of denominator = n2(n + 1) = n3 + n2

Now, {1 * 22 + 2 * 23 + …..n * (n + 1)2}/ {12 * 2 + 22 * 2 + …..n2 * (n + 1)}

=   a/ ak

= [  +  2  + ]/[   +  ]

Here,  +  2  +

= n2(n + 1)2/4 + 2{n(n + 1)(2n + 1)}/6 + n(n + 1)/2

= n(n + 1)/2[n(n + 1)/2 + 2(2n + 1)/3 + 1]

= n(n + 1)/2[3n2 + 3n + 8n + 4 + 6]/6

= n(n + 1)/2[3n2 + 11n + 10]/6

= n(n + 1)/2[3n2 + 6n + 5n + 10]/6

= n(n + 1)/2[3n(n + 2) + 5(n + 2)]/6

= [[n(n + 1)(n + 2)(3n + 5)]/12   ……………1

Again  +   = n2(n + 1)2/4 + {n(n + 1)(2n + 1)}/6

= n(n + 1)/2[n(n + 1)/2 + (2n + 1)/3]

= n(n + 1)/2[3n2 + 3n + 4n + 2]/6

= n(n + 1)/2[3n2 + 7n + 2]/6

= n(n + 1)[3n2 + 6n + n + 2]/12

= n(n + 1)[3n(n + 2) + 1(n + 2)]/12

= [n(n + 1)(n + 2)(3n + 1)]/12       ………….3

From equation 1, 2 and 3, we get

{1 * 22 + 2 * 23 + …..n * (n + 1)2}/ {12 * 2 + 22 * 2 + …..n2 * (n + 1)}

= [{n(n + 1)(n + 2)(3n + 5)}/12]/[{n(n + 1)(n + 2)(3n + 1)}/12]

= (3n + 5)/(3n + 1)

Thus, the given result is proved.

Question 27:

A farmer buys a used tractor for Rs 12000. He pays Rs 6000 cash and agrees to pay the balance in annual installments of Rs 500

plus 12% interest on the unpaid amount. How much will be the tractor cost him?

It is given that the farmer pays Rs 6000 in cash.

Therefore, unpaid amount = Rs 12000 – Rs 6000 = Rs 6000

According to the given condition, the interest paid annually is 12% of 6000, 12% of 5500, 12%

of 5000, …….., 12% of 500

Thus, total interest to be paid

= 12% of 6000 + 12% of 5500 + 12% of 5000 + … + 12% of 500

= 12% of (6000 + 5500 + 5000 + … + 500)

= 12% of (500 + 1000 + 1500 + … + 6000)

Now, the series 500, 1000, 1500 … 6000 is an A.P. with both the first term and common

difference equal to 500.

Let the number of terms of the A.P. be n.

So, 6000 = 500 + (n – 1) 500

⇒ 1 + (n – 1) = 12

⇒ n = 12

Sum of the A.P = (12/2)[2 * 500 + (12 - 1)500]

= 6[1000 + 5500]

= 6 * 6500

= 39000

Thus, total interest to be paid = 12% of (500 + 1000 + 1500 + … + 6000)

= 12% of 39000

= (12/100) * 39000

= Rs 4680

Thus, the cost of tractor = (Rs 12000 + Rs 4680) = Rs 16680

Question 28:

Shamshad Ali buys a scooter for Rs 22000. He pays Rs 4000 cash and agrees to pay the balance in annual installment of

Rs 1000 plus 10% interest on the unpaid amount. How much will the scooter cost him?

It is given that Shamshad Ali buys a scooter for Rs 22000 and pays Rs 4000 in cash.

So, Unpaid amount = Rs 22000 – Rs 4000 = Rs 18000

According to the given condition, the interest paid annually is

10% of 18000, 10% of 17000, 10% of 16000 … 10% of 1000

Thus, total interest to be paid

= 10% of 18000 + 10% of 17000 + 10% of 16000 + … + 10% of 1000

= 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of (1000 + 2000 + 3000 + … + 18000)

Here, 1000, 2000, 3000 … 18000 forms an A.P. with first term and common difference both

equal to 1000.

Let the number of terms be n.

So, 18000 = 1000 + (n – 1) (1000)

⇒ n = 18

Now, 1000 + 2000 + 3000 + ……….+ 18000

= (18/2)[2 * 1000 + (18 - 1)1000]

= 9[2000 + 17 * 1000]

= 9[2000 + 17000]

= 9 * 19000

= 171000

Total interest paid = 10% of (18000 + 17000 + 16000 + … + 1000)

= 10% of Rs 171000

= (10/100) * Rs 171000

= Rs 17100

Thus, the cost of scooter = Rs 22000 + Rs 17100 = Rs 39100

Question 29:

A person writes a letter to four of his friends.

He asks each one of them to copy the letter and mail to four different persons with instruction that they move the chain similarly.

Assuming that the chain is not broken and that it costs 50 paise to mail one letter.

Find the amount spent on the postage when 8th set of letter is mailed.

The numbers of letters mailed forms a G.P.: 4, 42, ………..,48

First term = 4

Common ratio = 4

Number of terms = 8

It is known that the sum of n terms of a G.P. is given by

Sn = a(rn - 1)/(r - 1)

So, S8 = 4(48 - 1)/(4 - 1)

= 4(65536 - 1)/3

= (4 * 65535)/3

= 4 * 21845

= 87380

It is given that the cost to mail one letter is 50 paisa.

So, cost of mailing 87380 letters = Rs 87380 * 50/100 = Rs 43690

Thus, the amount spent when 8th set of letter is mailed is Rs 43690.

Question 30:

A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually.

Find the amount in 15th year since he deposited the amount and also calculate the total amount after 20 years.

It is given that the man deposited Rs 10000 in a bank at the rate of 5% simple interest

annually.

Interest in first year = 5/100 * Rs 10000 = Rs 500

Amount in 15th year = 10000 + 500 + 500 +……….14 times

= Rs 10000 + 14 * Rs 500

= Rs 10000 + Rs 7000

= Rs 17000

Amount after 20 years = Rs 10000 + 500 + 500 + ……………20 times

= Rs 10000 + 20 * Rs 500

= Rs 10000 + Rs 10000

= Rs 20000

Question 31:

A manufacturer reckons that the value of a machine, which costs him Rs 15625, will depreciate each year by 20%.

Find the estimated value at the end of 5 years.

Cost of machine = Rs 15625

Machine depreciates by 20% every year.

Therefore, its value after every year is 80% of the original cost i.e.,4/5 of the original cost.

Value at the end of 5 years = 15625 * 4/5 * 4/5 * …….5 times

= 5 * 1024 = 5120

Thus, the value of the machine at the end of 5 years is Rs 5120.

Question 32:

150 workers were engaged to finish a job in a certain number of days. 4 workers dropped out on second day,

4 more workers dropped out on third day and so on. It took 8 more days to finish the work.

Find the number of days in which the work was completed.

Let x be the number of days in which 150 workers finish the work.

According to the given information,

150x = 150 + 146 + 142 + …. (x + 8) terms

The series 150 + 146 + 142 + …. (x + 8) terms is an A.P. with first term 146, common difference

–4 and number of terms as (x + 8)

=> 150x = {(x + 8)/2}[2 * 150 + (x + 8 - 1)(-4)]

=> 150x = (x + 8)[150 + (x + 7)(-2)]

=> 150x = (x + 8)[150 - 2x - 14]

=> 150x = (x + 8)[136 - 2x]

=> 75x = (x + 8)(68 – x)

=> 75x = 68x – x2 + 544 – 8x

=> x2 + 75x - 60x – 544 = 0

=> x2 + 15x  – 544 = 0

=> x2 + 32x - 17x – 544 = 0

=> x(x + 32) – 17(x + 32) = 0

=> (x + 32)(x - 17) = 0

=> x = -32, 17

However, x cannot be negative.

So, x = 17

Therefore, originally, the number of days in which the work was completed is 17.

Thus, required number of days = (17 + 8) = 25