Exercise 9.1

Write the first five terms of each of the sequences in Exercises 1 to 6 whose nth terms are:

1. a_n = n (n + 2) 

Solution: –

Given,

n^th term of a sequence a_n = n (n + 2)

On substituting n = 1, 2, 3, 4, and 5, we get the first five terms

a₁ = 1(1 + 2) = 3

a₂ = 2(2 + 2) = 8

a₃ = 3(3 + 2) = 15

a₄ = 4(4 + 2) = 24

a₅ = 5(5 + 2) = 35

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

2. a_n = n/n+1

Solution: –

Given n^th term, a_n = n/n+1

On substituting n = 1, 2, 3, 4, 5, we get

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

3. a_n = 2ⁿ

Solution: –

Given n^th term, a_n = 2ⁿ

On substituting n = 1, 2, 3, 4, 5, we get

a₁ = 2¹ = 2

a₂ = 2² = 4

a₃ = 2³ = 8

a₄ = 2⁴ = 16

a₅ = 2⁵ = 32

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

4.  a_n = (2n – 3)/6

Solution: –

Given n^th term, a_n = (2n – 3)/6

On substituting n = 1, 2, 3, 4, 5, we get

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

5. a_n = (-1)^n-1 5ⁿ⁺¹

Solution: –

Given n^th term, a_n = (-1)^n-1 5ⁿ⁺¹

On substituting n = 1, 2, 3, 4, 5, we get

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

6.

Solution: –

On substituting n = 1, 2, 3, 4, 5, we get first 5 terms

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

Find the indicated terms in each of the sequences in Exercises 7 to 10 whose n^th terms are:

7. a_n = 4n – 3; a₁₇, a₂₄

Solution: –

Given,

n^th term of the sequence is a_n = 4n – 3

On substituting n = 17, we get

a₁₇ = 4(17) – 3 = 68 – 3 = 65

Next, on substituting n = 24, we get

a₂₄ = 4(24) – 3 = 96 – 3 = 93

8. a_n = n²/2ⁿ ; a⁷

Solution: –

Given,

n^th term of the sequence is a_n = n²/2ⁿ

Now, on substituting n = 7, we get

a₇ = 7²/2⁷ = 49/ 128

9. a_n = (-1)^n-1 n³; a₉

Solution: –

Given,

n^th term of the sequence is a_n = (-1)^n-1 n³

On substituting n = 9, we get

a₉ = (-1)^9-1 (9)³ = 1 x 729 = 729

10.

Solution: –

On substituting n = 20, we get

Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series:

11. a₁ = 3, a_n = 3a_n-1 + 2 for all n > 1

Solution: –

Given, a_n = 3a_n-1 + 2 and a₁ = 3

Then,

a₂ = 3a₁ + 2 = 3(3) + 2 = 11

a₃ = 3a₂ + 2 = 3(11) + 2 = 35

a₄ = 3a₃ + 2 = 3(35) + 2 = 107

a₅ = 3a₄ + 2 = 3(107) + 2 = 323

Thus, the first 5 terms of the sequence are 3, 11, 35, 107 and 323.

Hence, the corresponding series is

3 + 11 + 35 + 107 + 323 …….

12. a₁ = -1, a_n = a_n-1/n, n ≥ 2

Solution: –

Given,

a_n = a_n-1/n and a₁ = -1

Then,

a₂ = a₁/2 = -1/2

a₃ = a₂/3 = -1/6

a₄ = a₃/4 = -1/24

a₅ = a₄/5 = -1/120

Thus, the first 5 terms of the sequence are -1, -1/2, -1/6, -1/24 and -1/120.

Hence, the corresponding series is

-1 + (-1/2) + (-1/6) + (-1/24) + (-1/120) + …….

13. a₁ = a₂ = 2, a_n = a_n-1 – 1, n > 2

Solution: –

Given,

a₁ = a₂, a_n = a_n-1 – 1

Then,

a₃ = a₂ – 1 = 2 – 1 = 1

a₄ = a₃ – 1 = 1 – 1 = 0

a₅ = a₄ – 1 = 0 – 1 = -1

Thus, the first 5 terms of the sequence are 2, 2, 1, 0 and -1.

The corresponding series is

2 + 2 + 1 + 0 + (-1) + ……

14. The Fibonacci sequence is defined by

1 = a₁ = a₂ and a_n = a_{n – 1} + a_{n – 2}, n > 2

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

Solution: –

Given,

1 = a₁ = a₂

a_n = a_{n – 1} + a_{n – 2}, n > 2

So,

a₃ = a₂ + a₁ = 1 + 1 = 2

a₄ = a₃ + a₂ = 2 + 1 = 3

a₅ = a₄ + a₃ = 3 + 2 = 5

a₆ = a₅ + a₄ = 5 + 3 = 8

Thus,

Exercise 9.2

1. Find the sum of odd integers from 1 to 2001.

Solution: –

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

It clearly forms a sequence in A.P.

Where, the first term, a = 1

Common difference, d = 2

Now,

a + (n -1)d = 2001

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

2n – 2 = 2000

2n = 2000 + 2 = 2002

n = 1001

We know,

S_n = n/2 [2a + (n-1)d]

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

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

Solution: –

The natural numbers lying between 100 and 1000, which are multiples of 5, are 105, 110, … 995.

It clearly forms a sequence in A.P.

Where, the first term, a = 105

Common difference, d = 5

Now,

a + (n -1)d = 995

105 + (n – 1)(5) = 995

105 + 5n – 5 = 995

5n = 995 – 105 + 5 = 895

n = 895/5

n = 179

We know,

S_n = n/2 [2a + (n-1)d]

Therefore, the sum of all natural numbers lying between 100 and 1000, which are multiples of 5, is 98450.

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 20^th term is –112.

Solution: –

Given,

The first term (a) of an A.P = 2

Let’s assume d be the common difference of the A.P.

So, the A.P. will be 2, 2 + d, 2 + 2d, 2 + 3d, …

Then,

Sum of first five terms = 10 + 10d

Sum of next five terms = 10 + 35d

From the question, we have

10 + 10d = ¼ (10 + 35d)

40 + 40d = 10 + 35d

30 = -5d

d = -6

a₂₀ = a + (20 – 1)d = 2 + (19) (-6) = 2 – 114 = -112

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

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

Solution: –

Let’s consider the sum of n terms of the given A.P. as –25.

We known that,

S_n = n/2 [2a + (n-1)d]

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

So here, a = –6

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

Thus, we have

5. In an A.P., if p^th term is 1/q and q^th term is 1/p, prove that the sum of first pq terms is ½ (pq + 1) where p ≠ q.  

Solution: –

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

Solution: –

Given A.P.,

25, 22, 19, …

Here,

First term, a = 25 and

Common difference, d = 22 – 25 = -3

Also given, sum of certain number of terms of the A.P. is 116

The number of terms be n

So, we have

S_n = n/2 [2a + (n-1)d] = 116

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

116 x 2 = n [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

(3n – 29) (n – 8) = 0

Hence,

n = 29/3 or n = 8

As n can only be an integral value, n = 8

Thus, 8^th term is the last term of the A.P.

a₈ = 25 + (8 – 1)(-3)

= 25 – 21

= 4

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

Solution: –

Given, the k^th term of the A.P. is 5k + 1.

k^th term = a_k = a + (k – 1)d

And,

a + (k – 1)d = 5k + 1

a + kd – d = 5k + 1

On comparing the coefficient of k, we get d = 5

a – d = 1

a – 5 = 1

⇒ a = 6

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

Solution: –

We know that,

S_n = n/2 [2a + (n-1)d]

From the question we have,

On comparing the coefficients of n² on both sides, we get

d/2 = q

Hence, d = 2q

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

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

Solution: –

Let a₁, a₂, and d₁, d₂ be the first terms and the common difference of the first and second arithmetic progression respectively.

Then, from the question we have

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.

Solution: –

Let’s take a and d to be the first term and the common difference of the A.P. respectively.

Then, it given that

Therefore, the sum of (p + q) terms of the A.P. is 0.

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

Prove that 

Solution: –

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

Then according to the question, we have

Now, subtracting (2) from (1), we get

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

Solution: –

Let’s consider that a and b to be the first term and the common difference of the A.P. respectively.

Then from the question, we have

Hence, the given result is proved.

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

Solution: –

Let’s consider a and b to be the first term and the common difference of the A.P. respectively.

a_m = a + (m – 1)d = 164 … (1)

We the sum of the terms is given by,

S_n = n/2 [2a + (n-1)d]

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

Solution: –

Let’s assume A₁, A₂, A₃, A₄, and A₅ to be five numbers between 8 and 26 such that 8, A₁, A₂, A₃, A₄, A₅, 26 are in an A.P.

Here we have,

a = 8, b = 26, n = 7

So,

26 = 8 + (7 – 1) d

6d = 26 – 8 = 18

d = 3

Now,

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

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

15. If

 is the A.M. between a and b, then find the value of n.

Solution: –

The A.M between a and b is given by, (a + b)/2

Then according to the question,

Thus, the value of n is 1.

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 7^th and (m – 1)^th numbers is 5: 9. Find the value of m.

Solution: –

Let’s consider a₁, a₂, … a_m be m numbers such that 1, a₁, a₂, … a_m, 31 is an A.P.

And here,

a = 1, b = 31, n = m + 2

So, 31 = 1 + (m + 2 – 1) (d)

30 = (m + 1) d

d = 30/ (m + 1) ……. (1)

Now,

a₁ = a + d

a₂ = a + 2d

a₃ = a + 3d …

Hence, a₇ = a + 7d

a_m–1 = a + (m – 1) d

According to the question, we have

Therefore, the value of m is 14.

17. A man starts repaying a loan as first instalment of Rs. 100. If he increases the instalment by Rs 5 every month, what amount he will pay in the 30^th instalment?

Solution: –

Given,

The first instalment of the loan is Rs 100.

The second instalment of the loan is Rs 105 and so on as the instalment increases by Rs 5 every month.

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

And the, A.P. is 100, 105, 110, …

Where, first term, a = 100

Common difference, d = 5

So, the 30^th term in this A.P. will be

A₃₀  = a + (30 – 1)d

= 100 + (29) (5)

= 100 + 145

= 245

Therefore, the amount to be paid in the 30^th instalment will be Rs 245.

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.

Solution: –

It’s understood from the question that, the angles of the polygon will form an A.P. with common difference d = 5° and first term a = 120°.

And, we know that the sum of all angles of a polygon with n sides is 180° (n – 2).

Thus, we can say

Thus, a polygon having 9 and 16 sides will satisfy the condition in the question.

Exercise 9.3

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

Solution: –

Given G.P. is 5/2, 5/4, 5/8, ………

Here, a = First term = 5/2

r = Common ratio = (5/4)/(5/2) = ½

Thus, the 20^th term and n^th term

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

Solution: –

Given,

The common ratio of the G.P., r = 2

And, let a be the first term of the G.P.

Now,

a₈ = ar ^8–1 = ar⁷

ar⁷ = 192

a(2)⁷ = 192

a(2)⁷ = (2)⁶ (3)

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

Solution: –

Let’s take a to be the first term and r to be the common ratio of the G.P.

Then according to the question, we have

a₅ = a r^5–1 = a r⁴ = p … (i)

a₈ = a r^8–1 = a r⁷ = q … (ii)

a₁₁ = a r^11–1 = a r¹⁰ = s … (iii)

Dividing equation (ii) by (i), we get

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

Solution: –

Let’s consider a to be the first term and r to be the common ratio of the G.P.

Given, a = –3

And we know that,

a_n = ar^n–1

So, a₄ = ar³ = (–3) r³

a₂ = a r¹ = (–3) r

Then from the question, we have

(–3) r³ = [(–3) r]²

⇒ –3r³ = 9 r²

⇒ r = –3

a₇ = a r ^7–1 = a r⁶ = (–3) (–3)⁶ = – (3)⁷ = –2187

Therefore, the seventh term of the G.P. is –2187.

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 ?

Solution: –

(a) The given sequence, 2, 2√2, 4,…

We have,

a = 2 and r = 2√2/2 = √2

Taking the n^th term of this sequence as 128, we have

Therefore, the 13^th term of the given sequence is 128.

(ii) Given sequence, √3, 3, 3√3,…

We have,

a = √3 and r = 3/√3 = √3

Taking the n^th term of this sequence to be 729, we have

Therefore, the 12^th term of the given sequence is 729.

(iii) Given sequence, 1/3, 1/9, 1/27, …

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

Taking the n^th term of this sequence to be 1/19683, we have

Therefore, the 9^th term of the given sequence is 1/19683.

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

Solution: –

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

Common ratio = x/(-2/7) = -7x/2

Also, common ratio = (-7/2)/x = -7/2x

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

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

Solution: –

Given G.P., 0.15, 0.015, 0.00015, …

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

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

Solution: –

The given G.P is √7, √21, 3√7, ….

Here,

a = √7 and

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

Solution: –

The given G.P. is 1, -a, a², -a³ ….

Here, the first term = a₁ = 1

And the common ratio = r = – a

We know that,

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

Solution: –

Given G.P. is x³, x⁵, x⁷, …

Here, we have a = x³ and r = x⁵/x³ = x²

11. Evaluate:

Solution: –

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.

Solution: –

Let a/r, a, ar be the first three terms of the G.P.

a/r + a + ar = 39/10 …… (1)

(a/r) (a) (ar) = 1 …….. (2)

From (2), we have

a³ = 1

Hence, a = 1 [Considering real roots only]

Substituting the value of a in (1), we get

1/r + 1 + r = 39/10

(1 + r + r²)/r = 39/10

10 + 10r + 10r² = 39r

10r² – 29r + 10 = 0

10r² – 25r – 4r + 10 = 0

5r(2r – 5) – 2(2r – 5) = 0

(5r – 2) (2r – 5) = 0

Thus,

r = 2/5 or 5/2

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

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

Solution: –

Given G.P. is 3, 3², 3³, …

Let’s consider that n terms of this G.P. be required to obtain the sum of 120.

We know that,

Equating the exponents we get, n = 4

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

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.

Solution: –

Let’s assume the G.P. to be a, ar, ar², ar³, …

Then according to the question, we have

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

a (1 + r + r²) = 16 … (1) and,

ar³(1 + r + r²) = 128 … (2)

Dividing equation (2) by (1), we get

r³ = 8

r = 2

Now, using r = 2 in (1), we get

a (1 + 2 + 4) = 16

a (7) = 16

a = 16/7

Now, the sum of terms is given as

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

Solution: –

Given,

a = 729 and a₇ = 64

Let r be the common ratio of the G.P.

Then we know that, a_n = a r^n–1

a₇ = ar^7–1 = (729)r⁶

⇒ 64 = 729 r⁶

r⁶ = 64/729

r⁶ = (2/3)⁶

r = 2/3

And, we know that

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.

Solution: –

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

Given, S₂ = -4

Then, from the question we have

And,

a₅ = 4 x a₃

ar⁴ = 4ar²

r² = 4

r = ± 2

Using the value of r in (1), we have

Therefore, the required G.P is

-4/3, -8/3, -16/3, …. Or 4, -8, 16, -32, ……

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

Solution: –

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

According to the given condition,

a₄ = a r³ = x … (1)

a₁₀ = a r⁹ = y … (2)

a₁₆ = a r¹⁵ = z … (3)

On dividing (2) by (1), we get

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

Solution: –

Given sequence: 8, 88, 888, 8888…

This sequence is not a G.P.

But, it can be changed to G.P. by writing the terms as

S_n = 8 + 88 + 888 + 8888 + …………….. to n terms

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.

Solution: –

The required sum = 2 x 128 + 4 x 32 + 8 x 8 + 16 x 2 + 32 x ½

= 64[4 + 2 + 1 + ½ + 1/2²]

Now, it’s seen that

4, 2, 1, ½, 1/2² is a G.P.

With first term, a = 4

Common ratio, r =1/2

We know,

Therefore, the required sum = 64(31/4) = (16)(31) = 496

20. Show that the products of the corresponding terms of the sequences a, ar, ar², …ar^n-1 and A, AR, AR², … AR^n-1 form a G.P, and find the common ratio.

Solution: –

To be proved: The sequence, aA, arAR, ar²AR², …ar^n–1AR^n–1, forms a G.P.

Now, we have

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

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 4^th by 18.

Solution: –

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

Then,

a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³

From the question, we have

a₃ = a₁ + 9

ar² = a + 9 … (i)

a₂ = a₄ + 18

ar = ar³ + 18 … (ii)

So, from (1) and (2), we get

a(r² – 1) = 9 … (iii)

ar (1– r²) = 18 … (iv)

Now, dividing (4) by (3), we get

-r = 2

r = -2

On substituting the value of r in (i), we get

4a = a + 9

3a = 9

∴ a = 3

Therefore, 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.

22. If the p^th, q^th and r^th terms of a G.P. are a, b and c, respectively. Prove that a^q-r b^r-p c^p-q = 1

Solution: –

Let’s take A to be the first term and R to be the common ratio of the G.P.

Then according to the question, we have

AR^p–1 = a

AR^q–1 = b

AR^r–1 = c

Then,

a^q–r b^r–p c^p–q

= A^q–r × R^{(p–1) (q–r)} × A^r–p × R^{(q–1) (r–p)} × A^p–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

Hence proved.

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

Solution: –

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

Thus,

The G.P. is a, ar, ar², ar³, … ar^n–1, where r is the common ratio.

Then,

b = ar^n–1  … (1)

P = Product of n terms

= (a) (ar) (ar²) … (ar^n–1)

= (a × a ×…a) (r × r² × …r^n–1)

= aⁿ r ^{1 + 2 +…(n–1)}  … (2)

Here, 1, 2, …(n – 1) is an A.P.

24. Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from .

Solution: –

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

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

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

a ^n +1 = ar ^{n + 1 – 1} = arⁿ

Thus, required ratio =

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 ^.

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

Solution: –

Given, a, b, c, d are in G.P.

So, we have

bc = ad  … (1)

b² = ac  … (2)

c² = bd  … (3)

Taking the R.H.S. we have

R.H.S.

= (ab + bc + cd)²

= (ab + ad + cd)²  [Using (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 (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)

= L.H.S.

Thus, L.H.S. = R.H.S.

Therefore,

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

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

Solution: –

Let’s assume G₁ and G₂ to be two numbers between 3 and 81 such that the series 3, G₁, G₂, 81 forms a G.P.

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

Now, we have the 1^st term as 3 and the 4^th term as 81.

81 = (3) (r)³

r³ = 27

∴ r = 3 (Taking real roots only)

For r = 3,

G₁ = ar = (3) (3) = 9

G₂ = ar² = (3) (3)² = 27

Therefore, the two numbers which can be inserted between 3 and 81 so that the resulting sequence becomes a G.P are 9 and 27.

27. Find the value of n so that may be the geometric mean between a and b.

Solution: –

We know that,

The G. M. of a and b is given by √ab.

Then from the question, we have

By squaring both sides, we get

28. The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio.

Solution: –

Consider the two numbers be a and b.

Then, G.M. = √ab.

From the question, we have

29. If A and G be A.M. and G.M., respectively between two positive numbers, prove that the

numbers are .

Solution: –

Given that A and G are A.M. and G.M. between two positive numbers.

And, let these two positive numbers be a and b.

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 2^nd hour, 4^th hour and n^th hour?

Solution: –

Given, the number of bacteria doubles every hour. Hence, the number of bacteria after every hour will form a G.P.

Here we have, a = 30 and r = 2

So, a₃ = ar² = (30) (2)² = 120

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

And, a₅ = ar⁴ = (30) (2)⁴ = 480

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

a_n +1 = arⁿ = (30) 2ⁿ

Therefore, the number of bacteria at the end of n^th hour will be 30(2)ⁿ.

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?

Solution: –

Given,

The amount deposited in the bank is Rs 500.

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

At the end of 2^nd year, amount = Rs 500 (1.1) (1.1)

At the end of 3^rd year, amount = Rs 500 (1.1) (1.1) (1.1) and so on….

Therefore,

The amount at the end of 10 years = Rs 500 (1.1) (1.1) … (10 times)

= Rs 500(1.1)¹⁰

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

Solution: –

Let’s consider the roots of the quadratic equation to be a and b.

Then, we have

We know that,

A quadratic equation can be formed as,

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

x² – x (a + b) + (ab) = 0

x² – 16x + 25 = 0 [Using (1) and (2)]

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

Exercise 9.4

Find the sum to n terms of each of the series in Exercises 1 to 7.

1. 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

Solution: –

Given series is 1 × 2 + 2 × 3 + 3 × 4 + 4 × 5 + …

It’s seen that,

n^th term, a_n = n ( n + 1)

Then, the sum of n terms of the series can be expressed as

2. 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

Solution: –

Given series is 1 × 2 × 3 + 2 × 3 × 4 + 3 × 4 × 5 + …

It’s seen that,

n^th term, a_n = n ( n + 1) ( n + 2)

= (n² + n) (n + 2)

= n³ + 3n² + 2n

Then, the sum of n terms of the series can be expressed as

3. 3 × 1² + 5 × 2² + 7 × 3² + …

Solution: –

Given series is 3 ×1² + 5 × 2² + 7 × 3² + …

It’s seen that,

n^th term, a_n = (2n + 1) n² = 2n³ + n²

Then, the sum of n terms of the series can be expressed as

4. Find the sum to n terms of the series 

Solution: –

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

Solution: –

Given series is 5² + 6² + 7² + … + 20²

It’s seen that,

n^th term, a_n = ( n + 4)² = n² + 8n + 16

Then, the sum of n terms of the series can be expressed as

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

Solution: –

Given series is 3 × 8 + 6 × 11 + 9 × 14 + …

It’s found out that,

a_n = (n^th term of 3, 6, 9 …) × (n^th term of 8, 11, 14, …)

= (3n) (3n + 5)

= 9n² + 15n

Then, the sum of n terms of the series can be expressed as

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

Solution: –

Given series is 1² + (1² + 2²) + (1² + 2² + 3² ) + …

Finding the n^th term, we have

a_n = (1² + 2² + 3² +…….+ n²)

Now, the sum of n terms of the series can be expressed as

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

Solution: –

Given,

a_n = n (n + 1) (n + 4) = n(n² + 5n + 4) = n³ + 5n² + 4n

Now, the sum of n terms of the series can be expressed as

9. Find the sum to n terms of the series whose n^th terms is given by n² + 2ⁿ

Solution: –

Given,

n^th term of the series as:

a_n = n² + 2ⁿ

Then, the sum of n terms of the series can be expressed as

10. Find the sum to n terms of the series whose n^th terms is given by (2n – 1)²

Solution: –

Given,

n^th term of the series as:

a_n = (2n – 1)² = 4n² – 4n + 1

Then, the sum of n terms of the series can be expressed as