NCERT Solutions Class 11 Mathematics Chapter 9 Sequences and Series

Exercise 9.1

1. Write the first five terms of each of the sequence defined by the following:
a_n = n (n + 2)
Answer :
a_n = n (n + 2)
For n = 1, a₁ = 1(1 + 2) = 3
For n = 2, a₂ = 2(2 + 2) = 8
For n = 3, a₃ = 3(3 + 2) = 15
For n = 4, a₄ = 4(4 + 2) = 24
For n = 5, a₅ = 5(5 + 2) = 35
Thus first five terms are 3, 8, 15, 24, 35.

2. Write the first 5 terms of the sequence defined by the following:
a_n = n/(n + 1)
Answer :
a_n = n/(n + 1)
for n = 1, a₁ = 1/(1 + 1) = 1/2
for n = 2, a₂ = 2/(2 +1 ) = 2/3
for n = 3, a₃ = 3/(3 + 1) = 3/4
for n = 4, a₄ = 4/(4 + 1) = 4/5
for n = 5, a₅ = 5/(5 + 1) = 5/6
Hence, first 5 terms of the given sequence are
1/2, 2/3, 3/4, 4/5 and 5/6.

3. Write the first five terms of the sequence whose n^th term is: a_n = 2ⁿ
Answer :
a_n = 2ⁿ, Putting n = 1, 2, 3, 4, 5
a₁ = 2¹ = 2, a₂ = 2² = 4, a₃ = 2³ = 8
a ₄ = 2⁴ = 16, a₅ = 2⁵ = 32
Required first five term of sequences are 2, 4, 8, 16, 32.

4. Write the first five terms of the sequence whose n ^th term is: a_n = (2n – 3)/6.
Answer :
Here a_n = (2n – 3)/6
Putting n = 1, 2, 3, 4, 5, we get
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
and a₅ = (2 × 5 – 3)/6 = (10 – 3)/6 = 7/6
∴ the first five terms are –(1/6), 1/6, 1/2, 5/6 and 7/6.

5. Write the first five terms of the sequence whose n ^th term is a_n = (–1)^{n – 1} 5 ^{n + 1} 
Answer :
Putting n = 1, 2, 3, 4, 5
a₁ = (–1)⁰. 5 ^{1 + 1} = 5² = 25
a₂ = (–1)¹. 5 ^{2 + 1} = – 5³ = –125
a₃ = (–1)². 5 ^{3 + 1} = 5⁴ = 625
a₄ = (–1)³. 5 ^{4 + 1} = – 5⁵ = – 3125
a₅ = (–1)⁴. 5 ^{5 + 1} = 5⁶ = 15625

6. Write the first 5 terms of the sequence defined by the following:
a_n = (n(n² + 5))/4
Answer :
for n = 1, a₁ = (1(1² + 5))/4 = 6/4 = 3/2
for n =2 , a₂ = (2(2² + 5))/4 = (2 × 9)/4 = 9/2
for n = 3, a₃ = (3(3² + 5))/4 = (3 × 14)/4 = 21/2
for n = 4, a₄ = (4(4² + 5))/4 = (4 × 21)/4 = 21
for n = 5, a₅ = (5(5² + 5))/4 = (5 × 30)/4 = 75/2
hence the first 5 terms are 3/2, 9/2, 21/2, 21 and 75/2

7. Find the indicated terms in the following sequence whose n ^th term is:
a_n = 4n – 3, a₁₇, a₂₄
Answer :
a_n = 4n – 3
For n = 17, a₁₇ = 4 × 17 – 3 = 68 – 3 = 65
For n = 24, a₂₄ = 4 × 24 – 3 = 96 – 3 = 93
Hence a₁₇ = 65 and a₂₄ = 93

8. Find the indicated terms in the following sequence whose n ^th term is:
a_n = n²/2ⁿ ; a₇
Answer :
a_n = n²/2ⁿ
putting n = 7
a₇ = 7²/2⁷ = 49/128

9. Find the indicated term in the following sequence whose n ^th term is:
a_n = (–1)^{n – 1} n ³, a₉
Answer :
a_n = (–1)^{n – 1} n³, Putting n = 9
a₉ = (–1)^{9 – 1} 9³ = 729

10. Find the indicated terms in the sequence whose n ^th term is a_n = (n(n -2))/(n + 3) ; a₂₀
Answer :
a_n = (n(n -2))/(n + 3),
Putting n = 20,
a₂₀ = (20(20 – 2))/(20 + 3) = (20 × 18)/23 = 360/23

11.  Write the first five terms of each of the sequences in Exercises 11 to 13 and obtain the corresponding series. a₁ = 3, a_n = 3, a_{n – 1} + 2 for all n > 1.
Answer :
a₁ = 3, a_n = 3 a_{n –1} + 2. Putting n = 2, 3, 4, 5
a₂ = 3.a₁ + 2 = 3. 3 + 2 = 9 + 2 = 11
a₃ = 3.a₂ + 2 = 3. 11 + 2 = 33 + 2 = 35
a₄ = 3.a₃ + 2 = 3. 35 + 2 = 105 + 2 = 107
a₅ = 3.a₄ + 2 = 3. 107 + 2 = 321 + 2 = 323
The first five terms of the sequences are
3, 11, 35, 107, 323
∴ The corresponding series is
3 + 11 + 35 + 107 + 323 +....

12. a₁ = -1, a_n = (a_{n -1})/n , n ≥ 2
Answer :
Here a₁ = -1, a_n = (a_{n – 1})/n , n ≥ 2
Putting n = 2, 3, 4, 5 and 6
a₂ = (a₂ – 1)/2 = a₁/2 = -1/2
a₃ = (a₃ – 1)/3 = a₂/2 = -(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
The first five terms of the sequences are
-1, -(1/2), -(1/6), -(1/24), -1/120
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.
Answer :
Given: a₁ = a₂ = 2
and a_n = a_{n– 1} – 1, n > 2.
For n = 3, a₃ = a_{2 – 1} = 2 – 1 = 1
For n = 4, a₄ = a_{3 – 1} = 1 – 1 = 0
For n = 5, a₅ = a_{4 – 1} = 0 – 1 = –1
Hence first five terms are 2, 2, 1, 0, –1 and the corresponding series = 2 + 2 + 1 + 0 + (–1) +....

14. The Fibonacci sequence is defined by
1 = a₁ = a₂ and an = a_{n – 1} + a_{n – 2}, n > 2
Find (a_{n+ 1})/a_n , for n = 1, 2, 3, 4, 5
Answer :
For n = 1, (a_n+1)/a_n = a₂/a₁ = 1/1 = 1 a₁ = a₂ = 1
and a_n = a_{n – 1} + a_{n – 2}, n > 2 …(A)
n = 3 in eqn (A) a₃ = a₂ + a₁ = 1 + 1 = 2
n = 4 in eqn (A) a₄ = a₃ + a₂ = 2 + 1 = 3
n = 5 in eqn (A) a₅ = a₄ + a₃ = 3 + 2 = 5
n = 6 in eqn (A) a₆ = a₅ + a₄ = 5 + 3 = 8
for n = 2, (a_n+1)/a_n = a₃/a₂ = 2/1 = 2;
for n = 3, (a_n+1)/a_n = a₄/a₃ = 3/2;
for n = 4, (a_n+1)/a_n = a₅/a₄ = 5/3;
for n = 5, (a_n+1)/a_n = a₆/a₅ = 8/5
The values of (a_n+1)/a_n = for n = 1, 2, 3, 4, 5 are 1, 2, 3/2 , 5/3 , 8/5

Exercise 9.2

1. Find the sum of odd integers from 1 to 2001.
Answer :
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
Here, a + (n-1)d = 2001 
⇒ 1 + (n - 1)(2) = 2001 
⇒ 2n - 2 = 2000
⇒ n = 1001 

= 1001 × 1001
= 1002001
Thus, 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.
Answer :
The natural numbers lying between 100 and 1000, which are multiples of  5, are 105, 110, .... 995. 
Here, a = 105 and d = 5 
a + (n - 1)d = 995 
⇒ 105 + (n - 1)5  = 995 
⇒ (n- 1)5 = 995 - 105 = 890 
⇒ n - 1 = 178 
⇒ n = 179 

= 179[2(105) + (178)(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.

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.
Answer :
First term = 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 = (1/4)(10 + 35d) 
⇒ 40 + 40d = 10 + 35d 
⇒ 30 = -5d 
⇒ d = -6 
∴ a₂₀ = a + (20 - 1)d = 2 + (19)(-6) = 2 - 114 = -112 
Thus, the 20th 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?
Answer :
Let the sum of n terms of the given A.P. be  -25. 
It is known that, S_n = (n/2) [2a + (n - 1)d], where n = number of terms, a = first term, and d = common difference 
Here, a = -6

⇒-100 = n(-25 + n) 
⇒ n² - 25n + 100 = 0 
⇒ n² - 5n - 20n + 100 = 0 
⇒ n(n -5) - 20(n - 5) = 0 
⇒ n = 20 or 5

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 1/2 (pq + 1) where p ≠ q. 
Answer :
It is known that the general term of an A.P. is a_n = a + (n – 1)d
∴ According to the given information,
p^th  term = a_p = a + (p - 1)d = 1/q             ...(1)
q^th  term = a_q = a + (q - 1)d = 1/p             ...(2)
Subtracting (2) from (1), we obtain 
(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 
Putting the value of d in (1), we obtain 

6. If the sum of a certain number of terms of the A.P. 25, 22, 19, … is 116. Find the last term
Answer :
Let the sum of n terms of the given A.P. be 116. 
S_n = (n/2) [2a + (n - 1)d] 
Here, a = 25 and d = 22 - 25 = -3 
∴ S_n = (n/2) [2× 25 + (n - 1)(-3)]
⇒  116 = n/2[50 - 3n + 3] 
⇒  232 = n (53 -3n) = 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 or n = 29/3 
However, n cannot be equal to 29/3 . Therefore, n = 8 
∴ 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.

7. Find the sum to n terms of the A.P., whose k^th term is 5k + 1.
Answer :
It is given that the k^th term of the A.P. is 5k + 1.
k^th term = a_k = a + (k – 1)d
∴ a + (k – 1)d = 5k + 1
a + kd – d = 5k + 1
Comparing the coefficient of k, we obtain 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.
Answer :
It is known that, S_n = (n/2) [2a + (n - 1)d]
According to the given condition,
n/2 [2a + (n - 1)d] = pn + qn²
⇒ n/2 [2a + nd - d] = pn + qn²
⇒ na + n² (d/2) - n .(d/2) = pn + qn²
Comparing the coefficients of n² on both sides, we obtain
d/2 = q
∴ d = 2q
Thus, 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
Answer :
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,

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.
Answer :
Let a and d be the first term and the common difference of the A.P. respectively. 
Here,  
S_p = p/2 [2a + (p -1)d] 
S_q = q/2 [2a + (q -1)d]  
According to the given condition,  
p/2 [2a + (p - 1)d] = q/2 [2a + (q - 1)d] 
⇒ p[2a + (p - 1)q] = q[2a + (q - 1)d] 
⇒ 2ap + pd(p -1) = 2aq + qd(q - 1)
⇒ 2a(p -q) + d[p(p -1) - q(q - 1)] = 0 
⇒ 2a(p - q) +d [p² - p - q² + q] = 0 
⇒ 2a(p - q) +d [(p -q)(p +q) - (p -q)] = 0 
⇒ 2a(p - q) +d [(p -q)(p + q - 1)] = 0 
⇒ 2a + d(p + q - 1) = 0 

11. Sum of the first p, q and r terms of an A.P. are a, b and c, respectively.
Prove that a/q (q - r) + b/q (r - p) + c/r (p - q) = 0 
Answer :
Let a₁ and d be the first term and the common difference of the A.P. respectively. 
According to the given information,  

Subtracting (2) from (1), we obtain 
(p - 1)d - (q - 1)d = 2a/p - 2b/q 
⇒ d(p - 1 - q + 1) = (2aq - 2bp)/pq 
⇒ d(p -q) = (2aq - 2bp)/pq 
⇒ d = 2(aq - bp)/pq(p -q)  ...(4) 
Subtracting (3) from (2), we obtain 

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).
Answer :
Let a and d be the first term and the common difference of the A.P. respectively.
According to the given condition,

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.
Answer :
Let a and b be the first term and the common difference of the A.P. respectively.
a_m = a + (m – 1)d = 164 …(1)
Sum of n terms, 
S_n = (n/2) [2a + (n - 1)d]  
Here,  
n/2 [2a + nd - d] = 3n² + 5n 
⇒ na + (d/2)n²  - (d/2)n = 3n² + 5n 
⇒ (d/2)n² + (a - d/2)n = 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 (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.

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

15. If  (aⁿ + bⁿ )/(a^n-1 + b^n-1 ) is the A.M. between a and b, then find the value of n. 
Answer :
A.M. of a and b = (a + b)/2 
According to the given condition,  
(a + b)/2 = (aⁿ + bⁿ)/(a^n-1 + b^n-1) 
⇒ (a + b)(a^n-1 + b^n-1) = 2(aⁿ + bⁿ) 
⇒ aⁿ + ab^n-1 + ba^n-1 + bⁿ = 2aⁿ + 2bⁿ 
⇒ ab^n-1 + a^n-1 b = aⁿ + bⁿ 
⇒ ab^n-1 - bⁿ = aⁿ - a^n-1 b 
⇒ b^n-1 (a - b) = a^n-1 (a - b) 
⇒ b^n-1 = a^n-1 
⇒ (a/b)^n-1 = 1 = (a/b)ⁿ 
⇒ n - 1 = 0 
⇒ n = 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.
Answer :
Let A₁, A₂, … A_m be m numbers such that 1, A₁, A₂, … A_m, 31 is an A.P.
Here, a = 1, b = 31, n = m + 2
∴ 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
A_m–1 = a + (m – 1) d
According to the given condition,

⇒ 9m + 1899 = 155m - 145 
⇒ 155m - 9m = 1899 + 145 
⇒ 146m = 2044
⇒ m = 14 
Thus, the value of m is  14.

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 30^th installment?
Answer :
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 30^th installment is 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.
Answer :
The angles of the polygon will form an A.P. with common difference d as 5° and first term a as 120°.
It is known that the sum of all angles of a polygon with n sides is 180° (n – 2). 
∴ S_n = 180° (n - 2) 
⇒  n/2 [2a + (n - 1)d]  = 180° (n - 2) 
⇒  n/2 [2a + (n - 1)d]  = 180° (n - 2)  
⇒  n [240 + (n - 1)5]  = 360(n - 2)
⇒ 240 n + 5n² - 5n = 360n - 720
⇒ 5n² + 235n - 360n + 720 = 0
⇒ 5n² - 125n + 720 = 0
⇒ n² - 25n + 144 = 0
⇒ n² - 16n - 9n + 144 = 0
⇒ n(n - 16) - 9(n - 16) = 0
⇒ (n - 9)(n - 16) = 0 
⇒ n = 9 or 16

Exercise 9.3

1. Find the 20th and nth terms of the G.P. 5/2, 5/4, 5/8, ...... 
Answer :
The given G.P. is  5/2, 5/4,  5/8 , ..... 
Here, a = First term = 5/2 

2. Find the 12^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.
Answer :
Common ratio, r = 2
Let a be the first term of the G.P.
∴ 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.
Answer :
Let a be the first term and r be the common ratio of the G.P.
According to the given condition,
a₅ = a r^5–1 = a r⁴ = p                     …(1)
a₈ = a r^8–1 = a r⁷ = q                     …(2)
a₁₁ = a r^11–1 = a r¹⁰ = s                …(3)
Dividing equation (2) by (1), we obtain 

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.
Answer :
Let a be the first term and r be the common ratio of the G.P.
∴ a = –3
It is known that, a_n = ar^n–1
∴a₄ = ar³ = (–3) r³
a₂ = a r¹ = (–3) r
According to the given condition,
(–3) r³ = [­(–3) r]²
⇒ –3r³ = 9 r²
⇒ r = –3
a₇ = a r ^7–1 = a r⁶ = (–3) (–3)⁶ = – (3)⁷ = –2187
Thus, 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 ? 
Answer :
(a) The given sequence is 2, 2√2, 4, ..... 
Here, a = 2 and r = 2√2/2 = √2 
Let the nth term of the given sequence be  128. 

Thus, the 13th term of the given sequence is  128. 
(b) The given sequence is √3, 3, 3√3, ..... 
Here, a = √3 and r = 3/√3 = √3 
Let the nth term of the given sequence be 729. 
a_n = ar^n-1 
∴ ar^n-1 = 729 
⇒ (√3)(√3)^n-1 = 729 

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

7. Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …
Answer :
The given G.P. is 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 .....
Answer :
The given G.P. is √7, √21, 3√7 
Here, a = √7 

9. Find the sum to n terms in the geometric progression 1, -a, a² , -a³ ..... (if a ≠ -1) 
Answer :
The given G.P. is 1, -a , a² , -a³ , ..... 
Here, first term = a₁ = 1 
Common ratio = r = -a 

10. Find the sum to n terms in the geometric progression x³ , x⁵ , x⁷ ..... (if x ≠ ± 1) 
Answer :
The given G.P. is x³ , x⁵ , x⁷ ..... 
Here, a = x³ and r = x² 

11. Evaluate

Answer :

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.
Answer :
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 obtain
a³ = 1
⇒ a = 1 (Considering real roots only)
Substituting a = 1 in equation (1), we obtain
1/r + 1 + r = 39/10 
⇒ 1 + r + r² = (39/10)r 
⇒ 10 + 10r + 10r² - 39r = 0 
⇒ 10r² - 29r + 10 = 0 
⇒ 10r² - 25r - 4r + 10 = 0 
⇒ 5r(2r - 5) - 2(2r - 5) = 0 
⇒ (5r - 2)(2r - 5) = 0 
⇒ r = 2/5 or 5/2 
Thus, the three terms of 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?
Answer :
The given G.P. is 3, 3², 3³, …
Let n terms of this G.P. be 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.
Answer :
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 obtain

15. Given a G.P. with a = 729 and 7^th term 64, determine S₇.
Answer :
a = 729
a₇ = 64
Let r be the common ratio of the G.P.
It is known that, a_n = a r^n–1
a₇ = ar^7–1 = (729)r⁶
⇒ 64 = 729 r⁶ 

16. Find a G.P. for which sum of the first two terms is – 4 and the fifth term is4 times the third term.
Answer :
Let a be the first term and r be the common ratio of the G.P.
According to the given conditions,

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.
Answer :
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)
Dividing (2) by (1), we obtain 

18. Find the sum to n terms of the sequence, 8, 88, 888, 8888…
Answer :
The given sequence is 8, 88, 888, 8888…
This sequence is not a G.P. However, 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
Answer :
Required sum  = 2 × 128 + 4 × 32 + 8 × 8 + 16 × 2 + 32 × 1/2 
= 64[4 + 2 + 1 + 1/2 + 1/2²]. 
Here, 4, 2, 1, 1/2, 1/2² is  a G.P. 
First term, a = 4 
Common ratio, r = 1/2 
It is known that, 

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
Answer :
It has to be proved that the sequence, aA, arAR, ar²AR², …ar^n–1AR^n–1, forms a G.P.
Second term/First term = arAR/aA = rR 
Third term/Second term = ar² AR² /ar AR = rR 
Thus, 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.
Answer :
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 (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 (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.

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 
Answer :
Let A be the first term and R be the common ratio of the G.P.
According to the given information,
AR^p–1 = a
AR^q–1 = b
AR^r–1 = c
a^q–rb^r–pc^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
Thus, the given result is 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.
Answer :
The first term of the G.P is a and the last term is b.
Therefore, the G.P. is a, ar, ar², ar³, … ar^n–1, where r is the common ratio.
b = ar^n–1 … (1)
P = Product of n terms
= (a) (ar) (ar²) … (ar^n–1)
= (a × a ×…a) (r × r² × …r^n–1)
= aⁿr ^{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 (n +1)^th  to (2n)^th  term is 1/rⁿ . 
Answer :
Let a be the first term and r be the common ratio of the G.P. 
Sum of first n terms a(l - rⁿ )/(1 - r) 
Since there are n terms from (n + 1)^th to (2n)^th term, 

25. If a, b, c and d are in G.P. show that (a² + b² + c² ) (b² + c² + d² ) =(ab + bc + cd)² . 
Answer :
a, b, c, d are in G.P.
Therefore,
bc = ad           ​​​​​​​…(1)
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 (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.
∴ L.H.S. = R.H.S.
(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.
Answer :
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.
∴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
Thus, the required two numbers are 9 and 27.

27. Find the value of n so that (aⁿ⁺¹ + bⁿ⁺¹)/(aⁿ + bⁿ) may be the geometric mean between a and b. 
Answer :
G. M. of a and b is √ab. 

28. The sum of two numbers is 6 time their geometric mean, show that numbers are in the ratio (3 + 2√2) : (3 - 2√2) .
Answer :
Let the two numbers be a and b.  
G.M. = √ab 
According to the given condition, 
a + b = 6√ab  ...(1)
⇒ (a + b)² = 36(ab) 
Also, 
(a - b)² = (a + b)² - 4ab = 36ab - 4ab = 32ab 
⇒ a - b = √32√ab 
= 4√2√ab  ...(2) 
Adding (1) and (2), we obtain  
2a = (6 + 4√2)√ab 
⇒ a = (3 + 2√2)√ab 
Substituting the value of a in (1), we obtain  
b = 6√ab - (3 + 2√2)√ab 
⇒ b = (3 - 2√2)√ab 

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)
Answer :
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.
∴ AM = A = (a + b)/2            ...(1)
GM = G = √(ab)                   ...(2)
From (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 obtain
(a – b)² = 4A² – 4G² = 4 (A²–G²)
(a – b)² = 4 (A + G) (A – G)

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^thhour?
Answer :
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
∴ a₃ = ar² = (30) (2)² = 120
Therefore, the number of bacteria at the end of 2^nd hour will be 120.
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
Thus, 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?
Answer :
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
∴ 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.
Answer :
Let the root of the quadratic equation be a and b.  
According to the given condition, 
A.M. = (a + b)/2 = 8 ⇒ a + b = 16 ...(1)
G.M. = √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 (1) and (2)]
Thus, the required quadratic equation is x² – 16x + 25 = 0