NCERT Solutions Class 11 Maths Chapter 8 - Sequence And Series

Q1: Find the 20^th and n^thterms of the G.P.
Ans:  The given G.P. is  
Here, a = First term =  
r = Common ratio =  

Q2: Find the 12^th term of a G.P. whose 8^th term is 192 and the common ratio is 2.
Ans:  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)

Q3: The 5^th, 8^th and 11^th terms of a G.P. are p, q and s, respectively. Show that q² = ps.
Ans:  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

Dividing equation (3) by (2), we obtain

Equating the values of r³ obtained in (4) and (5), we obtain

Thus, the given result is proved.

Q4: 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.
Ans:  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.

Q5: Which term of the following sequences:
(a)
(b)
(c)
Ans:  (a) The given sequence is  
Here, a = 2 and r =  
Let the n^th term of the given sequence be 128.

Thus, the 13^th term of the given sequence is 128.
(b) The given sequence is  
Here,  
Let the n^th term of the given sequence be 729.

Thus, the 12^th term of the given sequence is 729.
(c) The given sequence is  
Here,  
Let the n^th term of the given sequence be   .

Thus, the 9^th term of the given sequence is   .

Q6: For what values of x, the numbers   are in G.P?
Ans:  The given numbers are   .
Common ratio =   
Also, common ratio =  

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

Q7: Find the sum to 20 terms in the geometric progression 0.15, 0.015, 0.0015 …
Ans:  The given G.P. is 0.15, 0.015, 0.00015, …
Here, a = 0.15 and   

Q8: Find the sum to n terms in the geometric progression
Ans:  The given G.P. is  
Here,  

Q9: Find the sum to n terms in the geometric progression
Ans:  The given G.P. is  
Here, first term = a₁ = 1
Common ratio = r = – a

Q10: Find the sum to n terms in the geometric progression
Ans:  The given G.P. is  
Here, a = x³ and r = x²

Q11: Evaluate
Ans:  

       
The terms of this sequence 3, 3², 3³, … forms a G.P.

Substituting this value in equation (1), we obtain

Q12: The sum of first three terms of a G.P. is ³⁹/₁₀ & their product is 1. Find the common ratio and the terms.
Ans:  Let   be the first three terms of the G.P.

From (2), we obtain

a³ = 1
⇒ a = 1 (Considering real roots only)
Substituting a = 1 in equation (1), we obtain

Thus, the three terms of G.P. are   .

Q13: How many terms of G.P. 3, 3², 3³, … are needed to give the sum 120?
Ans:  The given G.P. is 3, 3², 3³, …
Let n terms of this G.P. be required to obtain the sum as 120.

Here, a = 3 and r = 3

∴ n = 4
Thus, four terms of the given G.P. are required to obtain the sum as 120.

Q14: 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.
Ans:  Let the G.P. be a, ar, ar², ar³, …
According to the given condition,
a +  ar +  ar² = 16 and ar^{3 +}  ar⁴  + ar⁵ = 128
⇒ a (1 + r  + r²) = 16 … (1)
ar³(1 + r  + r²) = 128 … (2)
Dividing equation (2) by (1), we obtain

Substituting r = 2 in (1), we obtain
a (1 +2 +4) = 16
⇒ a (7) = 16

Q15: Given a G.P. with a = 729 and 7^th term 64, determine S₇.
Ans:  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

Also, it is known that,  

Q16: Find a G.P. for which sum of the first two terms is –4 and the fifth term is 4 times the third term.
Ans:  Let a be the first term and r be the common ratio of the G.P.
According to the given conditions,

a₅ = 4 × a₃
ar⁴ = 4ar²
⇒ r² = 4
∴ r = ± 2
From (1), we obtain

Thus, the required G.P. is

  4, –8, 16, –32, …

Q17: 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.
Ans:  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

Dividing (3) by (2), we obtain

∴      Thus, x, y, z are in G. P.

Q18: Find the sum to n terms of the sequence, 8, 88, 888, 8888…
Ans:  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

Q19: Find the sum of the products of the corresponding terms of the sequences 2, 4, 8, 16, 32 and 128, 32, 8, 2, ½ .
Ans  Required sum =  

Here, 4, 2, 1,   is a G.P.
First term, a = 4
Common ratio, r =
It is known that,  

∴ Required sum =  

Q20: Show that the products of the corresponding terms of the sequences    form a G.P, and find the common ratio.
Ans:  It has to be proved that the sequence, aA, arAR, ar²AR², …ar^n–1AR^n–1, forms a G.P.

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

Q21: 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.
Ans:  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

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.

Q22: If the   terms of a G.P. are a, b and c, respectively. Prove that
Ans:  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–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
Thus, the given result is proved.

Q23: 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.
Ans:  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)
= aⁿ r ^{1 +2 …(n–1)} … (2)
Here, 1, 2, …(n – 1) is an A.P.
∴ 1+ 2 ………. +(n – 1)

Q24: Show that the ratio of the sum of first n terms of a G.P. to the sum of terms from 
Ans:  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   ^.

Q25: If a, b, c and d are in G.P. show that   .
Ans:  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.
∴

Q26: Insert two numbers between 3 and 81 so that the resulting sequence is G.P.
Ans:  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.

Q27: Find the value of n so that   may be the geometric mean between a and b.
Ans:  G. M. of a and b is   .
By the given condition,  
Squaring both sides, we obtain

Q28: The sum of two numbers is 6 times their geometric mean, show that numbers are in the ratio  .
Ans:  Let the two numbers be a and b.
G.M. =  
According to the given condition,

Also,

Adding (1) and (2), we obtain

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

Thus, the required ratio is  .

Q29: If A and G be A.M. and G.M., respectively between two positive numbers, prove that the numbers are .
Ans:  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.

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)

From (3) and (5), we obtain

Substituting the value of a in (3), we obtain

Thus, the two numbers are  .

Q30: 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?
Ans:  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.

Q31: What will Rs 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?Ans:  The amount deposited in the bank is Rs 500.
At the end of first year, amount =   = 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

Q32: If A.M. and G.M. of roots of a quadratic equation are 8 and 5, respectively, then obtain the quadratic equation.
Ans:  Let the root of the quadratic equation be a and b.
According to the given condition,

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