NCERT Solutions Class 11 Maths Chapter 8 - Sequence And Series

Question 1: If f is a function satisfying    such that    , find the value of n.

ANSWER :  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 (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

∴ 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,  

It is given that,  

Thus, the value of n is 4.

Question 2: 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.

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

It is known that,  

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

∴Last term of the G.P = 6^th term = ar^{6 – 1} = (5)(2)⁵ = (5)(32) = 160

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

Question 3: The first term of a G.P. is 1. The sum of the third term & fifth term is 90. Find the common ratio of G.P.

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

∴ a = 1

a₃ = ar² = r²

a₅ = ar⁴ = r⁴

∴ r² +  r⁴ = 90

⇒ r⁴ +  r² – 90 = 0

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

Question 4: 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.

ANSWER :  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.

∴(ar – 7) – (a – 1) = (ar² – 21) – (ar – 7)

⇒ ar – a – 6 = ar² – ar – 14

⇒ar² – 2ar  + a = 8

⇒ar² – ar – ar  + a = 8

⇒a(r^{2 +} 1 – 2r) = 8

⇒ a (r – 1)² = 8 … (2)

⇒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

When r = 2, a = 8

When  

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

When   , 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 5: 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.

ANSWER :  Let the G.P. be T₁, T₂, T₃, T₄, … T_2n.

Number of terms = 2n

According to the given condition,

T₁ + T₂ + T₃  …+ T_2n = 5 [T₁+  T₃  …+ T_2n–1]

⇒ T₁ + T₂  +T₃  …+ T_2n – 5 [T₁ + T₃  …+ T_2n–1] = 0

⇒ T₂ + T₄  … +T_2n = 4 [T₁+  T₃  …+ T_2n–1]

Let the G.P. be a, ar, ar², ar³, …

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

Question 6: 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.

ANSWER :  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

∴ 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 7: If  , then show that a, b, c and d are in G.P.

ANSWER :  It is given that,

From (1) and (2), we obtain

Thus, a, b, c, and d are in G.P.

Question 8: 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

ANSWER :  Let the G.P. be a, ar, ar², ar³, … ar^{n – 1}…

According to the given information,

Hence, P² Rⁿ = Sⁿ

Question 9: The p^th, q^th and r^th terms of an A.P. are a, b, c respectively. Show that

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

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

Therefore,

a_p = t + (p – 1) d = a … (1)

a_q = t + (q – 1)d = b … (2)

a_r = 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

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

(q – 1 – r + 1) d = b – c

⇒ (q – r) d = b – c

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

Thus, the given result is proved.

Question 10: If a  are in A.P., prove that a, b, c are in A.P.

ANSWER :  It is given that a   are in A.P.

Thus, a, b, and c are in A.P.

Question 11: If a, b, c, d are in G.P, prove that    are in G.P.

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

∴b² = ac … (1)

c² = bd … (2)

ad = bc … (3)

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)

Consider L.H.S.

(bⁿ + cⁿ)² = b^{2n +} 2bⁿcⁿ +  c²ⁿ

= (b²)ⁿ+ 2bⁿcⁿ + (c²) ⁿ

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

= aⁿ cⁿ +  bⁿcⁿ+  bⁿ cⁿ +  bⁿ dⁿ

= aⁿ cⁿ +  bⁿcⁿ + aⁿ dⁿ +  bⁿ dⁿ [Using (3)]

= cⁿ (aⁿ  + bⁿ)  +dⁿ (aⁿ +  bⁿ)

= (aⁿ +  bⁿ) (cⁿ  + dⁿ)

= R.H.S.

∴ (bⁿ  + cⁿ)² = (aⁿ +  bⁿ) (cⁿ  + dⁿ)

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

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

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

∴ a +  b = 3 and ab = p … (1)

Also, c and d are the roots of  

∴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 (1) and (2), we obtain

x  + xr = 3

⇒ x (1 + r) = 3

xr²   +xr³ =12

⇒ xr² (1 + r) = 12

On dividing, we obtain

Case I:

When r = 2 and x =1,

ab = x²r = 2

cd = x²r⁵ = 32

Case II:

When r = –2, x = –3,

ab = x²r = –18

cd = x²r⁵ = – 288

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

Question 13: The ratio of the A.M and G.M. of two positive numbers a and b, is m: n. Show that 

ANSWER :  Let the two numbers be a and b.

A.M    and G.M. =  

According to the given condition,

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

Adding (1) and (2), we obtain

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

Question 14: If a, b, c are in A.P,; b, c, d are in G.P and   are in A.P. prove that a, c, e are in G.P.

ANSWER :  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,   are in A.P.

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

From (1), we obtain

From (2), we obtain

Substituting these values in (3), we obtain

Thus, a, c, and e are in G.P.

Question 15: Find the sum of the following series up to n terms:

(i) 5+ 55 +555 … 

(ii) .6 + .66 + .666 …

ANSWER :   (i) 5 +55 +555 …

Let S_n = 5+ 55+555 ….. to n terms

(ii) .6 +.66 +.666 …

Let S_n = 06. +0.66 +0.666 … to n terms

Question 16: Find the 20^th term of the series 2 × 4+ 4 × 6 +6 × 8 …  n terms.

ANSWER :  The given series is 2 × 4 + 4 × 6 + 6 × 8 … n terms

∴ n^th term = a_n = 2n × (2n + 2) = 4n² + 4n

a₂₀ = 4 (20)² + 4(20) = 4 (400) + 80 = 1600 + 80 = 1680

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

Question 17: Find the sum of the first n terms of the series: 3 + 7 + 13 + 21 + 31 …

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

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

S = 3 + 7 + 13 + 21 ….  +a_{n – 2}  + a_{n – 1}  + a_n

On subtracting both the equations, we obtain

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

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

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

a_n = 3 + [4 + 6 + 8 … +(n –1) terms]

Question 18: If S₁, S₂, S₃ are the sum of first n natural numbers, their squares and their cubes, respectively, show that  

ANSWER :  From the given information,

Thus, from (1) and (2), we obtain

Question 19: Find the sum of the following series up to n terms:

ANSWER :  The n^th term of the given series is

Question 20: Show that  

ANSWER :  n^th term of the numerator = n(n + 1)² = n³ + 2n² +  n

n^th term of the denominator = n²(n + 1) = n³  + n²

From (1), (2), and (3), we obtain

Thus, the given result is proved.

Question 21: 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?

ANSWER :  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.

∴ 6000 = 500 + (n – 1) 500

⇒ 1 + (n – 1) = 12

⇒ n = 12

∴Sum of the A.P  

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

= 12% of 39000 = Rs 4680

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

Question 22: 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?

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

∴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.

∴ 18000 = 1000 + (n – 1) (1000)

⇒ n = 18

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

= 10% of Rs 171000 = Rs 17100

∴Cost of scooter = Rs 22000 + Rs 17100 = Rs 39100

Question 23: 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 8^th set of letter is mailed.

ANSWER :  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

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

∴Cost of mailing 87380 letters   = Rs 43690

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

Question 24: A man deposited Rs 10000 in a bank at the rate of 5% simple interest annually. Find the amount in 15^th year since he deposited the amount and also calculate the total amount after 20 years.

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

∴ Interest in first year  

∴Amount in 15^th year = Rs  

= Rs 10000 + 14 × Rs 500

= Rs 10000 + Rs 7000

= Rs 17000

Amount after 20 years =  

= Rs 10000 20 × Rs 500

= Rs 10000 + Rs 10000

= Rs 20000

Question 25: 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.

ANSWER :  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 =    = 5 × 1024 = 5120

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

Question 26: 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.

ANSWER :  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)

However, x cannot be negative.

∴x = 17

Therefore, originally, the number of days in which the work was completed is 17.

Thus, required number of days = (17 + 8) = 25