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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3

Question 1.
Find the sum of the first 20-terms of the arithmetic progression having the sum of first 10 terms as 52 and the sum of the first 15 terms as 77.
Solution:
Let ‘a’ be the first term and d be the common difference of A.P.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 1 1
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 2

Question 2.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 233
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 234
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 235

Question 3.
Compute the sum of first n terms of the following series:
(i) 8 + 88 + 888 + 8888 + ……
(i) 6 + 66 + 666 + 6666 + …..
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 7
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 8

Question 4.
Compute the sum of first n terms of 1 + (1 + 4) + (1 + 4 + 42) + (1 + 4 + 42 + 43) + ……..
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 9

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3

Question 5.
Find the general term and sum to n terms of the sequence \(1, \frac{4}{3}, \frac{7}{9}, \frac{10}{27}, \ldots \ldots \ldots\)
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 10
Denominator 1, 3, 9, 27, which is a G.P. a = 1, r = 3
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 11
It is an arithmetico Geometric series. Here the nth term is tn = [a + (n – 1)d]rn – 1 where a = 1, d = 3 and r = 1/3
Now the sum to n terms is
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 12

Question 6.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 13
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 14

Question 7.
Show that the sum of (m + n)th and (m – n)th term of an A.P. is equal to twice the mth term.
Solution:
Let the A.P. be a, a + d, a + 2d, ……..
tm + n = a + (m + n – 1)d
tm – n = a + (m – n – 1)d
tm = a + (m – 1)d
2tm = 2[a + (m – 1)d]
To prove tm + n + tm – n = 2tm
LHS tm + n + tm – n = a + (m + n – 1)d + a + (m – n – 1)d
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 15
= 2a + d [2m – 2]
= 2[a + (m – 1)d) = 2 tm = RHS

Question 8.
A man repays an amount of ₹ 3250 by paying ₹ 20 in the first month and then increases the payment by ₹ 15 per month. How long will it take him to clear the amount?
Solution:
a = 20, d = 15, Sn = 3250 to find n.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 16
3n2 + 5n – 1300 = 0
3n2 – 60n + 65n – 1300 = 0
3n (n – 20) + 65 (n – 20) = 0
(3n + 65) (n – 20) = 0
n = – 65/3 or 20
n = – 65/3 is not possible
∴ n = 20
So he will take 20 months to clear the amount.

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3

Question 9.
In a race, 20 balls are placed in a line at intervals of 4 meters with the first ball 24 meters away from the starting point. A contestant is required to bring the balls back to the starting place one at a time. How far would the contestant run to bring back all balls?
Solution:
t1 = 24 × 2 = 48, t2 =48 + 8 = 56 or (24 + 4)2, t3 =(28 + 4)2 = 64 which is an A.P.
Here a = 48,
d = 56 – 48 = 8
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 20
The contestant has to run 2480 m to bring all the balls.

Question 10.
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?
Solution:
Number of bacteria at the beginning = 30
Number of bacteria after 1 hour = 30 × 2 = 60
Number of bacteria after 2 hours = 30 × 22 = 120
Number of bacteria after 4 hours = 30 × 24 = 30 × 16 = 480
∴ Number of bacteria after nth hour = 30 × 2n

Question 11.
What will ₹ 500 amounts to in 10 years after its deposit in a bank which pays annual interest rate of 10% compounded annually?
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 21

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3

Question 12.
In a certain town, a viral disease caused severe health hazards upon its people disturbing their normal life. It was found that on each day, the virus which caused the disease spread in Geometric Progression. The amount of infectious virus particle gets doubled each day, being 5 particles on the first day. Find the day when the infectious virus particles just grow over 1,50,000 units?
Solution:
a = 5, r = 2, tn >150000, To find ‘n’
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 22
On the 15th day it will grow over 150000 units.

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 Additional Questions

Question 1.
If the ratio of the sums of m terms and n terms of an A.P. be m2 : n2, prove that the ratio of its mth and nth terms is (2m – 1) : (2n – 1).
Solution:
Let ‘a’ be the first term and d be the common difference of A.P.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 23
⇒ 2an + (mn – n)d = 2am + (mn – m)d
⇒ 2an – 2am = (mn – m – mn + n)d
⇒ 2a(n – m) = (n – m)d ⇒ d = 2a, [n – m ≠ 0, as n ≠ m]
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 232
Hence, the required ratio is (2m – 1) : (2n – 1)

Question 2.
Determine the number of terms of geometric progression {an} if a1 = 3, an = 96, Sn = 189.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 222
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 25

Question 3.
The sum of first three terms of a G.P. is to the sum of the first six terms as 125 : 152. Find the common ratio of the G.P.
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 26
Hence, the common ratio of the G.P. is \(\frac{3}{5}\)

Question 4.
Find the sum to n terms the series: (x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + …
Solution:
(x + y) + (x2 + xy + y2) + (x3 + x2y + xy2 + y3) + … to n terms
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 27
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 28

Question 5.
Sum the series: (1 + x) + (1 + x + x2) + (1 + x + x2 + x3) + … up to n terms
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 29

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3

Question 6.
Sum up to n terms the series: 7 + 77 + 777 + 7777 + …
Solution:
S = 1 + 77 + 777 + 7777 + … to n terms
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 30

Question 7.
If S1, S2, S3 be respectively the sums of n, 2n, 3n, terms of a G.P. , then prove that S1(S3 – S2) = (S2 – S1)2.
Solution:
Let a be the first term and r be the common ratio of G.P.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 31
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 32

Question 8.
If sum of the n terms of a G.P be S, their product P and the sum of their reciprocals R, the prove that \(P^{2}=\left(\frac{S}{R}\right)^{n}\)
Solution:
Let a be the first term and r be the common ratio of the G.P.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.3 33

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