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Tamilnadu Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5

Choose the correct or the most suitable answer:

Question 1.
The value of 2 + 4 + 6 + … + 2n is …..
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 1
Solution:
(d) n(n + 1)
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 2

Question 2.
The coefficient of x6 in (2 + 2x)10 is ……….
(a) 10C6
(b) 26
(c) 10C626
(d) 10C6210
Solution:
(d) 10C6210
Hint.
tr + 1 = 210(nCr)
To find coefficient of x6 put r = 6
∴ coefficient of x6 = 210 [10C6]

Question 3.
The coefficient of x8y12 in the expansion of (2x + 3y)20 is …….
(a) 0
(b) 28312
(c) 28312 + 21238
(d) 20C828312
Solution:
(d) 20C828312
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 3

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5

Question 4.
If nC10 > nCr for all possible r, then a value of n is ……..
(a) 10
(b) 21
(c) 19
(d) 20
Solution:
(d) 20
Hint.
Out of 10C10, 21C10, 19C10 and 20C10, 20C10 is larger.

Question 5.
If a is the arithmetic mean and g is the geometric mean of two numbers, then ……..
(a) a ≤ g
(b) a ≥ g
(c) a = g
(d) a > g
Solution:
(b) a ≥ g
Hint. AM ≥ GM
∴ a ≥ g

Question 6.
If (1 + x2)2 (1 + x)n = a0 + a1x + a2x2 + …. + xn + 4 and if a0, a1, a2 are in AP, then n is ……
(a) 1
(b) 2
(c) 3
(d) 4
Solution:
(b or c)n = 2 or 3
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 20

Question 7.
If a, 8, b are in A.P, a, 4, b are in G.P, if a, x, b are in HP then x is ……
(a) 2
(b) 1
(c) 4
(d) 16
Solution:
(a) 2
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 21

Question 8.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 22
(a) AP
(b) GP
(c) HP
(d) AGP
Solution:
(c) HP

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5

Question 9.
The HM of two positive numbers whose AM and GM are 16, 8 respectively is ………
(a) 10
(b) 6
(c) 5
(d) 4
Solution:
(d) 4
Hint.
Let the two numbers be a and b
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 23

Question 10.
If Sn denotes the sum of n terms of an AP whose common difference is d, the value of \(\mathrm{S}_{n}-2 \mathrm{S}_{n-1}+S_{n-2}\) is ……
(a) d
(b) 2d
(c) 4d
(d) d2
Solution:
(a) d
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 24

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5

Question 11.
The remainder when 3815 is divided by 13 is …….
(a) 12
(b) 1
(c) 11
(d) 5
Solution:
(a) 12
Hint.
3815 = (39 – 1)15 = 3915 – 15C1 3914(1) + 15C2 (39)13(1)2 – 15C3 (39)12(1)3 ….. + 15C14 (39)1(1) – 15C15(1)
Except -1 all other terms are divisible by 13.
∴ When 1 is added to it the number is divisible by 13. So the remainder is 13 – 1 = 12.

Question 12.
The nth term of the sequence 1, 2, 4, 7, 11, …… is
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 25
Solution:
(d) \(\frac{n^{2}-n+2}{2}\)

Question 13.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 26
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 27
Solution:
(d) \(\frac{\sqrt{2 n+1}-1}{2}\)
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 277

Question 14.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 28
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 299
Solution:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 29

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5

Question 15.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 30
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 31
Solution:
(c) \(\frac{n(n+1)}{\sqrt{2}}\)
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 32

Question 16.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 33
(a) 14
(b) 7
(c) 4
(d) 6
Solution:
(a) 14
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 34

Question 17.
The sum of an infinite GP is 18. If the first term is 6, the common ratio is ………
(a) \(\frac{1}{3}\)
(b) \(\frac{2}{3}\)
(c) \(\frac{1}{6}\)
(d) \(\frac{3}{4}\)
Solution:
(b) \(\frac{2}{3}\)
Hint:
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 35
18r = 18 – 6 = 12
r = 12/18 = 2/3

Question 18.
The coefficient of x5 in the series e-2x is ………
(a) \(\frac{2}{3}\)
(b) \(\frac{3}{2}\)
(c) \(\frac{-4}{15}\)
(d) \(\frac{4}{15}\)
Answer:
(c) \(\frac{-4}{15}\)
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 50

Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5

Question 19.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 51
Answer:
(c) \(\frac{(e-1)^{2}}{2 e}\)
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 52

Question 20.
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 53
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 54
Solution:
(b) \(\frac{3}{2} \log \left(\frac{5}{3}\right)\)
Samacheer Kalvi 11th Maths Solutions Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 55

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