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

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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 …..

Solution:
(d) n(n + 1)

Question 2.
The coefficient of x⁶ in (2 + 2x)¹⁰ is ……….
(a) ¹⁰C₆
(b) 2⁶
(c) ¹⁰C₆2⁶
(d) ¹⁰C₆2¹⁰
Solution:
(d) ¹⁰C₆2¹⁰
Hint.
t_{r + 1} = 2¹⁰(ⁿC_r)
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

Question 4.
If ⁿC₁₀ > ⁿC_r 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 + …. + x^{n + 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

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

Question 8.

(a) AP
(b) GP
(c) HP
(d) AGP
Solution:
(c) HP

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

Question 10.
If S_n 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

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 n^th term of the sequence 1, 2, 4, 7, 11, …… is

Solution:
(d) \(\frac{n^{2}-n+2}{2}\)

Question 13.

Solution:

Solution:
(d) \(\frac{\sqrt{2 n+1}-1}{2}\)

Question 14.


Solution:

Question 15.


Solution:
(c) \(\frac{n(n+1)}{\sqrt{2}}\)

Question 16.

(a) 14
(b) 7
(c) 4
(d) 6
Solution:
(a) 14

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:

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) \(\frac{-4}{15}\)

Question 19.

Answer:
(c) \(\frac{(e-1)^{2}}{2 e}\)

Question 20.


Solution:
(b) \(\frac{3}{2} \log \left(\frac{5}{3}\right)\)