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

Tamilnadu State Board New Syllabus Samacheer Kalvi 11th Maths Guide Pdf Chapter 5 Binomial Theorem, Sequences and Series Ex 5.5 Text Book Back Questions and Answers, Notes.

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
(1) \(\frac{n(n-1)}{2}\)
(2) \(\frac{n(n+1)}{2}\)
(3) \(\frac{2 n(2 n+1)}{2}\)
(4) n(n + 1)
Answer:
(4) n(n + 1)

Explaination:
2 + 4 + 6 + ……… + 2n = 2(1 + 2 + 3 + ………….. + n)
= 2 × \(\frac{n(n+1)}{2}\)
= n(n + 1)

Question 2.
The coefficient of x⁶ in (2 + 2x)¹⁰ is
(1) 10C₆
(2) 2₆
(3) 10C₆2⁶
(4) 10C₆2¹⁰
Answer:
(4) 10C₆2¹⁰

Explaination:

Question 3.
The coefficient of x⁸ y¹² in the expansion of (2x + 3y)²⁰ is
(1) 0
(2) 2⁸ 3¹²
(3) 2⁸ 3¹² + 2¹² 3⁸
(4) 20C₈ 2⁸ 3¹²
Answer:
(4) 20C₈ 2⁸ 3¹²

Explaination:

Question 4.
If nC₁₀ > nC_r for all possible r then a value of n is
(1) 10
(2) 21
(3) 19
(4) 20
Answer:
(4) 20

Explaination:
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
(1) a ≤ g
(2) a ≥ g
(3) a = g
(4) a > g
Answer:
(2) a ≥ g

Explaination:
Given Arithmetic mean = a,
Geometric mean = g
We have A. M ≥ G. M
∴ a ≥ g

Question 6.
If (1 + x²)² (1 + x)ⁿ = a₀ + a₁x + a₂x² + ………… + x^{n + 4} and if a₀, a₁, a₂ are in A. P then n is
(1) 1
(2) 2
(3) 3
(4) 4
Answer:
(3) 3

Explaination:

n² – 5n + 6 = 0
(n – 2) (n – 3) = 0
n = 2 or n = 3

Question 7.
If a, 8, b are in A .P , a, 4 , b are in G. P and if a, x ,b are in H . P then x is
(1) 2
(2) 1
(3) 4
(4) 16
Answer:
(1) 2

Explaination:
Given a, 8, b are in A. P ∴ 2 × 8 = a + b ⇒ a + b = 16 ——— (1)
Also a, 4, b are in G.P ∴ 4² = a . b ⇒ ab = 16 ——- (2)
Also a, x, b are in H.P. ∴ \(\frac{1}{a}, \frac{1}{x}, \frac{1}{b}\) are in A.P

Question 8.

(1) A. P
(2) G.P
(3) H.P
(4) AGP
Answer:
(3) H.P

Explaination:

Question 9.
The H.M of two positive numbers whose A.M and G.M are 16,8 respectively is
(1) 10
(2) 6
(3) 5
(4) 4
Answer:
(4) 4

Explaination:
Let a, b be the two numbers. Given A. M = \(\frac{a+b}{2}\) = 16
G.M = \(\sqrt{a b}\) = 8
a+b = 16 × 2 = 32
ab = 8² = 64
H.M = \(\frac{2 a b}{a+b}\)
= \(\frac{2 \times 64}{32}\) = 2 × 2 = 4

Question 10.
If S denote the sum of n terms of an A. P whose common difference is d, the value of S_n – 2S_{n- 1} + S_{n – 2} is
(1) d
(2) 2d
(3) 4d
(4) d²
Answer:
(1) d

Explaination:

= a + (n – 1) d – (a + (n – 2)d)
= a + (n – 1) d – a – (n – 2)d
= a + nd – d – a – nd + 2d = d

Question 11.
The remainder when 38¹⁵ is divided by 13 is
(1) 12
(2) 1
(3) 11
(4) 5
Answer:
(1) 12

Explaination:
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
(1) n² + 3n² + 2n
(2) n³ – 3n² + 3n
(3) \(\frac{n(n+1)(n+2)}{3}\)
(4) \(\frac{n^{2}-n+2}{2}\)
Answer:
(4) \(\frac{n^{2}-n+2}{2}\)

Explaination:

Question 13.
The sum up to n terms of the series

(1) \(\sqrt{2 n+1}\)
(2) \(\frac{\sqrt{2 n+1}}{2}\)
(3) \(\sqrt{2 n+1}-1\)
(4) \(\frac{\sqrt{2 n+1}-1}{2}\)
Answer:
(4) \(\frac{\sqrt{2 n+1}-1}{2}\)

Explaination:

Question 14.
The n^th term of the sequence

(1) 2ⁿ – n – 1
(2) 1 – 2^-n
(3) 2^-n + n – 1
(4) 2^n-1
Answer:
(2) 1 – 2^-n

Explaination:

Question 15.
The sum up to n terms of the series

(1) \(\frac{\mathbf{n}(\mathbf{n}+1)}{2}\)
(2) 2n (n + 1)
(3) \(\frac{\mathbf{n}(\mathbf{n}+1)}{\sqrt{2}}\)
(4) 1
Answer:
(3) \(\frac{\mathbf{n}(\mathbf{n}+1)}{\sqrt{2}}\)

Explaination:

Question 16.
The value of the series

(1) 14
(2) 7
(3) 4
(4) 6
Answer:
(1) 14

Explaination:

Question 17.
The sum of an infinite G.P is 18. If the first term is 6 the common ratio is
(1) \(\frac{1}{3}\)
(2) \(\frac{2}{3}\)
(3) \(\frac{1}{6}\)
(4) \(\frac{3}{4}\)
Answer:
(2) \(\frac{2}{3}\)

Explaination:
Let the geometric series be a, ar, ar², …………… ar^n-1
Given a = 6, S_∞ = 18

Question 18.
The coefficient of x⁵ in the series e^-2x is
(1) \(\frac{2}{3}\)
(2) \(\frac{3}{2}\)
(3) \(-\frac{4}{15}\)
(4) \(\frac{4}{15}\)
Answer:
(3) \(-\frac{4}{15}\)

Explaination:

Question 19.
The value of

(1) \(\frac{e^{2}+1}{2 e}\)
(2) \(\frac{(e+1)^{2}}{2 e}\)
(3) \(\frac{(e-1)^{2}}{2 e}\)
(4) \(\frac{e^{2}+1}{2 e}\)
Answer:
(3) \(\frac{(e-1)^{2}}{2 e}\)

Explaination:

Question 20.
The value of

(1) log \(\left(\frac{5}{3}\right)\)
(2) \(\frac{3}{2}\) log \(\left(\frac{5}{3}\right)\)
(3) \(\frac{5}{3}\) log \(\left(\frac{5}{3}\right)\)
(4) \(\frac{2}{3}\) log \(\left(\frac{2}{3}\right)\)
Answer:
(2) \(\frac{3}{2}\) log \(\left(\frac{5}{3}\right)\)

Explaination: