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 x^{6} in (2 + 2x)^{10} is

(1) 10C_{6}

(2) 2_{6}

(3) 10C_{6}2^{6}

(4) 10C_{6}2^{10}

Answer:

(4) 10C_{6}2^{10}

Explaination:

Question 3.

The coefficient of x^{8} y^{12} in the expansion of (2x + 3y)^{20} is

(1) 0

(2) 2^{8} 3^{12}

(3) 2^{8} 3^{12} + 2^{12} 3^{8}

(4) 20C_{8} 2^{8} 3^{12}

Answer:

(4) 20C_{8} 2^{8} 3^{12}

Explaination:

Question 4.

If nC_{10} > 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 ^{10}C_{10}, ^{21}C_{10}, ^{19}C_{10} and ^{20}C_{10}, ^{20}C_{10} 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 + x^{2})^{2} (1 + x)^{n} = a_{0} + a_{1}x + a_{2}x^{2} + ………… + x^{n + 4} and if a_{0}, a_{1}, a_{2} are in A. P then n is

(1) 1

(2) 2

(3) 3

(4) 4

Answer:

(3) 3

Explaination:

n^{2} – 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 ∴ 4^{2} = 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 = 8^{2} = 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) d^{2}

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 38^{15} is divided by 13 is

(1) 12

(2) 1

(3) 11

(4) 5

Answer:

(1) 12

Explaination:

38^{15} = (39 – 1)^{15} = 39^{15} – 15C_{1} 39^{14}(1) + 15C_{2} (39)^{13}(1)^{2} – 15C_{3} (39)^{12}(1)^{3} ….. + 15C_{14} (39)^{1}(1) – 15C_{15}(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) n^{2} + 3n^{2} + 2n

(2) n^{3} – 3n^{2} + 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) 2^{n} – 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, ar^{2}, …………… ar^{n-1}

Given a = 6, S_{∞} = 18

Question 18.

The coefficient of x^{5} 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: