Sequences and Series Online Objective Test | Sequences and Series Online Objective Test

# Sequences and Series # CBSE 11th Mathematics Prepare / Learn

Q1. the ratio of first to the last of n A.m.’s between 5 and 35 is 1 : 4. The value of n is

A. 19

B. 10

C. 11

D. 9

Answer : Option D
Explaination / Solution:

the sequence is $5, A_{1}, A_{2}, . . . . . A_{n}, 35$

here a=5 and b=35

we know that, $d = \frac{b - a}{n + 1} = \frac{35 - 5}{n + 1} = \frac{30}{n + 1}$

According to question,

# Sequences and Series # CBSE 11th Mathematics Prepare / Learn

Q2. The next term of the sequence 1, 3, 6, 10, …. Is

A. 15

B. 16

C. 14

D. 13

Answer : Option A
Explaination / Solution:

In this sequence the nth term is, $a_{n} = 1 + 2 + 3 + . . . . . . + n$

so, for n = 5, $a_{5} = 1 + 2 + 3 + 4 + 5 = 15$

# Sequences and Series # CBSE 11th Mathematics Prepare / Learn

Q3. If a, b, c, d are in H.P., then ab + bc + cd is

A. ( a + b)( c + d)

B. 3bd

C. 3ac

D. 3ad

Answer : Option D
Explaination / Solution:

using the property of H.M

$\begin{array}{l} \frac{2 a c}{a + c} = b \\ \frac{2 b d}{b + d} = c \end{array}$

multiplying the above equations, we get

$\begin{array}{l} \frac{4 a b c d}{(a + c) (b + d)} = b c \\ \Rightarrow 4 a b c d = b c (a b + b c + a d + c d) \\ \Rightarrow 4 a b c d = a b^{2} c + (b c)^{2} + a b c d + b c^{2} d \\ \Rightarrow 3 a b c d = a b^{2} c + (b c)^{2} + b c^{2} d \end{array}$