Question

Task 15
The fourth term of A.P is equal to 3 times the first term term and seventh term exceeds
twice the third term by 1. Find the common difference
(A) 2
(B) 3
(C) 1
(D) 4​

Answer 1

Required Answer

The common difference of the A.P is 2. Hence, Option - A is the correct option.

Solution

Given things

According to the question, we are given that:

$\Longrightarrow a_4 = 3 \times a$

$\Longrightarrow a_7 = 2 \times a_3 + 1$

How to do it?

In general, $n$th term of an A.P is calculated by:

$\rm{nth\:term = a+(n-1)d}$

By using the above formula, we will frame an equation and solve it by the method of substitution.

First Condition

$\Longrightarrow a_4 = 3 \times a$

$\Longrightarrow a+3d = 3a$

$\Longrightarrow 2a = 3d ---(i)$

Assume the above as $Eq.(i)$

Second Condition

$\Longrightarrow a_7 = 2 \times a_3 + 1$

$\Longrightarrow a+6d = 2(a+2d) + 1$

$\Longrightarrow a+6d = 2a+4d+1$

$\Longrightarrow a+1 = 2d$

Multiply 2 on the equation as a whole.

$\Longrightarrow 2(a+1) = 2(2d)$

$\Longrightarrow 2a+2 = 4d ---(ii)$

Assume the above as $Eq.(ii)$

Substitute the value of $2a = 3d$ in $Eq.(ii)$.

$\Longrightarrow 3d+2 = 4d$

$\Longrightarrow{\boxed{d = 2}}$

Hence, the common difference is 2.