Question

If $S_n=an+b{n}^2$, for an A. P. where $a$ and $b$ are constants, then the common difference of A. P. will be

• A. $2b$
• B. $a+b$
• C. $2a$
• D. $a-b$

Answer 1

The correct option is A

$2b$

Explanation for the correct option

Step 1: Given information

$S_n=an+b{n}^2$

By putting $n=1$, we get,

$S_1=a\left(1\right)+b{\left(1\right)}^2$

$S_1=a+b...\left(1\right)$

So, the first term of the A.P is $\left(a+b\right)$ $\left[\because a_1=S_1\right]$

Step 2: Put $n=2$,

$S_2=a\left(2\right)+b{\left(2\right)}^2$

$S_2=2a+4b$………………………$\left(2\right)$

Step 3: Subtract equation $\left(1\right)$ from equation $\left(2\right)$, we get,

$S_2-S_1=\left(2a+4b\right)-\left(a+b\right)$

Second term $=a+3b$ $\left[\because S_2-S_1=a_1+a_2-a_1\right]$

Step 4: Calculate the common difference

Common difference $=$ second term $-$ first term

Common difference $=a+3b-a-b$

Therefore, Common difference $=2b$

Hence, the correct option is ‘A’