Question

Given sum of the first $n$ terms of an AP is $2n+3n^2$. Another AP is formed with the same first terms and double of the common difference, the sum of $n$ terms of the new AP is

• A. $n+4n^2$
• B. $6n^2-n$
• C. $n^2+4n$
• D. $3n+2n^2$

Answer 1

The correct option is B $6n^2-n$

$2n+3n^2=\frac{n}{2}(4+6n)=\frac{n}{2}(10+6(n-1\left)\right)$

Clearly, the common difference is 6 for this AP

Now doubling common difference gives sum $=\frac{n}{2}(10+12(n-1\left)\right)=\frac{n}{2}(-2+12n)=6n^2-n$