Question

The sum of n terms of an AP is n² + n; the common difference will be:

• (A) n - 1
• (B) 1
• (C) 2
• (D) -4

Answer 1

The Correct Option is C

Explanation:
$S_n=n^2+n=\frac{n}{2}[2a+(n-1)d]$$\Rightarrow a+\frac{(n-1)d}{2}=n+1\dots (1)$Again, $S_n=n^2+n$$\therefore {\mathrm{S}}_{n-1}=(n-1{)}^2+(n-1)$$\therefore S_n-S_{n-1}=n^2+n-[(n-1{)}^2+(n-1)]$$=n^2+n-n^2-1+2n-n+1$or $T_n=2n$$\Rightarrow a+(n-1)d=2n\dots (2)$Subtracting(1) from (2), we get $[n-1-\left(\frac{n-1}{2}\right)]d=2n-(n+1)$Or d $\left(\frac{n-1}{2}\right)=n-1$$\therefore d=2$