Question

If the sum of $n$ terms of an AP is $\frac{n}{2}(7+5n)$, then its $n^{th}$ term is

• A. $5n+1$
• B. $7n+1$
• C. $2n+1$
• D. $5n+2$

Answer 1

The correct option is A $5n+1$

$S_n=\frac{n}{2}[7+5n]$

Then,

$S_{n-1}=\frac{(n-1)}{2}[7+5(n-1\left)\right]$

$\therefore S_{n-1}=\frac{(n-1)}{2}[2+5n]$

Now,

$T_n=S_n-S_{n-1}$

$=\frac{n}{2}[7+5n]-\frac{(n+1)}{2}[2+5n]$

$=\frac{1}{2}\left[n\right(7+5n)-(n-1\left)\right(2+5n\left)\right]$

$=\frac{1}{2}[7n+5n^2+3n-5n^2+2]$

$\therefore T_n=5n+1$