Question

Sum of the first 'n' terms of an A.P. having first term 'a' & last term 'l' is

• A. $\frac{n}{2}[2a-l]$
• B. $\frac{n}{2}[2a+l]$
• C. $\frac{n}{2}[a+l]$
• D. $\frac{n}{2}[a-l]$

Answer 1

The correct option is C $\frac{n}{2}[a+l]$

We know that sum first $n$ terms of an A.P. having first term $a$ and common difference $d$ is $\frac{n}{2}(2a+(n-1\left)d\right)$ ----(1)

The last term is the $n^{th}$ term which is given by $i=a+(n-1)d$ ----(2)

substituting (2) in (1) we get

Sum of $n$ terms of an A.P. is $\frac{n}{2}(a+l)$