Question

Prove that the sum of an odd number of terms in A.P. is equal to the middle term multiplied by the number of terms.

Answer 1

Let the AP be $a,a+d,a+2d,...$

The Sum of terms = $\frac{n}{2}(2a+(n-1\left)d\right)$

$=an+\frac{n(n-1)}{2}d$

Let there are $n$ terms in the AP and $n$ is an odd integer.

Hence, the middle term is $(\frac{n+1}{2}{)}^{\text{th}}\text{term}=a+(\frac{n+1}{2}-1)d=a+\frac{(n-1)}{2}d$

Hence, multiplying $n$ to the middle term would b equal to sum of terms.