Question

4. 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

We know the sum of first n terms of an AP.

$S_n=\dfrac {n}{2}[2a+(n-1)d]$

How featured the sum is if n is odd? Let,

$n=2k-1\quad\iff\quad k=\dfrac {n+1}{2}$

This is because, if $n=2k+1,$ then 'n' can accept all odd numbers but except 1. That's why.

In addition, since n is odd, $k=\dfrac {n+1}{2}$ is the middle position among the n terms, thus $a_k=a_{\frac {n+1}{2}}$ is the middle term.

So the sum becomes,

$S_n=\dfrac {n}{2}[2a+(2k-1-1)d]\\\\\\S_n=\dfrac {n}{2}[2a+(2k-2)d]\\\\\\S_n=\dfrac {n}{2}\cdot 2[a+(k-1)d]\\\\\\S_n=n[a+(k-1)d]\\\\\\S_n=n\cdot a_k\\\\\\\boxed {S_n=n\cdot a_{\frac {n+1}{2}}}$

Thus the sum to n terms of an AP is the middle term multiplied by the no. of terms if and only if n is odd.

Hence the Proof!

#answerwithquality

#BAL