Question

In the A.P $1,7,13,19,....415$ prove that the sum of the middle terms is equal to the sum of first and last terms.

Answer 1

We observe that $1,7,13,19,\dots ,415$ is an A.P. with first term $a=1$ and common difference $d=6.$ Let there be $n$ terms in the given A.P.

Then,
$n^{th}$ term $415$
$\Rightarrow a+(n-1)d=415$
$\Rightarrow 1+6(n-1)=415$
$\Rightarrow 6n=420$
$\Rightarrow n=70$

So, there are $70$ terms in the given A.P.

Therefore, ${\left(\frac{70}{2}\right)}^{th}={35}^{th}$ and ${(\frac{70}{2}+1)}^{th}={36}^{th}$ are the middle terms.

Now,
$\begin{array}{cc}{a}_{35}+a_{36}& =a+34d+a+35d\\ & =1+34\times 6+1+35\times 6\\ & =205+211\\ & =416\end{array}$

Also, $a_1+a_{70}=1+415=416$
Clearly, $a_{35}+a_{36}=a_1+a_{70}$.