Question

Consider the A.P. $2,5,8,11,.....,302.$ Show that twice of the middle term of the above A.P. is equal to the sum of its first and last term.

Answer 1

Clearly, $2,5,8,11,.....,302.$. is an A.P. with first term $a=2$ and common difference $d=3$. Let there be $n$ terms in the given A.P.

Then, $n^{th}$ term $=302$

$⟹a+(n-1)d=302$

$⟹2+4(n-1)=302$

$⟹3n=303$

$⟹n=101$
Clearly, $n$ is odd. Therefore, $\left(\frac{n+1}{2}\right)$ i.e. ${51}^{st}$ term is the middle term.
Now,
Middle term $=a_{51}=a+50d=2+50\times 3=152$

First term $+$ Last term $=2+302=304=2\times$ the last term
Clearly, twice the middle term is equal to the sum of the first and last term.