The first term of an A.P. is 5 and its 100th term is −292

In the given problem, we are given 1st and 100th term of an A.P.

We need to find the 50th term

Here,

$a=5$

$a_{100}=-292$

Now, we will find d using the formula 

So,

Also,

$a_{100}=a+(100-1) d$

$-292=a+99 d$

So, to solve for d

Substituting a = 5, we get

$-292=5+99 d$

$-292-5=99 d$

$\frac{-297}{99}=d$

$d=-3$

Thus,

$a=5$

$d=-3$

$n=50$

Substituting the above values in the formula, $a_{n}=a+(n-1) d$

$a_{50}=5+(50-1)(-3)$

$a_{50}=5-147$

$a_{50}=-142$

Therefore, $a_{50}=-142$