Question

How many terms of the series $26,21,16,...$ must be taken to amount to $74$?

Answer 1

26,21,16,............

$\therefore$ Common difference$=5$.

First term(a)$=26$

Sum of the AP series$=\frac{n(2a+(n-1\left)d\right)}{2}$

$=\frac{n}{2}(52+(n-1)-5)$

$\therefore \frac{n}{2}(52+(n-1\left)\right(-5\left)\right)=74$

$\Rightarrow n(52-5n+5)=148$

$\Rightarrow n(57-5n)=148$

$\therefore n=4$