Question

How many terms are need in the A.P. $24,20,\mathrm{16.....}$ to make the sum $72$?

Answer 1

Given A.P is $24,20,16,...$

so, the first term, $a=24$

the common difference, $d=20-24$

$=-4$.

The sum of first n terms of A.P is given by

$\frac{n}{2}[2a+(n-1\left)d\right]=\frac{n}{2}[48-(n-1\left)4\right]$

Given, the sum is $72$.

so, $\frac{n}{2}[48-(n-1\left)4\right]=72$

$\Rightarrow 24n-n(n-1)2=72$

$\Rightarrow 2n^2-26n+72=0\Rightarrow n^2-13n+36=0$

$\Rightarrow n=4$ or 9.

$\therefore$ 4 terms and 9 terms add upto 72.