Question

Consider the following AP:
24, 21, 18.....
How many terms of this AP must be taken so that their sum is 78?

Answer 1

Let the number of terms be n.

a = 24, d = -3, S = 78

$S=\frac{n}{2}(2a+(n-1\left)d\right)$

$\Rightarrow 78=\frac{n}{2}(48-3(n-1\left)\right)=\frac{n}{2}(51-3n)$

$\Rightarrow 3n^2-51n+156=0$

$\Rightarrow n^2-17n+52=0$

$\Rightarrow (n-4)(n-13)=0$

$\Rightarrow n=4,13$

The sum of the first 4 terms of this AP is 78. As you take more and more terms into the sum, the sum will increase but only up to a certain point, because after a point, the terms in the AP become negative. So, any more terms will decrease the sum. This way, the sum of the first 13 terms also comes out to 78.