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

Both 4 and 13

The correct answer is Option (3) → Both 4 and 13

Given AP: 24, 21, 18, ...

First term $a = 24$, common difference $d = 21 - 24 = -3$

Sum of $n$ terms of an AP: $S_n = \frac{n}{2} \left[2a + (n - 1)d \right]$

Given: $S_n = 78$

$\frac{n}{2} [2(24) + (n - 1)(-3)] = 78$

$\frac{n}{2} [48 - 3n + 3] = 78$

$\frac{n}{2} (51 - 3n) = 78$

$n(51 - 3n) = 156$

$51n - 3n^2 = 156$

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

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

Discriminant: $D = (-17)^2 - 4(1)(52) = 289 - 208 = 81$

$n = \frac{17 \pm \sqrt{81}}{2} = \frac{17 \pm 9}{2}$

$\Rightarrow n = \frac{26}{2} = 13$ or $n = \frac{8}{2} = 4$