How many terms of the AP: 24, 21, 18, ............. must be taken so that their sum is 78?
Answer & explanation
Correct answer: option 3
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 $n = \frac{8}{2} = 4$