How many terms of the A.P. :

24, 21, 18, ……….. must be taken so that their sum is 78 ?

In above A.P. a = 24 and d = -3.

We know that,

S = $\dfrac{n}{2}(2a + (n - 1)d)$

Let sum of n terms be 78.

$\Rightarrow 78 = \dfrac{n}{2}(2 \times 24 + (n - 1)(-3)) \\[1em] \Rightarrow 78 = \dfrac{n}{2}(48 - 3n + 3) \\[1em] \Rightarrow 78 \times 2 = n(51 - 3n) \\[1em] \Rightarrow 156 = 51n - 3n^2 \\[1em] \Rightarrow 3n^2 - 51n + 156 = 0 \\[1em] \Rightarrow 3(n^2 - 17n + 52) = 0 \\[1em] \Rightarrow n^2 - 17n + 52 = 0 \\[1em] \Rightarrow n^2 - 13n - 4n + 52 = 0 \\[1em] \Rightarrow n(n - 13) - 4(n - 13) = 0 \\[1em] \Rightarrow (n - 4)(n - 13) = 0 \\[1em] \Rightarrow n - 4 = 0 \text{ or } n - 13 = 0 \\[1em] \Rightarrow n = 4, 13.$

Hence, no. of terms = 4 or 13.