Question

Consider the following AP:

24, 21, 18.....

How many terms of this AP must be taken so that their sum is 78? [4 MARKS]

Answer 1

Formula: 1 Mark
Concept: 1 Mark
Application: 2 Marks

Let the number of terms be n.

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

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

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

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

$\Rightarrow 78=\frac{51n-3n^2}{2}$

$\Rightarrow 78\times 2=51n-3n^2$

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

dividing the equation by $3$,

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

factorize by splitting the middle term,

$n^2-13n-4n+52=0$

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

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

$\Rightarrow$ $n-4=0$ or $n-13=0$

$\Rightarrow n=4,13$

Since both the values are positive natural numbers, so, either of them can be taken.