Question

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

Answer 1

Given:$24,21,18,...$ are in A.P

$a=24,d=21-24=-3$

Sum$=\frac{n}{2}[2a+(n-1)d]$

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

$\Rightarrow 156=n[48-3n+3]$

$\Rightarrow 156=n[51-3n]$

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

$\Rightarrow 3n^2-12n-39n+156=0$

$\Rightarrow 3n(n-4)-39(n-4)=0$

$\Rightarrow (n-4)(3n-39)=0$

$\therefore n=4,n=\frac{39}{3}=13$

When $n=4,s_4=\frac{4}{2}[2\times 24+(4-1)(-3)]=2[48-9]=2\times 39=78$

When $n=13,s_{13}=\frac{13}{2}[2\times 24+(13-1)(-3)]=\frac{13}{2}[48-36]=\frac{13}{2}\times 12=78$

Hence number of terms $n=4$ or $n=13$