Question

Find the number of terms of an arithmetic progression whose third term is 20, 8th term is 10 and the sum of terms is 144.

• A. n=16 and n=9
• B. n=15 and n=8
• C. n=14 and n=7
• D. None of these

Answer 1

The correct option is A n=16 and n=9

Let the first term be 'a' and common difference be 'd'
$t_3=20,t_8=10$
$\Rightarrow a+2d=20$ ............(1)
$\Rightarrow a+7d=10$ ............(2)
Subtracting (1) from (2), we get:
$a+7d=10$
$a+2d=20$
____________
$5d=-10\Rightarrow d=-\frac{10}{5}$
$d=-2$
Substituting d = -2 in (1), we get:
$a+2d=20\Rightarrow a+2(-2)=20$
$\Rightarrow a=20+4\Rightarrow a=24$
Let the number of terms in given A.P. be 'n'
$S_n=144$
$\Rightarrow \frac{n}{2}(2a+(n-1\left)d\right)=144$
$\Rightarrow \frac{n}{2}(2\times 24+(n-1\left)\right(-2\left)\right)=144$
$\frac{n}{2}(48-2n+2)=144$
$\Rightarrow \frac{n}{2}(50-2n)=144$
$\Rightarrow \frac{n}{2}\times 2(25-n)=144$
$25n-n^2-144=0$
$\Rightarrow -(n^2-25n+144)=0$
$\Rightarrow n^2-25n+144=0$
$\Rightarrow n^2-16n-9n+144=0$
$\Rightarrow n(n-16)-9(n-16)=0$
$\Rightarrow (n-16)(n-9)=0$
$\Rightarrow n-16=0,n-9=0$
$\therefore n=16\therefore n=9$