Question

How many terms of the A.P.: $9,17,25,...$ must be taken to give a sum of $636$?

• A. $6$
  
• B. $8$
• C. $12$
• D. $16$

Answer 1

Answer: (C)

$12$

First term, $a=9$
Common difference, $d=17-9=25-17=8$ $d=8$
Sum $S_n=636$

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

$636=\frac{n}{2}[2\times 9+(n-1\left)8\right]$

$1272=n(18+8n-8)$

$1272=10n+8n^2$

$8n^2+10n-1272=0$

$4n^2+5n-636=0$

upon solving the quadratic equation, we get,

$n=12,-\frac{53}{4}$

n cannot be negative,

Therefore $n=12$