Question

If $3+5+7+9+$ ... upto $n$ terms $=288$, then $n=$ ____

• A. $12$
• B. $15$
• C. $16$
• D. $17$

Answer 1

Answer: (C)

$16$

Series is $3+5+7+9+.....$

which is in A.P. where,

First term $a=3$

and Common difference, $d=a_2-a_1=5-3=2$

Sum of n terms, $S_n=288$

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

So,

$\frac{n}{2}\{2\left(3\right)+(n-1)2\}=288$

$n\{6+2n-2\}=576$

$2n^2+4n=576$

$n^2+2n-288=0$

$n^2+18n-16n-288=0$

$(n+18)(n-16)=0$

$n=-18$ or $n=16$

$n$ is no. of terms which cannot be negative.

So, $n=16$