Question

The sum of the first $n$ term (n > 1) of an $A.P$. is $155$ and the common difference is $2$. If the first term is a positive integer, then-

• A. $n$ is an odd number
• B. $n$ is an even number
• C. $n=6$
• D. None of these

Answer 1

The correct option is A $n$ is an odd number

Let $S_n=155$ & $d=2$
$\Rightarrow \frac{n}{2}[2a+(n-1\left)d\right]=155$
$\Rightarrow \frac{n}{2}[2a+(n-1\left)2\right]=155$
$\Rightarrow \frac{n}{2}2[a+n-1]=155$
$\Rightarrow n[a+n-1]=155$ $...\left(1\right)$
Since the factor of $155$ is $31\ast 5$
Now, substitute $n=5$ in equation $\left(1\right)$; we get
$5(a+5-1)=155$
$\Rightarrow 5a+20=155$
$\Rightarrow 5a=135$
$\Rightarrow a=27$
Thus, $n$ is an odd number.