Question

If $1+6+11+..........+x=148$, then find the value of $x.$

• A. $22$
• B. $36$
• C. $42$
• D. $50$

Answer 1

The correct option is B $36$

Given series is $1+6+11+......+x=148$
Given series forms an AP. with first term $a=1$, common difference $d=6-1=5$ and sum of terms $S_n=148$
Since, $S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow 148=\frac{n}{2}[2\times 1+(n-1\left)5\right]$
$\Rightarrow 296=n[2+(n-1\left)5\right]$
$\Rightarrow 296=n[5n-3]$
$\Rightarrow 296=5n^2-3n$
$\Rightarrow 5n^2-3n-296=0$
$\Rightarrow 5n^2-40n+37n-296=0$
$\Rightarrow 5n(n-8)+37(n-8)=0$
$\Rightarrow (n-8)(5n+37)=0$
$\Rightarrow n-8=0$ and $5n+37\ne 0$ ....[$\because n$ can't be negative]
$\Rightarrow n=8$
Since, $x=a+(n-1)d$
$x=1+(8-1)5=36$
Hence, option B is correct.