Question

If $1+6+11+....+9x=148$ then $x$ is equal to

Answer 1

The given series is an arithmetic series in which $a=1,d=6-1=5$

Let it contain $n$ terms
Then $S_n=\frac{n}{2}[T_1+T_n]$
$\Rightarrow 148=\frac{n}{2}[1+9x]...\left(i\right)$

Also, $T_n=a+(n-1)d$
$\Rightarrow 9x=1+(n-1)5\Rightarrow n=\frac{9x+4}{5}$

By putting this value of n into equation $\left(i\right),$ we get
$\frac{1+9x}{2}\times \frac{9x+4}{5}=148\Rightarrow 9x^2+5x-164=0\Rightarrow x=4,-\frac{41}{9}$
As common difference is positive, so negative value of $x$ is not possible
Hence, $x=4$