Find the sum of the series

Note: The sum of the series is already provided in the question. The solution to find x is given below.

Let there be n terms in the series

$x=1+(n-1) 3$

$=3 n-2$

Let S be the sum of the series

$S=\frac{n}{2}[1+x]=715$

$\Rightarrow \mathrm{n}[1+3 \mathrm{n}-2]=1430$

$\Rightarrow \mathrm{n}+3 \mathrm{n}^{2}-2 \mathrm{n}=1430$

$\Rightarrow 3 \mathrm{n}^{2}-\mathrm{n}-1430=0$

Applying Sri Dhar Acharya formula, we get

$\mathrm{n}=\frac{1 \pm 131}{2 \times 3}$

$\mathrm{n}=\frac{132}{6}$ or $\frac{130}{6}$

⇒ n = 22 as n cannot be a fraction

Therefore $x=3 \times 22-2=64$

The value of $x$ is 64