Question

Solve:
$\left(ii\right)1+4+7+10+...+x=590$

Answer 1

Given: $1+4+7+10++x=590$
It is an A.P.
First term, $a=1$
Common difference, $d=4-1=3$
The sum of n term of an A. P. is
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\Rightarrow 590=2[2\times 1+(n-1)\times (3\left)\right]$
$\Rightarrow 590=\frac{n}{2}(3n-1)$
$\Rightarrow 3n^2-n-1180=0$ $\Rightarrow 3n^2-60n+59n-11800=0$
$\Rightarrow (n-20)(3n+59)=0$
$\Rightarrow n=-\frac{59}{3},20$
As $n\in N$, so $n=20$
The nth term of an A. P. is
$a_n=a+(n-1)d$
$\Rightarrow x=1+(20-1)\left(3\right)$
$\Rightarrow x=1+57$
$\Rightarrow x=58$