Question

Solve :- $1+4+7+10+....+x=590$

Answer 1

$1+4+7+10+....+x=590$

Its an A.P., where

$a=1d=4-1=3,S_n=590$

in order to find $x$, we need to find the number of terms i.e. $n$

$S_n=\frac{n}{2}(2a+(n-1\left)d\right)$

$590=\frac{n}{2}(2\times 1+(n-1\left)\right(3\left)\right)$

$590=\frac{n}{2}(2+n(3)-(3\left)\right)$

$590=\frac{n}{2}(2+3n-3)$

$590=\frac{n}{2}(3n-1)$

$590\times 2=n(3n-1)$

$1180=3n^2-n$

$0=3n^2-n-1180$

Here we have a quadratic equation i.e.

$3n^2-n-1180=0$

To find the value of n, we need to find the roots of the equation by the formula,

$n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

where $a=3,b=-1andc=1180$

$n=\frac{-(-1)\pm \sqrt{(-1{)}^2-4\times 3\times (-1180)}}{2\times 3}$

$n=\frac{1\pm \sqrt{1+14160}}{6}$

$n=\frac{1\pm \sqrt{14161}}{6}$

$n=\frac{1\pm 119}{6}$

$n=\frac{1+119}{6}$ or $n=\frac{1-119}{6}$

$n=\frac{120}{6}$ or $n=\frac{-118}{6}$

Since the number of terms i.e. $n$ cannot be negative therefore, $n=20$

For finding the last term,

$x=a+(n-1)d$

$x=1+(20-1)\left(3\right)$

$x=1+19\left(3\right)$

$x=1+57$

$x=58$

So, the last term of the series i.e. $x$ is $58$