Question

How many terms of the series $9,12,15,...$ must be taken to make $360$?

Answer 1

First term of series = $9$ and common difference = $3$

$n^{th}$ term = $T_n=9+(n-1)3$

To make $360$,

Sum = $9n+\frac{n(n-1)}{2}3$

$360=9n+\frac{n(n-1)}{2}3$

$720=18n+3(n^2-n)$

Dividing by $3$,

$240=6n+n^2-n$

$n^2+5n-240=0$

Solution to the quadratic equation is
$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$n=\frac{-5\pm \sqrt{25-4\left(1\right)(-240)}}{2\left(1\right)}$

$n=\frac{-5\pm \sqrt{985}}{2}$

$n=\frac{-5+31.38}{2}$ and $n=\frac{-5-31.38}{2}$

$n=13.19$ and $n=-18.19$

Since $n$ has to be an integer,
$n=14$ or $n=-19$