Question

How many terms of the AP : 9, 17, 25, ... must be taken to give a sum of 636?

Answer 1

9, 17, 25….

$a_1=9$

$a_2=17$

$d=a_2-a_1=17-9=18$

$S_n=636$

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

$636=\frac{n}{2}\left(2\times 9+\left(n-1\right)8\right)$

$1272=n\left(18+8x-8\right)$

$8n^2+10n-1272=0$

$4n^2+5n-636=0$

$D=b^2-4ac=25-4\times 4\times \left(-636\right)$

$=25+10176$

$=10201$

$x=\frac{-b\pm \sqrt{D}}{2a}$

$=\frac{-5\pm \sqrt{10201}}{2\times 4}$

$x=\frac{-5+101}{8},\frac{-5-101}{8}$

$=12,-51/4\left(\mathrm{Re}q.\right)$

$\therefore n=12$