Question

How many terms of $A.P$ $7,11,15,19,\mathrm{23.....}$ will add up to $900$?

Answer 1

Sum of $AP=\left(\frac{n}{2}\right)[2a+(n-1\left)d\right]$

$900=\left(\frac{n}{2}\right)[2\ast 7+(n-1\left)4\right]$

$1800=n(14+4n-4]$

$1800=n(10+4n)$

$4n^2+10n-1800=0$

$4n^2-80n+90n-1800=0$

$4n(n-20)+90(n-20)=0$

$(n-20)(4n+90)=0$

$n=20$