Question

IF $1+4+7+\cdots +x=287,$ find the value of x.

Answer 1

$S_n=\frac{n}{2}[2a+(n-1\left)d\right]Givena=1,d=4-1=3,Sn=287287=\frac{n}{2}(2\times 1+(n-1\left)3\right)287\times 2=n(2+3n-3)574=2n+3n^2-3n3n^2-n-574=0$

On solving the quadratic equation using formula $n=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-(-1)\pm \sqrt{(-1{)}^2-4(3\left)\right(-574)}}{2\times 3}=\frac{1\pm \sqrt{6889}}{6}=\frac{1\pm 83}{6}=\frac{1+83}{6}or\frac{1-83}{6}=\frac{84}{6}or\frac{-82}{6}=14or\frac{-41}{3}$

we get n = 14, $\frac{-41}{3}$

n can't be $\frac{-41}{3}$ because n is a positive number.

$n=14S_n=\frac{n}{2}[a+l]287=\frac{14}{2}(1+x)574=14(1+x)\frac{574}{14}=1+x41=1+xSo,x=41-1x=40$