Question

The sum of the first n terms of a sequence is $\frac{n(3n+1)}{2}$
i) when the sum of the first n terms is 155, show that $3n^2+n-310=0$
ii) Solve : $3n^2+n-310=0$

Answer 1

i)

$\frac{n(3n+1)}{2}=155$

$3n^2+n=310$

$\Rightarrow 3n^2+n-310=0$

ii)

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

$\Rightarrow n=\frac{-1\pm \sqrt{1^2-4\left(3\right)(-310)}}{2\left(3\right)}$

Solving the above equation, we get

$n=\frac{-1+\sqrt{1^2-4\left(3\right)(-310)}}{2\left(3\right)}=10$

$n=\frac{-1-\sqrt{1^2-4\left(3\right)(-310)}}{2\left(3\right)}=\frac{-31}{3}$