Question

The sum ${}^{\prime }{S}^{\prime }$ of the first $n$ natural numbers is given by the relation $S=\frac{(n+1)}{2}$. Find $n$, if the sum is $55$.

Answer 1

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

For $S=55\left(\text{Given}\right)$

$\therefore 55=\frac{n(n+1)}{2}$

$\Rightarrow 110=n^2+n$

$\Rightarrow n^2+n-110=0$

$\Rightarrow n^2+11n-10n-110=0$

$\Rightarrow (n+11)(n-10)=0$

$\because n\ne -ve$

$\therefore n=10$