Question

# The formula of the sum of first n natural numbers is $S = \dfrac{{n\left( {n + 1} \right)}}{2}$ . If the sum of first n natural number is 325 then find n.

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Hint: In this question, we have a formula of sum of first n natural numbers and we know the sum of first n natural numbers given in question. So, put the value of sum in formula and get value of n after solving the quadratic equation.

$\Rightarrow S = \dfrac{{n\left( {n + 1} \right)}}{2} \\ \Rightarrow 325 = \dfrac{{n\left( {n + 1} \right)}}{2} \\ \Rightarrow 650 = {n^2} + n \\ \Rightarrow {n^2} + n - 650 = 0 \\$
We can see quadratic equations in $n$ and solve the quadratic equation by using the Sridharacharya formula.
$\Rightarrow {n^2} + n - 650 = 0 \\ \Rightarrow n = \dfrac{{ - 1 \pm \sqrt {1 - 4 \times 1 \times \left( { - 650} \right)} }}{2} \\ \Rightarrow n = \dfrac{{ - 1 \pm \sqrt {1 + 2600} }}{2} \\ \Rightarrow n = \dfrac{{ - 1 \pm \sqrt {2601} }}{2} \\$
Now, use $\sqrt {2601} = 51$
$\Rightarrow n = \dfrac{{ - 1 \pm 51}}{2}$
$\Rightarrow n = \dfrac{{ - 1 + 51}}{2} \\ \Rightarrow n = \dfrac{{50}}{2} \\ \Rightarrow n = 25 \\$