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Question

Question: The formula of the sum of first n natural numbers is \(S = \dfrac{{n\left( {n + 1} \right)}}{2}\) . ...

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

Explanation

Solution

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.

Complete step-by-step answer:

We know the natural number form an A.P. and we have the sum of first n natural numbers is S=325.
Now, we apply the formula of sum of first n natural numbers mentioned in the question.
S=n(n+1)2 325=n(n+1)2 650=n2+n n2+n650=0  \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 nn and solve the quadratic equation by using the Sridharacharya formula.
n2+n650=0 n=1±14×1×(650)2 n=1±1+26002 n=1±26012  \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 2601=51\sqrt {2601} = 51
n=1±512\Rightarrow n = \dfrac{{ - 1 \pm 51}}{2}
We know the number of terms cannot be negative. So, we take only positive value.
n=1+512 n=502 n=25  \Rightarrow n = \dfrac{{ - 1 + 51}}{2} \\\ \Rightarrow n = \dfrac{{50}}{2} \\\ \Rightarrow n = 25 \\\
So, the value of n is 25.

Note: Whenever we face such types of problems we use some important points. We can see the formula of sum of first n natural numbers mentioned in the question is the same as the formula of sum of first n terms of A.P. So, put the value of sum in formula then after some calculation we can get the required answer.