Answer

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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.

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.

$

\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}$

We know the number of terms cannot be negative. So, we take only positive value.

$

\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.

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.

$

\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}$

We know the number of terms cannot be negative. So, we take only positive value.

$

\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.

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