
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.
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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