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
636k+ views
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.
Recently Updated Pages
Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Chemistry: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

