Questions & Answers

Question

Answers

Answer

Verified

150k+ views

Hint:- Check whether n numbers form an A.P or not and apply $ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ .

For finding the sum of first n positive integers.

As we know that the first positive integer is 1.

And the second positive integer will be 2.

And this sequence goes on till the last number is n.

So, first n positive numbers will be \[1,2,3,4......n - 1,n\].

As we see that these n numbers written above forms an A.P. With,

$ \Rightarrow $First number, $a = 1$.

$ \Rightarrow $Common difference,${\text{ }}d = 1$.

$ \Rightarrow $And total number of terms $ = n$.

As, we know that the sum of n terms of an A.P is given as ${S_n}$. Where,

$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ (1)

So, putting the values of a, d and n in equation 1. We get,

$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2 + \left( {n - 1} \right)} \right) = {\text{ }}\dfrac{{n\left( {n + 1} \right)}}{2}{\text{ }}$

Hence, sum of first n positive integers will be $\dfrac{{n\left( {n + 1} \right)}}{2}{\text{ }}$

Note:- Whenever we came up with this type of problem where we are asked to find the

Sum or product of some numbers then, first we had to check whether these number have any

relation with each other, because in most of the cases they form A.P or G.P. Then we can easily

find the sum or product of these numbers using A.P or G.P formula.

For finding the sum of first n positive integers.

As we know that the first positive integer is 1.

And the second positive integer will be 2.

And this sequence goes on till the last number is n.

So, first n positive numbers will be \[1,2,3,4......n - 1,n\].

As we see that these n numbers written above forms an A.P. With,

$ \Rightarrow $First number, $a = 1$.

$ \Rightarrow $Common difference,${\text{ }}d = 1$.

$ \Rightarrow $And total number of terms $ = n$.

As, we know that the sum of n terms of an A.P is given as ${S_n}$. Where,

$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$ (1)

So, putting the values of a, d and n in equation 1. We get,

$ \Rightarrow {S_n} = \dfrac{n}{2}\left( {2 + \left( {n - 1} \right)} \right) = {\text{ }}\dfrac{{n\left( {n + 1} \right)}}{2}{\text{ }}$

Hence, sum of first n positive integers will be $\dfrac{{n\left( {n + 1} \right)}}{2}{\text{ }}$

Note:- Whenever we came up with this type of problem where we are asked to find the

Sum or product of some numbers then, first we had to check whether these number have any

relation with each other, because in most of the cases they form A.P or G.P. Then we can easily

find the sum or product of these numbers using A.P or G.P formula.

Students Also Read