Answer

Verified

436.5k+ 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.

Recently Updated Pages

Assertion The resistivity of a semiconductor increases class 13 physics CBSE

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE

Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE

What are the possible quantum number for the last outermost class 11 chemistry CBSE

Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE

Trending doubts

What type of defect is shown by NaCl in a Stoichiometric class 12 chemistry CBSE

Difference Between Plant Cell and Animal Cell

Distinguish between tetrahedral voids and octahedral class 12 chemistry CBSE

Derive an expression for electric potential at point class 12 physics CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is BLO What is the full form of BLO class 8 social science CBSE

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Fill the blanks with proper collective nouns 1 A of class 10 english CBSE