Answer
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Hint: Use the formula ${S_{100}} = \dfrac{n}{2}[2a + (n - 1)d]$.
Complete step-by-step answer:
We have to find sum of first 100 natural numbers
Therefore , the natural numbers are 1,2,3,4,5……………..100.
Now ,if we see they form an A.P. with first term (a)=1 and common difference(d)=1 and number of terms(n) =100
Now using the formula of sum given below:
$
{S_{100}} = \dfrac{n}{2}[2a + (n - 1)d] \\
{\text{substituting the values}}{\text{. we get}} \\
{{\text{S}}_{100}} = \dfrac{{100}}{2}[2 \times 1 + (100 - 1) \times 1] \\
{S_{100}} = 50[2 + 99] \\
{S_{100}} = 50 \times 101 \\
{S_{100}} = 5050 \\
\therefore {\text{ The sum of first 100 natural numbers is 5050}}{\text{. }} \\
$
Note: Alternative solution
$
a = 1,d = 1,n = 100 \\
{\text{last term(l)}} = 100 \\
{S_{100}} = \dfrac{n}{2}[a + l] \\
{S_{100}} = \dfrac{{100}}{2}[1 + 100] \\
{S_{100}} = 50 \times [101] \\
{S_{100}} = 5050 \\
$
Complete step-by-step answer:
We have to find sum of first 100 natural numbers
Therefore , the natural numbers are 1,2,3,4,5……………..100.
Now ,if we see they form an A.P. with first term (a)=1 and common difference(d)=1 and number of terms(n) =100
Now using the formula of sum given below:
$
{S_{100}} = \dfrac{n}{2}[2a + (n - 1)d] \\
{\text{substituting the values}}{\text{. we get}} \\
{{\text{S}}_{100}} = \dfrac{{100}}{2}[2 \times 1 + (100 - 1) \times 1] \\
{S_{100}} = 50[2 + 99] \\
{S_{100}} = 50 \times 101 \\
{S_{100}} = 5050 \\
\therefore {\text{ The sum of first 100 natural numbers is 5050}}{\text{. }} \\
$
Note: Alternative solution
$
a = 1,d = 1,n = 100 \\
{\text{last term(l)}} = 100 \\
{S_{100}} = \dfrac{n}{2}[a + l] \\
{S_{100}} = \dfrac{{100}}{2}[1 + 100] \\
{S_{100}} = 50 \times [101] \\
{S_{100}} = 5050 \\
$
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