Sum of first n natural numbers

The natural numbers are 1,2,3,4,... so sum of first n natural numbers as
1 + 2 + 3 + 4 + ... + n
$S_{n}$ = 1 + 2 + 3 + 4 + ... + n
As we know that
$S_{n} = \frac{n}{2}[ 2a + (n-1)d]$

According to 'n' natural numbers
first term = a = 1 and common difference = d = 1
∴ $S_{n} = \frac{n}{2}[ 2 \times 1 + (n-1) \times1]$

$S_{n} = \frac{n}{2}[ 2 + n-1 ]$

$S_{n} = \frac{n}{2}[ n + 1 ]$

Examples on sum of first n natural numbers

1) Find the sum of first 20 terms of an A.P.
Solution: As we know that to find the sum of first 'n' natural numbers we use the following formula,
$S_{n} = \frac{n}{2}[ n + 1 ]$

Since here we have to find the sum of 20 terms so,
n = 20
$S_{20} = \frac{20}{2}[ 20 + 1 ]$

$S_{20} = 10 \times$ 21

$S_{20}$ = 210

2) Find the sum of first 55 terms of an A.P.
Solution: As we know that to find the sum of first 'n' natural numbers we use the following formula,
$S_{n} = \frac{n}{2}[ n + 1 ]$

Since here we have to find the sum of 55 terms so,
n = 55
$S_{55} = \frac{55}{2}[ 55 + 1 ]$

$S_{55} = \frac{55}{2} \times$ 56

$S_{55} = 28 \times$55

$S_{55}$ = 1540

3) Sum of first 'n' terms of an A.P is 120. Find n.
Solution: As we know that to find the sum of first 'n' natural numbers we use the following formula,
$S_{n} = \frac{n}{2}[ n + 1 ]$

But here the $S_{n}$ is given and we have to find 'n'(number of terms)
$S_{n}$ = 120
$S_{n} = \frac{n}{2}[ n + 1 ]$

120 = $\frac{n}{2}[ n + 1 ]$

240 = n (n + 1)
240 = $n^{2}$ + n
∴ $n^{2}$ + n - 240 = 0
Now this is a quadratic equation, so we will find the factors of this
The factors are
(n + 16)(n -15) = 0
∴ n = -16 or n = 15
But the number of terms never be negative so the value of n = 15.
So the sum for first 15 terms of an A.P is 120.