Question

Find the sum of first 123 even natural number

Answer 1

even numbers are
2,4,6,...

treating it as AP with n =123 , a=2 and d=2.
$a |n| = a + (n - 1)d \\ a |n| = 2 + (122)(2) \\ a |n| = 246$
now we know sum of AP
$s |n| = \frac{n}{2} (a + l) \\ = \frac{123}{2} (2 + 246) \\ = 123(1 + 123) \\ = 123 \times 124 \\ = 15252$

mark as brainliest if helped

Answer 2

Here is the answer

2 , 4 , 6 ..., 2n are the even natural numbers.

Here ,
$\tt{a = t_{1} = 2}$

$\tt{ t_{2} = 4}$

$\tt{t_{3} = 6...}$

$\tt{d = t_{2} - t_{1} = 4 - 2 = 2}$
n = 123

$\tt {s_{n} = \frac{n}{2} (2a + (n - 1)d)}$

$\tt {= \frac{123}{2}(2 \times 2 +(123 - 1) \times 2)}$

$\tt {= \frac{123}{2} (4 + 122 \times 2)}$

$\tt { = \frac{123}{2} (4 + 248)}$

$\frac{123}{2} \times 248$


$\tt {= 123 \times 124}$

$\tt {s_{123} = 15252}$

The sum of the first 123 even natural numbers is 15252.