Question

Find the sum of all natural numbers from 1 to 100, which are divisible by 2 or 5.​

Answer 1

$<b>Divisible by 2</b>$

Smallest natural number divisible by 2 from 1 to 100 = 2
Largest natural number divisible by 2 from 1 to 100 = 100

Required AP = 2, 4, 6, 8.....100

a = 2
d = 4 - 2
=2

$t_{n} = a + (n - 1)d \\ \\ = > 100 = 2 +( n - 1)2 \\ \\ = > \frac{100 - 2}{2} = n - 1 \\ \\ = > \frac{98}{2} = n - 1 \\ \\ = > 49 + 1 =n \\ \\ = > n = 50$

$S_{50} = \frac{n}{2} [2a + (n - 1)d] \\ \\ = \frac{50}{2} [2 \times 2 + (50 - 1)2] \\ \\ = 25[4 + 49 \times 2] \\ \\ = 25[4 + 98] \\ \\ = 25 \times 392 \\ \\ = 9800$

$<b>Divisible by 5</b>$

Smallest natural number divisible by 5 from 1 to 100 = 5
Largest natural number divisible by 5 from 1 to 100 = 100

Required AP = 5, 10, 15.....100

a = 5
d = 10 - 5
=5

$t_{n} = a + (n - 1)d \\ \\ = > 100 = 5 +( n - 1)5 \\ \\ = > \frac{100 - 5}{5} = n - 1 \\ \\ = > \frac{95}{5} = n - 1 \\ \\ = > 19 + 1 =n \\ \\ = > n = 20$

$S_{20} = \frac{n}{2} [2a + (n - 1)d] \\ \\ = \frac{20}{2} [2 \times 5 + (20 - 1)5] \\ \\ = 10[10 + 19 \times 5] \\ \\ = 10[10 + 95] \\ \\ = 10 \times 105 \\ \\ = 1050$

$\huge\boxed{Answer = 9800 \: and \: 1050}$

Answer 2

answer is 9800 and 1050

mark it brainliest..

its my order ...