Question

find the sum of all the naturl numbers between 100 and 500 which are divisble by 8​

Answer 1

Answer:-

• Sequence of natural numbers between 100 and 500 which are divisible by 8 is 104 , 112 , 120...496.

If we assume that this sequence is in AP,

• a(first term) = 104

• d(Common difference) = 112 - 104 = 8

• nth term = 496

We know that,

nth term of an AP $\sf(a_n)$ = a + (n - 1)d

→ 104 + (n - 1)(8) = 496

→ 104 + 8n - 8 = 496

→ 8n = 496 - 104 + 8

→ 8n = 400

→ n = 400/8

→ n = 50

We know that,

$\sf \:Sum\:of\:first\:n\:terms\:of\:an\:AP(S_n) = \frac{n}{2}[2a + (n - 1)d]$

$\sf \implies \: S _{n} = \frac{50}{2} [2(104) + (50 - 1)(8)] \\ \\ \sf \implies \: S _{n} = 25(208 + 392) \\ \\ \sf\implies \: S_{n} = 25(600) \\ \\ \implies\sf \large \: S _{n} = 15000.$

Hence, the sum of all the natural numbers between 100 and 500 that are divisible by 8 is 15000.