Question

find the sum of all natural numbers which are divisible by 5 between 91 and 180​

Answer 1

Solution

Let a be the first term and d be the common difference of the above AP

The AP would be :

95,100,...............,175

Here,

$\sf \: a = 95 \: and \: d = 5$

Firstly,we would need to find the number of terms

$\sf \: l = a + (n - 1)d \\ \\ \colon \implies \: \sf \: 175 = 95 + (n - 1)5 \\ \\ \colon \implies \: \sf \: 5n - 5 = 80 \\ \\ \colon \implies \: \sf \: 5n = 85 \\ \\ \colon \implies \: \boxed{ \boxed{ \sf \: n = 17}}$

Sum Of The Terms

$\sf \: s = \dfrac{n}{2} \times (a + l) \\ \\ \longrightarrow \: \sf \: s = \dfrac{17}{2} (95 + 175) \\ \\ \longrightarrow \: \sf \: s = \dfrac{17}{2} \times 270 \\ \\ \longrightarrow \: \boxed{ \boxed{ \sf \: s = 2295}}$

Sum of the terms divisible by 5,between 91 and 180 is 2295