Question

find the sum of first 50natural numbers which are divisible by 5
​

Answer 1

Therefore, 6375 is the sum of first 50 positive integers which are divisible by 5.

Answer 2

Answer:

S₅₀ = 6375

Step-by-step explanation:

A.P → 5, 10, 15, 20,....

Given:

• First-term = a = 5
• Common difference = d = 10 - 5 = 5
• Number of terms = n = 50

To find:

Sum of first 50 natural numbers divisible by 5

Solution:

$\sf{S_n = \dfrac{n}{2}\big(2a + (n-1)d\big)}$

$\sf{S_n = \dfrac{50}{2}\big(2\times5+(50-1)5\big)}$

$\Rightarrow \sf{S_n = 25\big(10+49(5)\big)}$

$\Rightarrow \sf{S_n = 25(10+245)}$

$\Rightarrow \sf{S_n = 25(255)}$

$\Rightarrow \sf{S_n = 6375}$

$\therefore \boxed{\bf{Sum \: of \: first \: 50 \: natural \: numbers \: divisible \: by \: 5 = 6375}}$