Question

find the sum of the first 100 even natural no. which is divisible by 5 ?

Answer 1

Hey MATE :

The first 100 natural even numbers divisible by 5 are:

10,20,30.....

We apply :

S = n/2[2a + (n-1)d]

S = 100/2 [2(10) + (100-1) 10]

S = 50[20 + 990]

S = 50 [1010]

S = 50500 Answer

Hope it helps

Hakuna Matata :))

Answer 2

first 100 even natural nos divisible by 5 = 10, 20,30 ,40 ,50,............


we can see an ap is being formed with
a= 10
d= 10

sum of
n= 100 terms

S= n/2(2a +(n-1)d)
= 50(20+ 99(10))
= 50 (20 +990)
=50 (1010)
= 50500




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