Question

find the sum of all3 digit numbers which are multiple of5

Answer 1

a = 100

d = 5

L = 995

=> a + (n-1)d = 995

=> 100 + (n-1)5 = 995

=> (n-1) 5 = 895

=> n - 1 = 179

=> n = 180

Now,

Sn = 180 /2 [ 100 + 995]

= 90 × 1095

= 98550

Answer 2

The first 3-digit number divisible by 5 is 100 and the last 3 digit number divisible by 5 is 995

So, you have to find the sum:- 100+105+110…………+995……………………………(equation 1 )

Clearly, 105–100=110–105=5

Therefore these numbers are in arithmetic progression where common difference is d=5

We know sum of the numbers in an A.P.(arithmetic progression)= n/2(2a+(n-1)d)

where n = number of terms in the A.P. = 180(995/5 - 100/5 = 199 - 20 = 179. But, 100 is also a three digit number divisible by 5. So n = 179 + 1 = 180)

a = First term of the A.P. = 100

d = common difference = 5

Therefore, required sum can be obtained by putting the required values in equation 1 = 180/2(2*100+(180–1)5) = 90(200+179(5)) = 90(200+895) = 90(1095)=98550

So, your answer is 98550

hope you find the answer convincing!!!