Question

Find the sum of all 3 -digit numbers which leave the remainder 2 when divided by 5

Answer 1

clearly, this will be an arithmetic progression with first term 102 and last term 997,(known by hit and trial method) with common difference 5, hence we can apply formulae for sum to n terms of an AP

Hoping it helped, please feel free to ask if there's any doubt

Answer 2

Heya !!!

This is Your Answer

We have to find the sum of all three digit numbers which leaves remainder 2 when divided by 5.

We can see the it forms AP ......
as...
102, 107, 112, ......., 992, 997.
Here..
a = 102.
l or an = 997, and
d = 5.

Using,
an = a + (n-1)d
997 = 102 + (n - 1)5
895 = 5 (n-1)
n -1 = 179
n = 180.

Now, Using 
[tex]S_{n} = \frac{n}{2} (a+l) \\ S_{180} = \frac{180}{2} (102+997) \\ S_{180} = 90 \times 1099 \\ S_{180} =98910. [/tex]

Hence, the required answer is 98910.

Hope It helps