Question

find the sum of all three digit no which leaves remainder 3 when divided by 5

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Answer 1

Answer:

99090

Step-by-step explanation:

The first three digit number which leaves remainder 3 when divided by 5 is 103 and the last number in 998. So, an Arithmetic progression can be formed in which the first number is 103, the common difference = 5.

SO, the number of terms in an AP = $(\frac{l-a}{d}+ 1)$

So, The number of terms = 180.

Sum of terms of an AP = $\frac{n}{2}(2a+(n-1)d$

So, Sum of terms = 99090