Question

Find the sum of all 3 digit numbers which leaves the same remainder 2 when divided by 5

Answer 1

Answer:98910

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Answer 2

Answer:

98910

Step-by-step explanation:

Three digit number which leaves remainder 2 when divided by 5 is

102,107,112,117......997

This form an AP whose first term is =102, d=5

Let 997 is the n th term of AP i.e. a(n)=997

a(n)=a+(n−1)d

997=102+(n−1)5

5n=900

n=180

Sum of all three digits which leaves remainder 2 when divided by 5 is

S (n)=n/2 [2a+(n−1)d]

= 180/2 [2×102(180−1)5]

=90[204+179×5]

=90[204+895]

=90×1099

=98910