Question

Find the sum of all four digit numbers which leave remainder 2 when divided by 5​

Answer 1

Answer:

ANSWER

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. an=997

an=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

Sn=2n[2a+(n−1)d]

=2180[2×102(180−1)5]

=90[204+179×5]

=90[204+895]

=90×1099

=98910

Answer 2

Answer:

hope it helps you......