Find the sum of all three …
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Yogita Ingle 5 years, 3 months ago
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=n/2 [2a+(n−1)d]
=180/2 [2×102(180−1)5]
=90[204+179×5]
=90[204+895]
=90×1099
=98910
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