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ramu31
@ramu31
September 2021
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find the sum of all three digit numbers which leave the remainder 3 when divided by5
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kunal0912
AP for these numbers is, 103,108,113,.................,998
a= 103, d = 5
Let there are n terms in this AP so an = 998
a+(n-1)d = 998
103 + (n-1)5 = 998
(n-1) = 179
so, n = 180
So, Sn = S180 = 180/2[a+l]
= 90[103+998]
= 90*1101
= 99090
3 votes
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Answers & Comments
a= 103, d = 5
Let there are n terms in this AP so an = 998
a+(n-1)d = 998
103 + (n-1)5 = 998
(n-1) = 179
so, n = 180
So, Sn = S180 = 180/2[a+l]
= 90[103+998]
= 90*1101
= 99090