find the sum of all 3 digit numbers which leaves the remainder 3 when divided by 5
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No. of 3 digits are,
103,108,.........,998
a=103
d=108_103=5
An=998
a+(n_1)×d=998
or,103+(n_1)×5=998
or,(n_1)×5=998_103=895
or,n_1=895/5=179
n=180
so,Sn=n/2(a+An)
S180=180/2(103+998)
=90×1101=10268ans.
103,108,.........,998
a=103
d=108_103=5
An=998
a+(n_1)×d=998
or,103+(n_1)×5=998
or,(n_1)×5=998_103=895
or,n_1=895/5=179
n=180
so,Sn=n/2(a+An)
S180=180/2(103+998)
=90×1101=10268ans.
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