Question
The sum of all the three digit numbers which when divided by 8 give a remainder of 5 is___
Answer:
:
For this condition,
Series is 109,117,125,..........997
a= 109 ,d=8 , n=112
Sn=n2[2a+(n−1)d]
=1122[2×109+(112−1)8]
=61936.
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:
For this condition,
Series is 109,117,125,..........997
a= 109 ,d=8 , n=112
Sn=n2[2a+(n−1)d]
=1122[2×109+(112−1)8]
=61936.
Was this answer helpful ?
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