find sum of all 3 digits no with remainder1 which is divisible by 4
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the no are =101,105,109 -------------997
a=101
d=4
for finding the value of 'n' the formula will be used
l=a+(n-1)d
997=101+(n-1)×4
896÷4=(n-1)
224+1=n
225=n
and for finding the sum the formula will used
Sn=n÷2(2a+(n-1)×d)
= 225÷2(2×101+(225-1)×4)
=225÷2(202+896)
=225÷2(1098)
=225×549
=123525 answer
a=101
d=4
for finding the value of 'n' the formula will be used
l=a+(n-1)d
997=101+(n-1)×4
896÷4=(n-1)
224+1=n
225=n
and for finding the sum the formula will used
Sn=n÷2(2a+(n-1)×d)
= 225÷2(2×101+(225-1)×4)
=225÷2(202+896)
=225÷2(1098)
=225×549
=123525 answer
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