Find the sum of first 100 natural numbers divisible by 5 answer 50500
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first term is 5
& common difference is 5
then sum of 100th terms is=
n/2(2a+(n-1)d)
=100/2(2*5+99*5)
=50(10+495)
=50*505
=25250.........(and)
& common difference is 5
then sum of 100th terms is=
n/2(2a+(n-1)d)
=100/2(2*5+99*5)
=50(10+495)
=50*505
=25250.........(and)
5U8H0J1T:
may be i am wrong...
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