Sum of the integers from 1 to 1000 inclusive that are divisible by 25
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this question is related to AP.
first nos. divisible by 25 between 1 and 1000 is 25 to 1000
here a=25
last term is 1000
d=25
n we have to find out
using an=a+(n-1)d
1000=25+(n-1)25
975=(n-1)25
n-1=39
n=40
now using sn=2a+(n-1)d
=2(25)+(40-1)25
=50+(39)25
=50+975
=1025
this the required sum.
first nos. divisible by 25 between 1 and 1000 is 25 to 1000
here a=25
last term is 1000
d=25
n we have to find out
using an=a+(n-1)d
1000=25+(n-1)25
975=(n-1)25
n-1=39
n=40
now using sn=2a+(n-1)d
=2(25)+(40-1)25
=50+(39)25
=50+975
=1025
this the required sum.
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