Find the sum of all natural 3 digit numbers divisible by 9
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TO FIND THE SUM OF ALL THREE DIGIT NUMBERS DIVISIBLE BY 9
a = 108
d = 9
last term = 999
last term = a + (n-1)d
999 = 108 + (n-1)9
891 = (n-1)9
==> n-1 = 99
==> n = 100
SUM OF 100 TERMS = 100/2 ( 108 + 999)
= 50 ( 1107)
==> 55350
a = 108
d = 9
last term = 999
last term = a + (n-1)d
999 = 108 + (n-1)9
891 = (n-1)9
==> n-1 = 99
==> n = 100
SUM OF 100 TERMS = 100/2 ( 108 + 999)
= 50 ( 1107)
==> 55350
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