Find the sum of the first 45 numbers which three digit numbers divisible by 6
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Three digit number divisible by 6
102,108................996
a=102. d=6. n=45.
Sn = n/2[2a+(n-1)d]
= 45/2[2*102+(45-1)6]
= 45/2 (234)
= 10530
102,108................996
a=102. d=6. n=45.
Sn = n/2[2a+(n-1)d]
= 45/2[2*102+(45-1)6]
= 45/2 (234)
= 10530
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