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Find the sum of multiples of 3 less than 100. Hence or otherwise find the sum of the...
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The multiples of 3, less than 100, are 1*3=3, 2*3=6, 3*3=9, ........ 33*3=99.
There are clearly 33 such terms which are actually in an arithmetic progression (AP) with common difference =3.
The summation of such a progression may be obtained by the general formula: S = n/2 (a1 + an) where a1 is the first term, an, the last term and n is the total number of terms.
Therefore, the sum of multiples of 3, less than 100, is = 1683.
In order to find the sum of the numbers less than 100 which are not multiples of 3, one has to find the sum of all the numbers, less than 100, then subtract from it, the sum of the numbers less than 100, which are multiples of 3.
S = S2-S1
By the same arguments, S2 = sum of all the numbers, less than 100 = 99/2 (1+99) = 4950.
And S1 = sum of the numbers less than 100, which are multiples of 3 = 1683.
Therefore, the required summation, S = 4950 - 1683 = 3267
Therefore, the sum of the numbers less than 100 which are not multiples of 3, is = 3267.
Posted by llltkl on May 8, 2013 at 10:00 AM (Answer #2)
The numbers are
which is an A.P with first term `a_1=3, d=3 , a_n=99` ,where `a_n` is nth term of the A.P. ,d is common difference of A.P.
But we know
Thus sum of n terms of an AP is `S_n=(n/2)(2a_1+(n-1)d)`
Sum of n natural numbers is=`(n(n+1))/2`
Thus sum of first 99 ,natural number is
Sum of numbers non divisible by 3 and less than 100=`S-S_33`
Posted by pramodpandey on May 8, 2013 at 1:08 PM (Answer #3)
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