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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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user584223 | eNotes Newbie

Posted May 8, 2013 at 8:48 AM via web

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Find the sum of multiples of 3 less than 100. Hence or otherwise find the sum of the numbers less than 100 which are not multiples of 3.

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llltkl | College Teacher | (Level 3) Valedictorian

Posted May 8, 2013 at 10:00 AM (Answer #2)

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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.

= 33/2(3+99)

= 1683

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.

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pramodpandey | College Teacher | (Level 3) Valedictorian

Posted May 8, 2013 at 1:08 PM (Answer #3)

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The numbers are

3,6,9,.....,96,99

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

`a_n=a_1+(n-1)d`

`99=3+(n-1)3`

`n=33`

Thus sum of n terms of an AP is `S_n=(n/2)(2a_1+(n-1)d)`

`S_33=(33/2)(2xx3+(33-1)xx3)`

`=(33/2)(102)=1683`

Sum of n  natural numbers is=`(n(n+1))/2`

Thus sum of first 99 ,natural number is

`S=(99xx(99+1))/2` =`4950`

Sum of numbers non divisible by 3 and less than 100=`S-S_33`

`=3267`

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