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by scoobydooby » Sun May 10, 2009 12:09 pm
multiples of 5 between 100 and 1000
=105, 110, 115,.........,995
=(5+100), (10+100), (15+100),.......,(895+100)
sum of all the multiples above=(5+10+15...+895)+100n
=5(1+2+3.......179)+100n (n=179, sum of 1 to n=n*(n+1)/2)
=5*179.180/2+100*179
=450*179+100*179
=179*550
=98450

hence, B
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by sureshbala » Sun May 10, 2009 8:48 pm
Multiples of 5 between100 and 1000 are

105, 110, .......995

i.e. 5 x 21, 5 x 22, ............., 5 x 199

So the sum is

5(21+22+.....+199)

=5[179/2(21+199)] (Sum of n terms in AP = n/2(first term+last term))
= 5 x 179 x 110 = 895 x 110 = 98450
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by Pranay » Mon May 11, 2009 1:19 am
sureshbala wrote:Multiples of 5 between100 and 1000 are

105, 110, .......995

i.e. 5 x 21, 5 x 22, ............., 5 x 199

So the sum is

5(21+22+.....+199)

=5[179/2(21+199)] (Sum of n terms in AP = n/2(first term+last term))
= 5 x 179 x 110 = 895 x 110 = 98450
Hi Suresh,

Nice explaination.

Instead,

105, 110, 115, ........, 995 are also in AP with d=5 and n = 178

Putting this in sum to n-terms formula => (178/2)[105+995] doesnot work out ???

Pranay
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