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AP-III

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by maihuna » Sun Aug 16, 2009 3:25 am
If the sum of m terms of an AP is to the sum of n terms of another AP as m^2 to n^2, what will be the mth term to nth terms.

m/n
1/2
2m+1/2n+3
2m-1/2n-1
m-1/n-1
Last edited by maihuna on Tue Aug 18, 2009 6:31 am, edited 1 time in total.
Charged up again to beat the beast :)
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Re: AP-III

by Morgoth » Sun Aug 16, 2009 3:56 am
IMO D


sum of m terms = m/2 [2a + (m-1)d]
sum of n terms = n/2 [2a + (n-1)d]

m/2 [2a + (m-1)d] / n/2 [2a + (n-1)d] = m^2 / n^2

you solve this to get 2a = d


mth term = a + (m-1)d

nth term = a + (n-1)d

substitute d=2a

a+ (m-1)2a / a+ (n-1)2a = (a + 2am - 2a )/ (a + 2an - 2a)

= 2m-1 / 2n-1

Hence D
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by imhimanshu » Mon Aug 17, 2009 7:00 am
@ Morgoth..How come you assumed that both the AP's has first term "a"
Pls explain
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by tohellandback » Mon Aug 17, 2009 11:59 pm
plugging numbers
assume the series {1} and {1,3}
here m=1,n=2
only option D gives you 1/3.
answer D
The powers of two are bloody impolite!!
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by maihuna » Tue Aug 18, 2009 6:30 am
tohellandback wrote:plugging numbers
assume the series {1} and {1,3}
here m=1,n=2
only option D gives you 1/3.
answer D
There is another way to look at it: Some of n odd numbers = n^2

so given ratio indicates the numbers are odd which can be represented using 2n-1
Charged up again to beat the beast :)
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