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
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AP-III
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Morgoth
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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
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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imhimanshu
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tohellandback
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plugging numbers
assume the series {1} and {1,3}
here m=1,n=2
only option D gives you 1/3.
answer D
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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maihuna
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There is another way to look at it: Some of n odd numbers = n^2tohellandback wrote:plugging numbers
assume the series {1} and {1,3}
here m=1,n=2
only option D gives you 1/3.
answer D
so given ratio indicates the numbers are odd which can be represented using 2n-1
Charged up again to beat the beast 
