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Question

Given that n numbers of A.Ms are inserted between two sets of numbers a,2b and 2a,b where a,bR.
Suppose further that the mth means between these sets of numbers are same, then the ratio a:b equals

A
nm+1:n
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B
n:nm+1
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C
nm+1:m
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D
m:nm+1
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Solution

The correct option is D m:nm+1
anA.Ms2b
So total number of terms in the series is n+2
So common difference
d=2ban+1
Am=a+m(2ban+1)(1)
simillarly
2anA.Msbd=b2an+1
Am=2a+m(b2an+1)(2)
equating (1) & (2)
a=mn+1(b+a)ab=mnm+1

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