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8. Sequences and Series
medium
Given that $n$ A.M.'s are inserted between two sets of numbers $a,\;2b$and $2a,\;b$, where $a,\;b \in R$. Suppose further that ${m^{th}}$ mean between these sets of numbers is same, then the ratio $a:b$ equals
A
$n - m + 1:m$
B
$n - m + 1:n$
C
$n:n - m + 1$
D
$m:n - m + 1$
Solution
(d) ${m^{th}}$ mean between $a,\;2b$ is $a + \frac{{m(2b – a)}}{{n + 1}}$ ……$(i)$
and ${m^{th}}$ mean between $2a,\;b$ is $2a + \frac{{m(b – 2a)}}{{n + 1}}$……$(ii)$
Accordingly, $a + \frac{{m(2b – a)}}{{n + 1}} = 2a + \frac{{m(b – 2a)}}{{n + 1}}$
$ \Rightarrow $ $m(2b – a) = a(n + 1) + m(b – 2a)$
$ \Rightarrow $ $a(n – m + 1) = bm$
$ \Rightarrow $ $\frac{a}{b} = \frac{m}{{n – m + 1}}$.
Standard 11
Mathematics
