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Between $1\:\:and\:\:31$ $m$ numbers are inserted so that the resulting sequence is an $A.P.$ and the ratio of $7^{th}\:\;and\:\:(m-1)^{th}$ numbers is $5:9$. find the value of $m$.

$\begin{array}{1 1}13 \\ 12 \\ 14 \\ 15 \end{array} $

1 Answer

  • $r^{th}A.M.$ between $a\:\;and\:\:b=A_r=a+rd$ where $d=\large\frac{b-a}{n+1}$$\:\:and\:\:n$ is no. of $A.M.^s$ inserted.
Given that $m\:\:A.M.^s$ are inserted between $1\:\:and\:\:31$
$\Rightarrow\:1,A_1,A_2,A_3,....A_m,31$ are in $A.P.$
Also given that $A_7:A_{m-1}=5:9$
We know that $A_r=a+rd$, $a=1,\:b=31$
and $d=\large\frac{b-a}{m+1}$
$A_7=a+7d$ and $A_{m-1}=a+(m-1)d$
and $d=\large\frac{31-1}{m+1}=\frac{30}{m+1}$
$\Rightarrow\:73m=1022$ or $m=14$
answered Feb 24, 2014 by rvidyagovindarajan_1

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