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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 $m$.

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

Toolbox:
• $r^{th}A.M.$ between $a\:\;and\:\:b=A_r=a+rd$ where $d=\large\frac{b-a}{n+1}$$\:\:and\:\:n is no. of A.M.^s inserted. Given that m\:\:A.M.^s are inserted between 1\:\:and\:\:31 \Rightarrow\:1,A_1,A_2,A_3,....A_m,31 are in A.P. Also given that A_7:A_{m-1}=5:9 We know that A_r=a+rd, a=1,\:b=31 and d=\large\frac{b-a}{m+1} A_7=a+7d and A_{m-1}=a+(m-1)d \Rightarrow\:\large\frac{A_7}{A_{m-1}}=\frac{a+7d}{a+(m-1)d}=\frac{5}{9} and d=\large\frac{31-1}{m+1}=\frac{30}{m+1} \Rightarrow\:\large\frac{1+7d}{1+(m-1)d}=\frac{5}{9} \Rightarrow\:9+63d=5+5(m-1)d \Rightarrow\:d(68-5m)=-4 \large\frac{30}{m+1}$$.(68-5m)=-4$
$\Rightarrow\:30(68-5m)=-4(m+1)$
$\Rightarrow\:15(68-5m)=-2m-2$
$\Rightarrow\:73m=1022$ or $m=14$