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(B) Let $A_{1}, A_{2}, \ldots, A_{m}$ be $m$ numbers such that $1, A_{1}, A_{2}, \ldots, A_{m}, 31$ is an $A.P.$Here,$a=1$,$b=31$,and the total number

Vedclass · Accepted Answer

$14$

Vedclass · Answer

(B) Let $A_{1}, A_{2}, \ldots, A_{m}$ be $m$ numbers such that $1, A_{1}, A_{2}, \ldots, A_{m}, 31$ is an $A.P.$Here,$a=1$,$b=31$,and the total number of terms is $n=m+2$.The common difference $d$ is given by $b = a + (n-1)d$,so $31 = 1 + (m+2-1)d$.$30 = (m+1)d \Rightarrow d = \frac{30}{m+1}$ (Equation $1$).The $k^{	ext{th}}$ inserted number is $A_{k} = a + kd = 1 + kd$.Given the ratio of the $7^{	ext{th}}$ and $(m-1)^{	ext{th}}$ inserted numbers is $5:9$:$\frac{1+7d}{1+(m-1)d} = \frac{5}{9}$.Substituting $d = \frac{30}{m+1}$:$\frac{1+7(\frac{30}{m+1})}{1+(m-1)(\frac{30}{m+1})} = \frac{5}{9}$.$\frac{m+1+210}{m+1+30m-30} = \frac{5}{9}$.$\frac{m+211}{31m-29} = \frac{5}{9}$.$9(m+211) = 5(31m-29)$.$9m + 1899 = 155m - 145$.$146m = 2044$.$m = \frac{2044}{146} = 14$.Thus,the value of $m$ is $14$.

Between $1$ and $31$,$m$ numbers have been inserted in such a way that the resulting sequence is an $A.P.$ and the ratio of the $7^{\text{th}}$ and $(m-1)^{\text{th}}$ inserted numbers is $5:9$. Find the value of $m$.

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