If $\mathrm{m}$ arithmetic means are inserted between 1 and 31 so that the ratio of the $7^{\text {th }}$ and $(\mathrm{m}-1)^{\text {th }}$ means is $5: 9$, then find the value of $m$.

Question

Aicharya · Accepted Answer

Let the means be $x_1, x_2, \ldots \ldots x_m$ so that $1, \mathrm{x}_1, \mathrm{x}_2, \ldots . \mathrm{x}_{\mathrm{m}}, 31$ is an A.P. of $(\mathrm{m}+2)$ terms. Now, $31=\mathrm{T}_{\mathrm{m}+2}=\mathrm{a}+(\mathrm{m}+1) \mathrm{d}=1+(\mathrm{m}+1) \mathrm{d}$ $\therefore \mathrm{d}=\frac{30}{\mathrm{~m}+1}$ Given : $\frac{\mathrm{x}_7}{\mathrm{x}_{\mathrm{m}-1}}=\frac{5}{9}$ $\begin{aligned} & \therefore \frac{T_8}{T_{m}}=\frac{a+7d}{a+(m-1)d}=\frac{5}{9} \ & \Rightarrow 9a+63d=5a+(5m-5)d \ & \Rightarrow 4.1=(5m-68) \frac{30}{m+1} \ & \Rightarrow 2m+2=75m-1020 \Rightarrow 73m=1022 \ & \therefore m=\frac{1022}{73}=14 \end{aligned}$

If \(\mathrm{m}\) arithmetic means are inserted between 1 and 31 so that the ratio of the \(7^{\text {th…

Solution