If (m) arithmetic means are inserted between 1 and 31 so that the...

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}) $\begin{aligned}& \therefore \frac{\mathrm{T}_8}{\mathrm{ T}_{\mathrm{m}}}=\frac{\mathrm{a}+7 \mathrm{ d}}{\mathrm{a}+(\mathrm{m}-1) \mathrm{d}}=\frac{5}{9}\\ & \Rightarrow 9 \mathrm{a}+63 \mathrm{ d}=5 \mathrm{a}+(5 \mathrm{ m}-5) \mathrm{d}\\ & \Rightarrow 4.1=(5 \mathrm{ m}-68) \frac{30}{\mathrm{ m}+1}\\ & \Rightarrow 2 \mathrm{ m}+2=75 \mathrm{ m}-1020 \Rightarrow 73 \mathrm{ m}=1022 \\ & \therefore \quad \mathrm{m}=\frac{1022}{73}=14 \end{aligned}$)