Q49 / 125 · BITSAT 2010 · Mathematics

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 \(m\).

  • Correct answer
    A
    14
  • B
    24
  • C
    10
  • D
    20

Solution

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 : \(\frac{\mathrm{x}_7}{\mathrm{x}_{\mathrm{m}-1}}=\frac{5}{9}\) \(\begin{aligned} & \therefore \frac{\mathrm{T}_8}{\mathrm{~T}_{\mathrm{m}}}=\frac{\mathrm{a}+7 \mathrm{~d}}{\mathrm{a}+(\mathrm{m}-1) \mathrm{d}}=\frac{5}{9} \\ & \Rightarrow 9 \mathrm{a}+63 \mathrm{~d}=5 \mathrm{a}+(5 \mathrm{~m}-5) \mathrm{d} \\ & \Rightarrow 4.1=(5 \mathrm{~m}-68) \frac{30}{\mathrm{~m}+1} \\ & \Rightarrow 2 \mathrm{~m}+2=75 \mathrm{~m}-1020 \Rightarrow 73 \mathrm{~m}=1022 \\ & \therefore \quad \mathrm{m}=\frac{1022}{73}=14 \end{aligned}\)

Hence, the correct answer is option: 'A'.

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