Q162 / 270 · NDA 2016 (Phase 2) · Mathematics

How many geometric progressions is/are possible containing 27,8 and 12 as three of its/their terms?

  • A
    One
  • B
    Two
  • C
    Four
  • Correct answer
    D
    Infinitely many

Solution

Let ' $a$ ' be the first term \& ' $x$ ' be the common ratio. Also, suppose $27,8 \& 12$ be the $\mathrm{p}^{\text {th }}, \mathrm{q}^{\text {th }} \& \mathrm{r}^{\text {th }}$ term of the G.P. $\begin{aligned} \therefore \quad \mathrm{ax}^{\mathrm{p}-1} & =27 \\ \mathrm{ax}^{\mathrm{q}-1} & =8 \\ \& \mathrm{ax}^{\mathrm{r}-1} & =12 \end{aligned}$ Now, $27 \times 8^2=12^3$ $\begin{aligned} & \Rightarrow \quad \mathrm{ax}^{\mathrm{p}-1} \mathrm{x}\left(\mathrm{ax}^{\mathrm{q}-1}\right)^2=\left(\mathrm{ax}^{\mathrm{r}-1}\right)^3 \\ & \Rightarrow \quad \mathrm{x}^{\mathrm{p}-1} \cdot \mathrm{x}^{2 \mathrm{q}-2}=\mathrm{x}^{2 \mathrm{r}-3} \\ & \Rightarrow \mathrm{p}-1+2 \mathrm{q}-2=3 \mathrm{r}-3\end{aligned}$ $\Rightarrow p+2 q-3 r=0$ ...(1) There are infinitely many solutions for the eq. (1).

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

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