How many Geometric Progressions exist containing 27,8 and 12 as three of its terms?

Question

Professor PC Thomas Classes · Answer

Let $27,8,12$ be $t_{m}, t_{n}, t_{p}$ of a G.P., whose first term is $a$ and common ratio $r$. Then $t_{m}=a r^{m-1}=27, t_{n}=a r^{n-1}=8$, $t_{p}=a r^{p-1}=12$.\ $	herefore r=\left(\frac{27}{8}ight)^{\frac{1}{m-n}}=\left(\frac{2}{3}ight)^{\frac{1}{n-p}}$\ $	herefore \quad \frac{3}{3 m-n} \cdot 3^{\frac{1}{n-p}}=\frac{1}{2^{n-p}} \cdot \frac{3}{2^{m-n}}$\ $	herefore \frac{\frac{3}{3^{m-n}+\frac{1}{n-p}}}{2^{\frac{1}{n-p}+\frac{3}{m-n}}}=1 \Rightarrow\left(\frac{3}{2}ight)^{\frac{3}{m-n}+\frac{1}{n-p}}=1$\ $	herefore \frac{2}{m-n}+\frac{1}{n-p}=0 \Rightarrow 2 n=3 p-m$\ $m=6 k, n=3 k, p=4 k, k \in N$.\ $	herefore \quad$ There exist infinite G.P's for which 27, 8,12 are three of its terms.\