Question

Five numbers are in continued proportion whose second term is 6 and fourth term is 54. Find the numbers.

Answer 1

$$\begin{gathered} \mathrm{Let}the\mathrm{five}\mathrm{numbers}\mathrm{in}\mathrm{continued}\mathrm{proportion}bea,ak,a{k}^2,a{k}^{3}anda{k}^4. \\ Given: \\ ak=6\mathrm{and}a{k}^3=54 \\ Thus,wehave: \\ \frac{a{k}^3}{ak}=\frac{54}{6} \\ \Rightarrow k^2=9 \\ \Rightarrow k=3 \\ \mathrm{From}\mathrm{the}\mathrm{given}\mathrm{term},\mathrm{we}\mathrm{get}: \\ ak=6 \\ Now, \\ 3a=6 \\ \therefore a=2 \\ Also, \\ a{k}^2=2\times 9=18 \\ \Rightarrow a{k}^4=2\times 81=162 \\ \mathrm{Hence},\mathrm{the}\mathrm{numbers}\mathrm{are}2,6,18,54\mathrm{and}162. \end{gathered}$$