Question

Five numbers are in continued proportion where the first term is 1 and the last term is 256. Find the numbers.

Answer 1

$$\begin{gathered} \mathrm{Let}the\mathrm{five}\mathrm{numbers}\mathrm{in}\mathrm{continued}\mathrm{proportion}bea,ak,a{k}^2,a{k}^{3}anda{k}^4. \\ Given: \\ a=1\mathrm{and}a{k}^4=256 \\ Thus,wehave: \\ \frac{a{k}^4}{a}=\frac{256}{1} \\ \Rightarrow k^4=256 \\ \Rightarrow k=4 \\ And, \\ ak=4 \\ \therefore a{k}^2=1\times 16=16 \\ \Rightarrow a{k}^3=1\times 64=64 \\ \mathrm{Thus},\mathrm{the}\mathrm{numbers}\mathrm{are}1,4,16,64\mathrm{and}256. \end{gathered}$$