Question

The sum of three numbers in G.P. is 14. If the first two terms are each increased by 1 and the third term decreased by 1, the resulting numbers are in A.P. Find the numbers.

Answer 1

Let the numbers be a, ar and ar².
$$\begin{gathered} \mathrm{Sum}=14 \\ \Rightarrow a+ar+a{r}^2=14 \\ \Rightarrow a(1+r+r^2)=14...\left(\mathrm{i}\right) \end{gathered}$$
According to the question, a + 1, ar + 1 and ar² − 1 are in A.P.
$$\begin{gathered} \therefore 2\left(ar+1\right)=a+1+a{r}^2-1 \\ \Rightarrow 2ar+2=a+a{r}^2 \\ \Rightarrow 2ar+2=14-ar\left[\mathrm{From}\left(\mathrm{i}\right)\right] \\ \Rightarrow 3ar=12 \\ \Rightarrow a=\frac{4}{r}...\left(\mathrm{ii}\right) \\ \mathrm{Putting}a=\frac{4}{r}\mathrm{in}\left(\mathrm{i}\right) \\ \Rightarrow \frac{4}{r}(1+r+r^2)=14 \\ \Rightarrow 4r^2-10r+4=0 \\ \Rightarrow 4r^2-8r-2r+4=0 \\ \Rightarrow \left(4r-2\right)\left(r-2\right)=0 \\ \Rightarrow r=\frac{1}{2},2 \end{gathered}$$
$$\begin{gathered} \mathrm{Putting}r=\frac{1}{2}\mathrm{in}\left(\mathrm{ii}\right),\mathrm{we}\mathrm{get}a=8. \\ S\mathrm{o},\mathrm{the}\mathrm{G}.\mathrm{P}.\mathrm{is}8,4\mathrm{and}2. \end{gathered}$$

Similarly putting r = 2 in (ii), we get a = 2.
So, the G.P is 2, 4 and 8.