Question

The sum of three numbers in G.P. is 21 and the sum of their squares is 189. Find the numbers.

Answer 1

Let the required numbers be $a\mathit{,}\mathit{}ar\mathrm{and}\mathit{}a{r}^{\mathit{2}}.$
Sum of the numbers = 21
$$\begin{gathered} \Rightarrow a+ar+a{r}^2=21 \\ \Rightarrow a(1+r+r^2)=21...\left(\mathrm{i}\right) \end{gathered}$$
Sum of the squares of the numbers = 189
$\Rightarrow a^2+(ar{)}^2+(a{r}^2{)}^2=189$
$$\begin{gathered} \Rightarrow a^2+(ar{)}^2+(a{r}^2{)}^2=189 \\ \Rightarrow a^2\left(1+r^2+r^4\right)=189...\left(\mathrm{ii}\right) \end{gathered}$$
$$\begin{gathered} \mathrm{Now},a(1+r+r^2)=21[\mathrm{From}(\mathrm{i}\left)\right] \\ \mathrm{Squaring}\mathrm{both}\mathrm{the}\mathrm{sides} \\ \Rightarrow a^2{\left(1+r+r^2\right)}^2=441 \\ \Rightarrow a^2\left(1+r^2+r^4\right)+2a^{2}r\left(1+r+r^2\right)=441 \\ \Rightarrow 189+2ar\left\{a\left(1+r+r^2\right)\right\}=441[\mathrm{Using}(\mathrm{ii}\left)\right] \\ \Rightarrow 189+2ar\times 21=441[\mathrm{Using}(\mathrm{i}\left)\right] \\ \Rightarrow ar=6 \\ \Rightarrow a=\frac{6}{\mathrm{r}}...\left(\mathrm{iii}\right) \\ \mathrm{Putting}a=\frac{6}{r}\mathrm{in}\left(\mathrm{i}\right) \\ \frac{6}{r}\left(1+r+r^2\right)=21 \\ \Rightarrow \frac{6}{r}+6+6r=21 \\ \Rightarrow 6r^2+6r+6=21r \\ \Rightarrow 6r^2-15r+6=0 \\ \Rightarrow 3(2r^2-5r+2)=0 \\ \Rightarrow 2r^2-5r+2=0 \\ \Rightarrow (2r-1)(r-2)=0 \\ \Rightarrow r=\frac{1}{2},2 \\ \mathrm{Putting}r=\frac{1}{2}\mathrm{in}a=\frac{6}{r},\mathrm{we}\mathrm{get}a=12. \\ \mathrm{So},\mathrm{the}\mathrm{numbers}\mathrm{are}12,6\mathrm{and}3. \\ \mathrm{Putting}\mathit{}r=2\mathrm{in}\mathit{}a=\frac{6}{r},\mathrm{we}\mathrm{get}\mathit{}a=3. \\ \mathrm{So},\mathrm{the}\mathrm{numbers}\mathrm{are}3,6\mathrm{and}12. \\ \mathrm{Hence},\mathrm{the}\mathrm{numbers}\mathrm{that}\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}\mathrm{are}3,6\mathrm{and}12. \end{gathered}$$