Question

Three numbers form a GP .if the third term is decreased by 128 then the three number,thus obtained ,will form an AP .if the second term of this AP is decreased by 16 a GP will be formed again.Determine the numbers

Answer 1

Dear student
$$\begin{gathered} \mathrm{Let}\mathrm{the}\mathrm{three}\mathrm{numbers}\mathrm{in}\mathrm{GP}\mathrm{be}\frac{\mathrm{a}}{\mathrm{r}},\mathrm{a},\mathrm{ar} \\ \mathrm{If}\mathrm{third}\mathrm{term}\mathrm{is}\mathrm{decreased}\mathrm{by}128\mathrm{then}\mathrm{it}\mathrm{will}\mathrm{become}\mathrm{ar}-128 \\ \mathrm{So},\frac{\mathrm{a}}{\mathrm{r}},\mathrm{a},\mathrm{ar}-128\mathrm{is}\mathrm{in}\mathrm{A}.\mathrm{P} \\ 2\mathrm{a}=\frac{\mathrm{a}}{\mathrm{r}}+\mathrm{ar}-128 \\ 2\mathrm{ar}=\mathrm{a}+{\mathrm{ar}}^2-128\mathrm{r}....\left(1\right) \\ \mathrm{Now},\mathrm{second}\mathrm{term}\mathrm{is}\mathrm{decreased}\mathrm{by}16\mathrm{i}.\mathrm{e}.,\mathrm{a}-16 \\ \mathrm{So},\frac{\mathrm{a}}{\mathrm{r}},\mathrm{a}-16,\mathrm{ar}-128\mathrm{is}\mathrm{in}\mathrm{G}.\mathrm{P} \\ {\left(\mathrm{a}-16\right)}^2=\frac{\mathrm{a}}{\mathrm{r}}\left(\mathrm{ar}-128\right) \\ {\mathrm{a}}^2+256-32\mathrm{a}={\mathrm{a}}^2-\frac{128\mathrm{a}}{\mathrm{r}} \\ 256-32\mathrm{a}=-\frac{128\mathrm{a}}{\mathrm{r}} \\ 16-2\mathrm{a}=-\frac{8\mathrm{a}}{\mathrm{r}} \\ 16\mathrm{r}-2\mathrm{ar}+8\mathrm{a}=-0 \\ 8\mathrm{r}-\mathrm{ar}+4\mathrm{a}=0 \\ \mathrm{r}\left(8-\mathrm{a}\right)+4\mathrm{a}=0 \\ \mathrm{r}=\frac{4\mathrm{a}}{8-\mathrm{a}}.....\left(2\right) \\ \mathrm{Put}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{r}\mathrm{in}\mathrm{eq}\left(1\right),\mathrm{we}\mathrm{get} \\ 2\mathrm{ar}=\mathrm{a}+{\mathrm{ar}}^2-128\mathrm{r} \\ 2\mathrm{a}\left(\frac{4\mathrm{a}}{8-\mathrm{a}}\right)=\mathrm{a}+\mathrm{a}{\left(\frac{4\mathrm{a}}{8-\mathrm{a}}\right)}^2-128\left(\frac{4\mathrm{a}}{8-\mathrm{a}}\right) \\ 8{\mathrm{a}}^2\left(8-\mathrm{a}\right)=\mathrm{a}{\left(8-\mathrm{a}\right)}^2+16{\mathrm{a}}^3-512\mathrm{a}\left(8-\mathrm{a}\right) \\ 64{\mathrm{a}}^2-8{\mathrm{a}}^3=64\mathrm{a}+{\mathrm{a}}^3-16{\mathrm{a}}^2+16{\mathrm{a}}^3-4096\mathrm{a}+512{\mathrm{a}}^2 \\ 25{\mathrm{a}}^3+432{\mathrm{a}}^2-4032\mathrm{a}=0 \\ (\mathrm{a}+24)(25\mathrm{a}-168)=0 \\ \mathrm{a}=-24\mathrm{or}\mathrm{a}=\frac{168}{25} \\ \mathrm{so},\mathrm{when}\mathrm{a}=-24\mathrm{then}\mathrm{r}=\frac{4\left(-24\right)}{8+24}=-3 \\ \mathrm{When}\mathrm{a}=\frac{168}{25}\mathrm{then}\mathrm{r}=\frac{4\left({\displaystyle \frac{168}{25}}\right)}{8-{\displaystyle \frac{168}{25}}}=21 \\ \mathrm{So},\mathrm{number}\mathrm{are}8,-24,72\mathrm{or}\frac{8}{25},\frac{168}{25},\frac{3528}{25} \end{gathered}$$
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