Question Number 186805 by pascal889 last updated on 10/Feb/23
Answered by Rasheed.Sindhi last updated on 10/Feb/23
$${r}=\frac{{a}_{\mathrm{4}} }{{a}_{\mathrm{10}} }=\frac{{a}_{\mathrm{1}} }{{a}_{\mathrm{4}} } \\ $$$${r}=\frac{{a}+\mathrm{3}{d}}{{a}+\mathrm{9}{d}}=\frac{{a}}{{a}+\mathrm{3}{d}} \\ $$$${r}−\mathrm{1}=\frac{{a}+\mathrm{3}{d}}{{a}+\mathrm{9}{d}}−\mathrm{1}=\frac{{a}}{{a}+\mathrm{3}{d}}−\mathrm{1} \\ $$$${r}−\mathrm{1}=\frac{−\mathrm{6}{d}}{{a}+\mathrm{9}{d}}=\frac{−\mathrm{3}{d}}{{a}+\mathrm{3}{d}} \\ $$$$\frac{{r}−\mathrm{1}}{−\mathrm{3}{d}}=\frac{\mathrm{2}}{{a}+\mathrm{9}{d}}=\frac{\mathrm{1}}{{a}+\mathrm{3}{d}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{a}+\mathrm{9}{d}=\mathrm{2}{a}+\mathrm{6}{d} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:{a}=\mathrm{3}{d} \\ $$$$\frac{{r}−\mathrm{1}}{−\mathrm{3}{d}}=\frac{\mathrm{1}}{{a}+\mathrm{3}{d}}\Rightarrow\frac{{r}−\mathrm{1}}{−{a}}=\frac{\mathrm{1}}{\mathrm{2}{a}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{1}−{r}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{r}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${first}\:{term}\:{of}\:{gp}:\:\mathrm{4} \\ $$$${common}\:{ratio}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${S}_{\mathrm{6}} =\mathrm{4}\left(\frac{\mathrm{1}−\left(\mathrm{1}/\mathrm{2}\right)^{\mathrm{6}} }{\mathrm{1}−\mathrm{1}/\mathrm{2}}\right)=\mathrm{4}\left(\frac{\mathrm{63}/\mathrm{64}}{\mathrm{1}/\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\mathrm{4}\left(\frac{\frac{\mathrm{63}}{\mathrm{64}}}{\frac{\mathrm{1}}{\mathrm{2}}}\right)=\mathrm{4}×\frac{\mathrm{63}}{\mathrm{64}}×\mathrm{2}=\frac{\mathrm{63}}{\mathrm{8}}=\mathrm{7}\frac{\mathrm{7}}{\mathrm{8}} \\ $$