Question

The first three of four given numbers are in G.P. and their last three are in A.P. with common difference 6. If first and fourth numbers are equal, then the first number is
(a) 2
(b) 4
(c) 6
(d) 8

Answer 1

(d) 8

$$\begin{gathered} \mathrm{The}\mathrm{first}\mathrm{and}\mathrm{the}\mathrm{last}\mathrm{numbers}\mathrm{are}\mathrm{equal}. \\ \mathrm{Let}\mathrm{the}\mathrm{four}\mathrm{given}\mathrm{numbers}\mathrm{be}p,q,r\mathrm{and}p. \\ \mathrm{The}\mathrm{first}\mathrm{three}\mathrm{of}\mathrm{four}\mathrm{given}\mathrm{numbers}\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \therefore {\mathrm{q}}^2=\mathrm{p}·\mathrm{r}........\left(\mathrm{i}\right) \\ \mathrm{And},\mathrm{the}\mathrm{last}\mathrm{three}\mathrm{numbers}\mathrm{are}\mathrm{in}\mathrm{A}.\mathrm{P}.\mathrm{with}\mathrm{common}\mathrm{difference}6. \\ \mathrm{We}\mathrm{have}: \\ \mathrm{First}\mathrm{term}=\mathrm{q} \\ \mathrm{Second}\mathrm{term}=\mathrm{r}=\mathrm{q}+6 \\ \mathrm{Third}\mathrm{term}=\mathrm{p}=\mathrm{q}+12 \\ \mathrm{Also},2\mathrm{r}=\mathrm{q}+\mathrm{p} \\ \mathrm{Now},\mathrm{putting}\mathrm{the}\mathrm{values}\mathrm{of}\mathrm{p}\mathrm{and}\mathrm{r}\mathrm{in}\left(\mathrm{i}\right): \\ {\mathrm{q}}^2=\left(\mathrm{q}+12\right)\left(\mathrm{q}+6\right) \\ \Rightarrow {\mathrm{q}}^2={\mathrm{q}}^2+18\mathrm{q}+72 \\ \Rightarrow 18\mathrm{q}+72=0 \\ \Rightarrow \mathrm{q}+4=0 \\ \Rightarrow \mathrm{q}=-4 \\ \mathrm{Now},\mathrm{putting}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{q}\mathrm{in}\mathrm{p}=\mathrm{q}+12: \\ \mathrm{p}=-4+12=8 \end{gathered}$$