Question

the 2nd , 3rd and 6th terms of an AP are consecutive terms of a GP. Find the common ratio of the GP.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{terms}\mathrm{in}\mathrm{AP}\mathrm{can}\mathrm{be}, \\ \mathrm{a}+\mathrm{d},\mathrm{a}+2\mathrm{d},\mathrm{a}+5\mathrm{d} \\ \mathrm{Now}\mathrm{these}\mathrm{are}\mathrm{in}\mathrm{GP}\mathrm{so}, \\ {\left(\mathrm{a}+2\mathrm{d}\right)}^2=\left(\mathrm{a}+5\mathrm{d}\right)\left(\mathrm{a}+\mathrm{d}\right) \\ {\mathrm{a}}^2+4{\mathrm{d}}^2+4\mathrm{ad}={\mathrm{a}}^2+6\mathrm{ad}+5{\mathrm{d}}^2 \\ -{\mathrm{d}}^2=2\mathrm{ad} \\ \mathrm{d}=-2\mathrm{a} \\ \mathrm{Now}\mathrm{common}\mathrm{ratio}=\frac{\mathrm{a}+2\mathrm{d}}{\mathrm{a}+\mathrm{d}}=\frac{\mathrm{a}-4\mathrm{a}}{\mathrm{a}-2\mathrm{a}}=\frac{3}{1}=3 \end{gathered}$$