Question

If pth, qth and rth terms of an A.P. are in G.P., then the common ratio of this G.P. is

(a) $\frac{p-q}{q-r}$

(b) $\frac{q-r}{p-q}$

(c) pqr

(d) none of these

Answer 1

(b) $\frac{q-r}{p-q}$

Let a be the first term and d be the common difference of the given A.P.
Then, we have:
$$\begin{gathered} {\mathrm{p}}^{\mathrm{th}}\mathrm{term},a_p=a+\left(p-1\right)d \\ {q}^{th}\mathrm{term},a_q=a+\left(q-1\right)d \\ {r}^{th}\mathrm{term},a_r=a+\left(r-1\right)d \\ \mathrm{Now},\mathrm{according}\mathrm{to}\mathrm{the}\mathrm{question}\mathrm{the}{p}^{\mathrm{th}},\mathrm{the}{q}^{\mathrm{th}}\mathrm{and}\mathrm{the}{r}^{\mathrm{th}}\mathrm{terms}\mathrm{are}\mathrm{in}\mathrm{G}.\mathrm{P}. \\ \therefore {\left(a+\left(q-1\right)d\right)}^2=\left(a+\left(p-1\right)d\right)\times \left(a+\left(r-1\right)d\right) \\ \Rightarrow a^2+2a\left(q-1\right)d+{\left(\left(q-1\right)d\right)}^2=a^2+ad\left(r-1+p-1\right)+\left(p-1\right)\left(r-1\right)d^2 \\ \Rightarrow ad\left(2q-2-r-p+2\right)+d^2\left(q^2-2q+1-pr+p+r-1\right)=0 \\ \Rightarrow a\left(2q-r-p\right)+d\left(q^2-2q-pr+p+r\right)=0\left(\because d\mathrm{cannot}\mathrm{be}0\right) \\ \Rightarrow a=-\frac{\left(q^2-2q-pr+p+r\right)d}{\left(2q-r-p\right)} \\ \therefore \mathrm{Common}\mathrm{ratio},r=\frac{a_q}{a_p} \\ =\frac{a+\left(q-1\right)d}{a+\left(p-1\right)d} \\ =\frac{\frac{\left(q^2-2q-pr+p+r\right)d}{\left(p+r-2q\right)}+\left(q-1\right)d}{\frac{\left(q^2-2q-pr+p+r\right)d}{\left(p+r-2q\right)}+\left(p-1\right)d} \\ =\frac{q^2-2q-pr+p+r+pq+rq-2q^2-p-r+2q}{q^2-2q-pr+p+r+p^2+pr-2pq-p-r+2q} \\ =\frac{pq-pr-q^2+qr}{p^2+q^2-2pq} \\ =\frac{p\left(q-r\right)-q\left(q-r\right)}{{\left(p-q\right)}^2} \\ =\frac{\left(p-q\right)\left(q-r\right)}{{\left(p-q\right)}^2} \\ =\frac{\left(q-r\right)}{\left(p-q\right)} \end{gathered}$$