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

The correct option is B

$\frac{q-r}{p-q}$

(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 :

$p^{th}$ term, $a_p=a+(p-1)d$

$q^{th}$ term, $a_q=a+(q-1)d$

$r^{th}$ term, $a_r=a+(r-1)d$

Now, according to the question the $p^{th}$, the $q^{th}$

and the $r^{th}$ terms are in G.P.

$\therefore (a+(q-1)d{)}^2=(a+(p-1)d)\times (a+(r-1)d)$

$\Rightarrow a^2+2a(q-1)d+\left(\right(q-1)d{)}^2=a^2+ad(r-1+p-1)+(p-1\left)\right(r-1)d^2$

$\Rightarrow ad(2q-r-p+2)+d^2(q^2-2q+1-pr+p+r-1)=0$

$\Rightarrow a(2q-r-p)+d(q^2-2q-pr+p+r)$

$=0$ ( $\because$ d cannot be 0 )

$\Rightarrow a=-\frac{(q^2-2q-pr+p+r)d}{(2q-r-p)}$

$\therefore$ Common ratio, $r=\frac{a_p}{a_q}$

$=\frac{a+(q-1)d}{a+(p-1)d}=\frac{(q^2-2q-pr+p+r)d+(q-1)d}{\frac{(q^2-2q-pr+p+r)d}{p+r-2q}+(p-1)d}$

$=\frac{q^2-2q-pr+p+r+pq+rq-2q^2-p-r+2q}{q^2-2q-pr+p+r+p^2+prq-2p{q}^2-p-r+2q}$

$=\frac{p(q-r)-q(q-r)}{(p-q{)}^2}=\frac{(p-q)(q-r)}{(p-q{)}^2}=\frac{(q-r)}{(p-q)}$