Question

If the pth ,qth and rth terms of an A.P. be in G P with the common ratio $k$, then the roots of the equation $(q-r)x^2+(r-p)x+(p-q)=0$ are

• A. $1$ and $k$
• B. $2$ and $k$
• C. $1$ and $\frac{1}{k}$
• D. None of these

Answer 1

The correct option is C $1$ and $\frac{1}{k}$

Let $a$ and $d$ be the first term and common difference of the A.P.
Then $pth$ term is given by $a+(p-1)d$
Similarly, $qth$ term is $a+(q-1)d$
and $rth$ term is $a+(r-1)d$
As given in the problem
$\frac{a+(q-1)d}{a+(p-1)d}=\frac{a+(r-1)d}{a+(q-1)d}=k$
On solving the above equation, we get $\frac{q-r}{p-q}=k-------\left(1\right)$
In the given equation , the sum of the coefficients is zero.
So, $1$ is clearly a root of the equation.
Product of the roots of the given equation is $\frac{p-q}{q-r}$
other root from $\left(1\right)$ is $\frac{p-q}{q-r}=\frac{1}{k}$