Question

Let $a_1,a_2,a_3.........$ are in A.P such that $a_p$, $a_q$,$a_r$ are in both A.P and GP then $a_q$:$a_p$ is equal to

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is C

Let the common ration of the G.P be $\alpha$ and $a_p$ = x

Then,

$a_q$ = x$\alpha$

$a_r$ = x${\alpha }^2$

We want to find $a_q$:$a_p$ or $\frac{a_q}{a_p}$

$\frac{a_q}{a_p}$ = $\frac{x\alpha }{x}$ = $\alpha$

We have to find $\alpha$ in terms of p,q and r,because the options are in terms of them.

$Since{a}_p$,$a_q$,$a_r$ in A.P,we can express them in terms of p,q and respectively.

$a_p$ = a + (p-1)d

aq = a + (q - 1)d

ar = a + (r - 1)d

It also has the common difference and the first term.

Now,if we can establish some relation between the $\alpha$ and $a_p$,$a_q$,$a_r$ without a and d,our job will be done.For that,

consider $\frac{{\alpha }_r-{\alpha }_q}{{\alpha }_q-{\alpha }_p}$

= $\frac{x{\alpha }^2-x\alpha }{x\alpha -x}$

= $\frac{\alpha (x\alpha -x)}{x\alpha -x}$

= $\alpha$ = $a_q$:$a_p$

So, $a_q$:$a_p$ = $\frac{{\alpha }_r-{\alpha }_q}{{\alpha }_q-{\alpha }_p}$

= $\frac{\alpha (r-1)d-(\alpha +(q-1\left)d\right)}{\alpha +(q-1)d-(\alpha +(p-1\left)d\right)}$

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

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