If the $p t h$ , $q t h$ , $r t h$ and $s t h$ terms of an $A . P .$ are in $G . P .$ , then $p - q, q - r, r - s$ are in

A . P .

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G . P .

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H . P .

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None of these

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Solution

The correct option is B $G . P .$
$p^{t h}, q^{t h}, r^{t h}$ and $s^{t h}$ term are of $A P$ is in $G P$
So, $\frac{a_{q}}{a_{p}} = \frac{a_{r}}{a_{q}} = \frac{a_{s}}{a_{r}} =$ common difference
$a_{p} = a + (p - 1) d - (i) a_{q} = a + (q - 1) d - (i i) a_{r} = a + (r - 1) d - (i i i) a_{s} = a + (s - 1) d - (i v)$
Taking , $\frac{a_{q}}{a_{p}} = \frac{a_{r}}{a_{q}}$
Subtracting $1$ from both sides.
$\frac{a_{q} - a_{p}}{a_{p}} = \frac{a_{r} - a_{q}}{a_{q}} \frac{a_{q}}{a_{p}} = \frac{a_{r} - a_{q}}{a_{q} - a_{p}}$
Putting value of $a_{r}, a_{q}$ and $a_{p}$ from $(i), (i i)$ and $(i i i)$ in $R . H . S$
$\frac{a_{q}}{a_{p}} = \frac{a + (r - 1) d - a - (q - 1) d}{a + (q - 1) d - a - (p - 1) d} = \frac{r d - a - q d + d}{q d - d - p d + d} \frac{a_{q}}{a_{p}} = \frac{d (r - q)}{d (q - p)}$
Similarly,
$\frac{a_{r}}{a_{q}} = \frac{(s - r)}{(r - q)} A s, \frac{a_{q}}{a_{p}} = \frac{a_{r}}{a_{q}} = \frac{r - q}{q - p} = \frac{s - r}{r - q} S o, (p - q), (q - r), (r - s)$
$S o, \frac{q - r}{p - q} = \frac{r - s}{q - r}$ (Hence proved)

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