Let $a_{1}, a_{2}, a_{3}, . . . .$ be in AP and $a_{p}, a_{q}, a_{r}$ be in GP. Then $a_{q} : a_{p}$ is equal to-

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

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

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

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Solution

The correct option is C $\frac{r - q}{q - p}$
Let common difference be $d$
$a_{p} = a_{1} + (p - 1) d, a_{q} = a_{1} + (q - 1) d, a_{r} = a_{1} + (r - 1) d$
as $a_{p}, a_{q}, a_{r}$ are in G.P.
$\frac{a_{q}}{a_{p}} = \frac{a_{r}}{a_{q}} = \frac{a_{q} - a_{r}}{a_{p} - a_{q}}$ (by law of proportions)
or $\frac{a_{q}}{a_{p}} = \frac{a_{r}}{a_{q}} = \frac{a_{1} + (q - 1) d - a_{1} - (r - 1) d}{a_{1} + (p - 1) d - a_{1} - (q - 1) d} = \frac{q - r}{p - q}$ or $\frac{a_{q}}{a_{p}} = \frac{q - r}{p - q} = \frac{r - q}{q - p}$

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Let a1,a2,a3,.... be in AP and ap,aq,ar be in GP. Then aq:ap is equal to-

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Let $a_{1}, a_{2}, a_{3}, . . . .$ be in AP and $a_{p}, a_{q}, a_{r}$ be in GP. Then $a_{q} : a_{p}$ is equal to-