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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none of these

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Solution

The correct option is D $\frac{r - q}{q - p}$
Given $a_{1}, a_{2},, a_{3}, . . . . . .$ are in A.P
And $a_{p}, a_{q}, a_{r}$ are in G.P
Also $a_{q} = a_{1} + (q - 1) d a_{p} = a_{1} + (p - 1) d a_{r} = a_{1} + (r - 1) d$ , where $d$ is the common difference of A.P
Let's subtract the terms to get :
$a_{r} - a_{q} = (r - q) d$ and $a_{q} - a_{p} = (q - p) d$
On dividing, we get $\frac{a_{r} - a_{q}}{a_{q} - a_{p}} = \frac{(r - q)}{(q - p)}$
But we know that
$a_{q} = a_{p} \times R a_{r} = a_{p} \times R^{2} = a_{q} \times R$
Where $R$ is the common ratio of the G.P
So putting the values we get
$\frac{a_{q} \times R - a_{q}}{a_{p} \times R - a_{p}} = \frac{(r - q)}{(q - p)}$
$\Rightarrow \frac{a_{q} (R - 1)}{a_{P} (R - 1)} = \frac{(r - q)}{(q - p)}$
$\Rightarrow \frac{a_{q}}{a_{P}} = \frac{(r - q)}{(q - p)}$

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