If pth, qth, rth and sth terms of A.P. are in G.P. then show that p-q, q-r, r-s are in G.P.

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Solution

Given,

$p t h, q t h, r t h, s t h$ terms are in AP

$a_{p} = a + (p - 1) d$

$a_{q} = a + (q - 1) d$

$a_{r} = a + (r - 1) d$

$a_{s} = a + (s - 1) d$

$a_{p}, a_{q}, a_{r}, a_{s}$ are in G.P.

$\frac{a_{q}}{a_{p}} = \frac{a_{r}}{a_{q}} = \frac{a_{s}}{a_{r}}$

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$\frac{a_{q}}{a_{p}} = \frac{a_{r}}{a_{q}}$

$\frac{a_{q}}{a_{p}} - 1 = \frac{a_{r}}{a_{q}} - 1$

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

$\frac{a + (q - 1) d - [a + (p - 1) d]}{a + (r - 1) d - [a + (q - 1) d]} = \frac{a + (p - 1) d}{a + (q - 1) d}$

$\frac{q - r}{p - q} = \frac{a_{p}}{a_{q}}$ ..........(1)

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$\frac{a_{r}}{a_{q}} = \frac{a_{s}}{a_{r}}$

$\frac{a_{r}}{a_{q}} - 1 = \frac{a_{s}}{a_{r}} - 1$

$\frac{a_{r} - a_{s}}{a_{q} - a_{r}} = \frac{a_{r}}{a_{q}}$

$\frac{a + (r - 1) d - [a + (s - 1) d]}{a + (q - 1) d - [a + (r - 1) d]} = \frac{a + (r - 1) d}{a + (q - 1) d}$

$\frac{r - s}{q - r} = \frac{a_{q}}{a_{p}}$ ..........(2)

From (1) and (2)

$\frac{a_{p}}{a_{q}} = \frac{a_{q}}{a_{p}}$

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

Hence $p - q, q - r, r - s$ are in GP

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