If $p^{t h}, q^{t h}, r^{t h}$ and $s^{t h}$ terms of an $A . P$ are in $G . P$ , then show that $(p - q) (q - r) (r - s)$ are also in $G . P$ .

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Solution

Given,

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

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

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

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

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

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

consider,

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

subtracting 1 on both sides, we get,

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

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

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

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

now consider,

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

subtracting 1 on both sides, we get,

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

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

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

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

From (1)

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

From (1) and (2), we have,

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

Therefore $(p - q), (q - r), (r - s)$ are in G.P

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