Question

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

Answer 1

The $p^{th}$ term will be,

$a+\left(p-1\right)d$

The $q^{th}$ term will be,

$a+\left(q-1\right)d$

The $r^{th}$ term will be,

$a+\left(r-1\right)d$

The $s^{th}$ term will be,

$a+\left(s-1\right)d$

Since these terms are given in G.P., so let $x$ be the first term and $y$ be the common ratio.

$a+\left(p-1\right)d=x$ (1)

$a+\left(q-1\right)d=xy$ (2)

$a+\left(r-1\right)d=x{y}^2$ (3)

$a+\left(s-1\right)d=x{y}^3$ (4)

Subtracting equation (2) from (1),

$\left(q-p\right)d=x\left(y-1\right)$ (5)

Subtracting (3) from (2),

$\left(r-q\right)d=xy\left(y-1\right)$ (6)

Subtracting (4) from (3),

$\left(s-r\right)d=x{y}^2\left(y-1\right)$ (7)

Dividing equation (6) by (5),

$\frac{r-q}{q-p}=y$ (8)

Dividing equation (7) by (6),

$\frac{s-r}{r-q}=y$ (9)

From equation (8) and (9),

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

Or,

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

This shows, that $\left(p-q\right)$, $\left(q-r\right)$ and $\left(r-s\right)$ are in G.P.