Question

If pth, qth, rth term of an A.P. are in G.P., prove that common ratio of G.P is $\frac{q-r}{p-q}$

Answer 1

In AP, first term be a

Difference be d

$\therefore p^{th}$ term $\Rightarrow a+(p-1)d$

$q^{th}$ term $a+(q-1)d$

$r^{th}$ term $a+(r-1)d$

since they are in GP

and let first term in G.P be A

and common ration be r

$\therefore$ As given

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

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

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

subtracting (1) from (2)

$\left[\right(q-1)-(p-1\left)\right]d=A[r-1]$

$(q-p)d=A(r-1)...\left(4\right)$

subtracting (2) form (3)

$(r-q)d=Ar(r-1)...\left(5\right)$

Dividing (5), we get

$\frac{Ar(r-1)}{A(r-1)}=\frac{(r-q)d}{(q-p)d}$

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

Hence proved.