Question

If there exists a geometric progression containing 27, 8 and 12 as three of its terms (not necessarily consecutive) then no. of progressions possible are

• A. $1$
• B. $2$
• C. infinite
• D. None of these

Answer 1

The correct option is C infinite

$8,12,29$

Let the first term be $8$

$\therefore a=8$ ...(i)

Let the $p^{th}$ ther be $12$

$\therefore a{r}^{p-1}=12$ ...(ii)

and $q^{th}$ term be $27$

$\therefore a{r}^{q-1}=27$ ...(iii)

(i) / (ii)

$\frac{a}{a{r}^{p-1}}=\frac{8}{12}$

$\Rightarrow r^{p-1}=\frac{3}{2}=1.5$ ...(iv)

(iii) / (i)

$r^{q-1}=\frac{27}{8}={\left(\frac{3}{2}\right)}^3$

$r^{q-1}=(1.5{)}^3$ ...(v)

(v) / (iv)

$\Rightarrow \frac{r^{q-1}}{r^{p-1}}=\frac{(1.5{)}^3}{1.5}$

$\Rightarrow r^{q-1-p+1}=(1.5{)}^2$

$\Rightarrow r^{q-p}=(1.5{)}^2$

$=(r^{p-1}{)}^2$ ...from (iv)

$\Rightarrow r^{q-p}=r^{2p-2}$

$\therefore q-p=2p-2$

$\Rightarrow q=3p-2$

for every distinct value of '$p$' there will be a district integer value of '$q$'

$\therefore$ Infinite no. of progression are possible.

option $C$