How many geometric progressions are possible containing 27, 8 and 12 as three of its/their terms

Options

A. One
B. Two
C. Four
D. Infinitely many

Answer

D

Explanation

Let 'a' be the first term & 'x' be the common ratio. Also, suppose 27, 8 & 12 be the `p^(th),q^(th)" & "r^(th)` term of the GP.
`:." "ax^(p-1)=27`
`ax^(q-1)=8`
&`ax^(r-1)=12`
Now, `27xx8^(2)=12^(3)`
`impliesax^(p-1)x(ax^(q-1))^(2)=(ax^(r-1))^(3)`
`impliesx^(p-1).x^(2q-2)=x^(2r-3)`
`impliesp-1+2q-2=3r-3`
`impliesp+2q-3r=0" "...(1)`
There are infinitely many solutions for the eq. (1).