Sol. Let ‘a’ be the first term and ‘r’ be the common ratio. Then if a G.P. exists there must exist distinct positive integers p, q and s such that

27 = ar p–1 ; 8 = ar q–1 ; 12 = ar s–1 …(1)

To determine whether or not these integers exist we analyze the system of equations (1)

We easily get = r p–q , = r q–s ⇒ = ⇒ =

⇒ 3(q – s) = q – p ⇒ 3s = 2q + p … (2)

Any three different positive integers satisfying (2) will satisfy the condition of the problem and as there are infinitely many triads (p, q, s) satisfying (2), the number of G.P.’s possible is infinite. Let us determine one of them. One of the solutions is

p = 1, q = 4, s = 3 ⇒ 27 = ar 0 , 8 = ar 3 , 12 = ar 3 .

From first two equations we easily get a = 27, r = 2/3

(Note that these values satisfy the third equation also).

Thus 27 + (27) + (27) + (27) +……. is a G.P. whose first term is 27, third term is 12 and fourth term is 8.