Let the first term of C.P. be ′a′ and the common ratio be ′r′ where −1 < r < 1

The G.P. is a, ar, ar² …

Therefore the sum of the infinite terms of the G.P. is \(\frac{a}{1−r}\) = 57 … (i)

If taking the cube of each terms the new G.P. is a³, a³r³, a³r⁶ , …

Therefore the sum of their cube is \(\frac{a^3}{1−r^3}\) = 9747 … (ii)

Taking the cube of the (i)

\(\frac{a^3}{1−r^3}\) = (57)³ ⇒ a³ = (57)³ (1 − r)³ … (iii)

Substituting the value of ′a′ in terms of r in (ii)

\(\frac{(57)^3 (1 − r)^3}{1 − r^3}\) = 9747

\(\frac{(1 − r)^3}{(1 − r)(1 − r^2 + r)}\) = \(\frac{9747}{ (57)^3}\)

\(\frac{(1 − r)^2}{(1 + r^2 + r)}\) = \(\frac{1}{19}\)

\(\frac{1 + r^2 − 2r}{1 + r^2 + r}\) = \(\frac{1}{19}\)

19(1 + r² − 2r) = 1 + r² + r

18 r² − 39r + 18 = 0

6 r² − 13r + 6 = 0

6 r² − 9r − 4r + 6 = 0

(2r − 3)(3r − 2) = 0

Therefore, r =\(\frac{3}{2}\) or r = \(\frac{2}{3}\) since r < 1.

∴ r = ^{\(\frac{2}{3}\)}

Substitute in eq. (i),

\(\frac{a}{1 − \frac{2}{3}}\) = 57

⇒ 3a = 57

⇒ a = \(\frac{57}{3}\) = 19

Thus, the first term of the GP. is 19 and the common ratio is \(\frac{2}{3}\).

The G.P. is 19, \(\frac {38}{3}\), \(\frac {76}{6}\) and so on.