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.
