In a G.P, the ratio of the sum of first 3 terms to that of the first 6

1.	In a G.P, the ratio of the sum of first 3 terms to that of the first 6 terms is 125 : 152. Find the common ratio of the G.P(a) \(\frac{2}{3}\) (b) \(\frac{3}{4}\)(c) \(\frac{4}{3}\) (d) \(\frac{3}{5}\)
Answer» (d) \(\frac{3}{5}\) Let a be the first term and r the common ratio of the G.P. Then, \(\frac{S_3}{S_6}\) = \(\frac{125}{152}\) ⇒ \(\frac{\frac{a(r^3-1)}{(r-1)}}{\frac{a(r^6-1)}{(r-1)}}\) = \(\frac{125}{152}\) \(\frac{(r^3-1)}{(r^6-1)}\) = \(\frac{125}{152}\) ⇒ \(\frac{(r^3-1)}{(r^3+1)(r^3-1)}\) = \(\frac{125}{152}\) ⇒ r³ + 1 = \(\frac{125}{152}\) ⇒ r³ = \(\frac{152}{125}\) - 1 = \(\frac{27}{125}\) ⇒ r³ = \(\big(\frac{3}{5}\big)^3\) ⇒ r = \(\frac{3}{5}.\)

In a G.P, the ratio of the sum of first 3 terms to that of the first 6 terms is 125 : 152. Find the common ratio of the G.P(a) \(\frac{2}{3}\) (b) \(\frac{3}{4}\)(c) \(\frac{4}{3}\) (d) \(\frac{3}{5}\)

Answer»

(d) \(\frac{3}{5}\)

Let a be the first term and r the common ratio of the G.P.

Then, \(\frac{S_3}{S_6}\) = \(\frac{125}{152}\) ⇒ \(\frac{\frac{a(r^3-1)}{(r-1)}}{\frac{a(r^6-1)}{(r-1)}}\) = \(\frac{125}{152}\)

\(\frac{(r^3-1)}{(r^6-1)}\) = \(\frac{125}{152}\) ⇒ \(\frac{(r^3-1)}{(r^3+1)(r^3-1)}\) = \(\frac{125}{152}\)

⇒ r³ + 1 = \(\frac{125}{152}\) ⇒ r³ = \(\frac{152}{125}\) - 1 = \(\frac{27}{125}\)

⇒ r³ = \(\big(\frac{3}{5}\big)^3\) ⇒ r = \(\frac{3}{5}.\)