JEE Mains · Maths · STD 11 - 8. sequence and series
If the sum and product of four positive consecutive terms of a \(G.P.\), are \(126\) and \(1296\), respectively, then the sum of common ratios of all such \(GPs\) is \(.........\).
- A \(7\)
- B \(\frac{9}{2}\)
- C \(3\)
- D \(14\)
Answer & Solution
Correct Answer
(A) \(7\)
Step-by-step Solution
Detailed explanation
\(a, a r, a r^2, a r^3(a, r > 0)\) \(a^4 r^6=1296\) \(a^2 r^3=36\) \(a=\frac{6}{r^{3 / 2}}\) \(a+a r+a r^2+a r^3=126\) \(\frac{1}{r^{3 / 2}}+\frac{r}{r^{3 / 2}}+\frac{r^2}{r^{3 / 2}}+\frac{r^3}{r^{3 / 2}}=\frac{126}{6}=21\)…
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