JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(a, b\) and \(c\) be in \(G.P\) with common ratio \(r,\) where \(a \ne 0\) and \(0\, < \,r\, \le \,\frac{1}{2}\). If \(3a, 7b\) and \(15c\) are the first three terms of an \(A.P.,\) then the \(4^{th}\) term of this \(A.P\) is
- A \(\frac{2}{3}a\)
- B \(\frac{7}{3}a\)
- C \(5a\)
- D \(a\)
Answer & Solution
Correct Answer
(D) \(a\)
Step-by-step Solution
Detailed explanation
\(b=ar\) \(c = a{r^2}\) \(3a.7b\) and \(15c\) are in\(A.P,\) \( \Rightarrow 14b = 3a + 15c\) \( \Rightarrow 14\left( {ar} \right) = 3a + 15a{r^2}\) \( \Rightarrow 14r = 3 + 15{r^2}\)…
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