JEE Mains · Maths · STD 11 - 8. sequence and series
Let \(a, b\) and \(c\) be the \(7^{th},\,11^{th}\) and \(13^{th}\) terms respectively of a non -constant \(A.P.\) If these are also the three consecutive terms of a \(G.P.\) then \(\frac {a}{c}\) is equal to
- A \(\frac {1}{2}\)
- B \(4\)
- C \(2\)
- D \(\frac {7}{13}\)
Answer & Solution
Correct Answer
(B) \(4\)
Step-by-step Solution
Detailed explanation
\(a=A+6d\) \(b=A+10d\) \(c=A+12d\) \(a,b,c\) are in \(G.P.\) \( \Rightarrow {\left( {A + 10d} \right)^2} = \left( {A + 6d} \right)\left( {a + 12d} \right)\) \( \Rightarrow \frac{A}{d} = - 14\)…
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