Question:
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 three consecutive terms of a G.P., then \(\frac{a}{c}\) is equal to :
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 three consecutive terms of a G.P., then \(\frac{a}{c}\) is equal to :
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Use \(b^2=ac\) with \(a=A+6d,\ b=A+10d,\ c=A+12d\) to find \(A\) in terms of \(d\).
Updated On: Nov 5, 2026
- \(2\)
- \(\frac{1}{2}\)
- \(\frac{7}{13}\)
- \(4\)
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The Correct Option is D
Solution and Explanation
Let the A.P. have first term \(A\) and common difference \(d\ne0\): \(a=A+6d\), \(b=A+10d\), \(c=A+12d\).
G.P. condition: \(b^2=ac\Rightarrow A^2+20Ad+100d^2=A^2+18Ad+72d^2\Rightarrow 2Ad+28d^2=0\Rightarrow A=-14d\).
Then \(a=-8d\), \(c=-2d\), so \(\frac{a}{c}=4\).
Answer: (4) \(4\)
G.P. condition: \(b^2=ac\Rightarrow A^2+20Ad+100d^2=A^2+18Ad+72d^2\Rightarrow 2Ad+28d^2=0\Rightarrow A=-14d\).
Then \(a=-8d\), \(c=-2d\), so \(\frac{a}{c}=4\).
Answer: (4) \(4\)
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