Question:
Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal, then the last term is :
Given a sequence of 4 numbers, first three of which are in G.P. and the last three are in A.P. with common difference six. If first and last terms of this sequence are equal, then the last term is :
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Take the terms as \(a, ar, ar^2, a\) and use the common difference 6 for the last three.
Updated On: Nov 4, 2026
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The Correct Option is B
Solution and Explanation
Let the first three terms be \(a, ar, ar^2\); the fourth term equals the first, \(a\).
Last three \(ar, ar^2, a\) are in A.P. with common difference 6: \(ar^2-ar=6\) and \(a-ar^2=6\).
Adding: \(a-ar=12\Rightarrow a(1-r)=12\). Also \(ar(r-1)=6\Rightarrow -r\cdot a(1-r)=6\Rightarrow -12r=6\Rightarrow r=-\frac12\).
Then \(a=\frac{12}{1+\frac12}=8\). The sequence is \(8,-4,2,8\); the last three \(-4,2,8\) have common difference 6.
Last term \(=8\).
Answer: (2) 8
Last three \(ar, ar^2, a\) are in A.P. with common difference 6: \(ar^2-ar=6\) and \(a-ar^2=6\).
Adding: \(a-ar=12\Rightarrow a(1-r)=12\). Also \(ar(r-1)=6\Rightarrow -r\cdot a(1-r)=6\Rightarrow -12r=6\Rightarrow r=-\frac12\).
Then \(a=\frac{12}{1+\frac12}=8\). The sequence is \(8,-4,2,8\); the last three \(-4,2,8\) have common difference 6.
Last term \(=8\).
Answer: (2) 8
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