Question:
If \(3^{2\sin2\alpha-1}\), \(14\) and \(3^{4-2\sin2\alpha}\) are the first three terms of an A.P. for some \(\alpha\), then the sixth term of this A.P. is:
If \(3^{2\sin2\alpha-1}\), \(14\) and \(3^{4-2\sin2\alpha}\) are the first three terms of an A.P. for some \(\alpha\), then the sixth term of this A.P. is:
Show Hint
Use \(2\times14=\) sum of the first and third terms with \(t=3^{2\sin2\alpha}\); reject the root that is out of range.
Updated On: Nov 5, 2026
- \(66\)
- \(81\)
- \(65\)
- \(78\)
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The Correct Option is A
Solution and Explanation
Let \(t=3^{2\sin2\alpha}\). Then the first term is \(\frac t3\) and the third term is \(\frac{81}{t}\).
A.P. condition: \(2(14)=\frac t3+\frac{81}t\Rightarrow t^2-84t+243=0\Rightarrow t=\frac{84\pm78}{2}=81\) or \(3\).
Since \(\sin2\alpha\in[-1,1]\), \(t=3^{2\sin2\alpha}\in\left[\frac19,9\right]\). So \(t=81\) is rejected and \(t=3\).
First term \(=\frac33=1\), third term \(=\frac{81}3=27\); the common difference \(d=14-1=13\) (check \(27-14=13\)).
Sixth term \(=1+5(13)=66\).
Answer: (1) \(66\)
A.P. condition: \(2(14)=\frac t3+\frac{81}t\Rightarrow t^2-84t+243=0\Rightarrow t=\frac{84\pm78}{2}=81\) or \(3\).
Since \(\sin2\alpha\in[-1,1]\), \(t=3^{2\sin2\alpha}\in\left[\frac19,9\right]\). So \(t=81\) is rejected and \(t=3\).
First term \(=\frac33=1\), third term \(=\frac{81}3=27\); the common difference \(d=14-1=13\) (check \(27-14=13\)).
Sixth term \(=1+5(13)=66\).
Answer: (1) \(66\)
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