The \({2^{\mathrm{nd}}}\), \({3^{\mathrm{rd}}}\) and \({4^{\mathrm{th}}}\) terms of a geometric progression are \({\cos\theta}\), \({\sqrt{2}\sin\theta}\) and \({\sqrt{3}\tan\theta}\), respectively, where \({0<\theta<\frac{\pi}{2}}\).
Show clearly that the sum of the first \({6}\) terms of the progression is
\({\frac{43}{12}\left(6+\sqrt{6}\right)}\).
Video solution locked
A free account saves your progress; this video needs tuition enrolment.
680+ questions here come with free worked solutions. A free account saves your questions and progress; this one needs tuition enrolment.
Create a free account to practise similar questions.
Need help? Join our JC Math tuition classes.
Learn more