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Mathematics

Find the seventh term of the G.P. :

3+1,1,3−12,\sqrt{3} + 1, 1, \dfrac{\sqrt{3} - 1}{2}, ………..

GP

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Answer

Rationalising the term, 3−12\dfrac{\sqrt{3} - 1}{2} we get,

⇒3−12×3+13+1=32−122(3+1)=3−12(3+1)=22(3+1)=13+1.\Rightarrow \dfrac{\sqrt{3} - 1}{2} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] = \dfrac{\sqrt{3}^2 - 1^2}{2(\sqrt{3} + 1)} \\[1em] = \dfrac{3 - 1}{2(\sqrt{3} + 1)} \\[1em] = \dfrac{2}{2(\sqrt{3} + 1)} \\[1em] = \dfrac{1}{\sqrt{3} + 1}.

So, Sequence = 3+1,1,13+1,\sqrt{3} + 1, 1, \dfrac{1}{\sqrt{3} + 1}, ………..

Common ratio(r) = 13+1\dfrac{1}{\sqrt{3} + 1}.

We know that nth term of G.P.,

an = arn - 1

⇒a7=(3+1)(13+1)7−1=(3+1)(13+1)6=(13+1)5=(13+1×3−13−1)5=(3−132−(1)2)5=(3−13−1)5=(3−12)5=132(3−1)5.\Rightarrow a_7 = (\sqrt{3} + 1)\Big(\dfrac{1}{\sqrt{3} + 1}\Big)^{7 - 1} \\[1em] = (\sqrt{3} + 1)\Big(\dfrac{1}{\sqrt{3} + 1}\Big)^{6} \\[1em] = \Big(\dfrac{1}{\sqrt{3} + 1}\Big)^{5} \\[1em] = \Big(\dfrac{1}{\sqrt{3} + 1} \times \dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}\Big)^{5} \\[1em] = \Big(\dfrac{\sqrt{3} - 1}{\sqrt{3}^2 - (1)^2}\Big)^5 \\[1em] = \Big(\dfrac{\sqrt{3} - 1}{3 - 1}\Big)^5 \\[1em] = \Big(\dfrac{\sqrt{3} - 1}{2}\Big)^5 \\[1em] = \dfrac{1}{32}(\sqrt{3} - 1)^5.

Hence, seventh term of the G.P. = 132(3−1)5.\dfrac{1}{32}(\sqrt{3} - 1)^5.

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