Question

In G.P., the first term and common ratio are both $\dfrac{1}{2}\left( \sqrt{3}+i \right)$, then the absolute value of its ${{n}^{th}}$ term is: -
(a) 1
(b) ${{2}^{n}}$
(c) ${{4}^{n}}$
(d) None

Answer 1

Assume ‘a’ as the first term of the G.P and ‘r’ as its common ratio. Apply the formula for ${{n}^{th}}$ term of G.P given as: - ${{T}_{n}}=a{{r}^{n-1}}$ to obtain the ${{n}^{th}}$ term. Here, {{T}_{n}}=$$$${{n}^{th}} term, a = first term and r = common ratio of the G.P. Now, assume this ${{n}^{th}}$ term obtained as $x+iy$ and compare the real and imaginary part to obtain the values of x and y. Use the formula: - Absolute value = $\sqrt{{{x}^{2}}+{{y}^{2}}}$ to get the answer.

Complete step-by-step solution
Here, we have been provided with a G.P whose first term and common ratio is given as $\dfrac{1}{2}\left( \sqrt{3}+i \right)$. We have to find the absolute value of ${{n}^{th}}$ term.
Now, let us assume the first term of the given G.P as ‘a’ and the common ratio as ‘r’. So, the formula for ${{n}^{th}}$ term of a G.P. is given as: -
${{T}_{n}}=a{{r}^{n-1}}$, here ${{T}_{n}}$ is the ${{n}^{th}}$ term.
Substituting the given values of a and r, we get,
${{T}_{n}}=\dfrac{1}{2}\left( \sqrt{3}+i \right)\times {{\left[ \dfrac{1}{2}\left( \sqrt{3}+i \right) \right]}^{n-1}}$
Applying the formula: - ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$, we get,

& \Rightarrow {{T}_{n}}={{\left[ \dfrac{1}{2}\left( \sqrt{3}+i \right) \right]}^{n}} \\\ & \Rightarrow {{T}_{n}}=\dfrac{1}{{{2}^{n}}}{{\left( \sqrt{3}+i \right)}^{n}} \\\ \end{aligned}$$ Assuming $$\left( \sqrt{3}+i \right)=\left( x+iy \right)$$, we get, on comparing real and imaginary part: - $$x=\sqrt{3},y=1$$ Now, we know that absolute value of a complex number of the form $${{\left( x+iy \right)}^{n}}$$ is given as: - $${{\left( \sqrt{{{x}^{2}}+{{y}^{2}}} \right)}^{n}}$$, so we have, $$\Rightarrow $$ Absolute value of $${{T}_{n}}=\dfrac{1}{{{2}^{n}}}\times {{\left( \sqrt{{{\left( \sqrt{3} \right)}^{2}}+{{1}^{2}}} \right)}^{n}}$$ $$\Rightarrow $$ Absolute value of $${{T}_{n}}=\dfrac{1}{{{2}^{n}}}\times {{\left( \sqrt{4} \right)}^{n}}$$ $$\Rightarrow $$ Absolute value of $${{T}_{n}}=\dfrac{1}{{{2}^{n}}}\times {{2}^{n}}$$ $$\Rightarrow $$ Absolute value of $${{T}_{n}}=1$$ **Hence, option (a) is the correct answer.** **Note:** One may note that while calculating the absolute value of $${{T}_{n}}$$ we have taken $$\dfrac{1}{{{2}^{n}}}$$ out of the bracket. This is because $$\dfrac{1}{{{2}^{n}}}$$ is common for both x and y and it is a constant. Now, note that there can be another method to solve the question. We can assume $$\dfrac{1}{2}\left( \sqrt{3}+i \right)={{e}^{i\theta }}$$, ,where ‘$$\theta $$’ will be $$\dfrac{\pi }{6}$$. This is because $$\left( \dfrac{\sqrt{3}}{2}+\dfrac{1}{2}i \right)=\left( \cos \dfrac{\pi }{6}+i\sin \dfrac{\pi }{6} \right)={{e}^{i\dfrac{\pi }{6}}}$$. From here also $${{n}^{th}}$$ term and the absolute value can be determined. The absolute value will be $$\sqrt{{{\cos }^{2}}\dfrac{n\pi }{6}+{{\sin }^{2}}\dfrac{n\pi }{6}}=1$$.