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

Hint: 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,
\[\begin{align}
  & \Rightarrow {{T}_{n}}={{\left[ \dfrac{1}{2}\left( \sqrt{3}+i \right) \right]}^{n}} \\
 & \Rightarrow {{T}_{n}}=\dfrac{1}{{{2}^{n}}}{{\left( \sqrt{3}+i \right)}^{n}} \\
\end{align}\]
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 \[\sqrt{{{\cos }^{2}}\dfrac{n\pi }{6}+{{\sin }^{2}}\dfrac{n\pi }{6}}=1\].