MHT CET · Maths · Complex Number
The value of \(\frac{\mathrm{i}^{248}+\mathrm{i}^{246}+\mathrm{i}^{244}+\mathrm{i}^{242}+\mathrm{i}^{240}}{\mathrm{i}^{249}+\mathrm{i}^{247}+\mathrm{i}^{245}+\mathrm{i}^{243}+\mathrm{i}^{241}}\), \((\mathrm{i}=\sqrt{-1})\) is
- A \(\mathrm{i}\)
- B 1
- C -1
- D \(-\mathrm{i}\)
Answer & Solution
Correct Answer
(D) \(-\mathrm{i}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \frac{i^{245}+i^{246}+i^{244}+i^{242}+i^{240}}{i^{243}+i^{247}+i^{245}+i^{243}+i^{241}} \\ & =\frac{i^{240}\left(i^8+i^6+i^4+i^2+1\right)}{i^{241}\left(i^8+i^6+i^4+i^2+1\right)} \\ & =\frac{i^{200}}{i^{241}} \\ & =\frac{1}{i} \\ & =\frac{i}{i^2}=-i\end{aligned}\)
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