Question
Which term of the sequence $12+8\text{i},\ 6\text{i},\ 10+4\text{i},\ ...$ is (a) real (b) purely imaginary?

Answer

The given sequence is $12+8\text{i},\ 6\text{i},\ 10+4\text{i},\ ...$
Here, $​​\text{a}=12+8\text{i}$ $\text{d}=-1-2\text{i}$ Then, $​​\text{a}_\text{n}=​​\text{a}(​​\text{n}-1)​​\text{d}$
$=12+8\text{i}+(​​\text{n}-1)(-1-2\text{i})$
$=(13-\text{n})+\text{i}(10-2\text{i})$ Let $n^{th}$ term be purely real the $(10-2\text{n})=0$ or $\text{n}=5$
So, 5th term is purely real. Let $n^{th}$ term be purely im againarym than, $13-\text{n}=0$ $\therefore\text{n}=13$ So, 13th term is purely inaginary.

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