Question

Which term of the sequence $\left( -8+18i \right),\left( -6+15i \right),\left( -4+12i \right)\ldots \ldots $is purely imaginary?
A. \[{{5}^{th}}\]
B. \[{{7}^{th}}\]
C. \[{{8}^{th}}\]
D. \[{{6}^{th}}\]

Answer 1

Hint: A purely imaginary term means that term which has only an imaginary part and its real part would be equal to 0. So, we will find the term that has the real part exactly equal to 0 and then select the correct option accordingly.

Complete step-by-step answer:

In this question, we have been given a sequence, $\left( -8+18i \right),\left( -6+15i \right),\left( -4+12i \right)\ldots \ldots $ and we have been asked to find a term which is purely imaginary. To solve this question, we should know that purely imaginary numbers are those numbers that have their real part exactly equal to 0. Like, if we take the first term, $\left( -8+18i \right)$, then we have the real part as -8 and the imaginary part as 18i, but for a purely imaginary number, the real part must be equal to 0. Now, if we observe the given sequence, we come to know that there is a pattern, the real part is increasing by 2 and the imaginary part is decreasing by 3i respectively with every preceding term of the sequence. So, we can say that the fourth term of the sequence will be,
$\begin{align}
  & =\left\{ \left( -4+2 \right)+\left( 12i-3i \right) \right\} \\
 & \Rightarrow \left( -2+9i \right) \\
\end{align}$
Similarly, the fifth term in the sequence will be,
$\begin{align}
  & =\left\{ \left( -2+2 \right)+\left( 9i-3i \right) \right\} \\
 & \Rightarrow \left( 0+6i \right) \\
\end{align}$
So, if we observe the fifth term, that is, 6i, then we can see that the real part is 0 and the imaginary part 6i is independent of the real part. Therefore, it is a purely imaginary number. Hence, the fifth position of the given sequence is purely imaginary.
Therefore, option A is the correct answer.

Note: Many students believe that it is not possible to form a number without a real part and they consider 6i as 0 + 6i, where they consider 0 as the real part and so do not consider this as purely imaginary. We can also consider it as an AP series with common differences as $\left( 2-3i \right)$ . Then using the formula for ${{n}^{th}}$ term of series, i.e. ${{T}_{n}}=a+\left( n-1 \right)d$ , where a is the first term, d is the common difference; we can find the \[{{5}^{th}},{{6}^{th}},{{7}^{th}},{{8}^{th}}\] terms.