A) 9th term
B) 2nd term
C) 4th term
D) 8 term
Correct Answer: A
Solution :
Given sequence \[8-6i,\,\,\,\,\,\,\,\,\,7-4i,\,\,\,\,\,\,6-2i,....\] Then the given series form an Ap. with common difference \['d'=-1+2i\] \[\therefore \] \[{{T}_{n}}=(8-6i)+(n-1)\times (-1+2)\] \[=8-6i+-n+1+2ni-2i\] \[=(8-n+1)+i(-6+2n-2)\] \[=(9-n)+(2n-8)i\] For the term which have purely imaginary number, put \[9-n=0\Rightarrow n=9\] So, 9th term is a purely imaginary term.
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