Question

13th term of 1+i, 2i, 1+3i

Answer 1

Step-by-step explanation:

ap: 1+i , 2i, 1+ 3i

a= 1+i

d= 2i-(1+i)

= 2i-1-i= i-1

$a(13) = a + (n - 1)d \\ \: \: \: \: \: \: \: \: \: \: \: \: = 1 + i + (13 - 1)(i - 1) \\ = 1 + i + 12i - 12 \\ = 13i - 11$

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