Write a polynomial equation of degree 3 such that two of its roots are 2 and an imaginary number.
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Answer:
= 6n - 4
Step-by-step explanation:
There is a common difference between consecutive terms in the sequence, that is
8 - 2 = 14 - 8 = 6
This indicates the sequence is arithmetic with n th term
= a₁ + (n - 1)d
where a₁ is the first term and d the common difference
Here a₁ = 2 and d = 6 , thus
= 2 + 6(n - 1) = 2 + 6n - 6 = 6n - 4
Thus the formula representing the sequence is
= 6n - 4