Answer :

Given,


a1 = 3 and an = 3an–1 + 2, for all n > 1


We can find the first four terms of a sequence by putting values of n


from 1 to 4


When n = 1:


a1 = 3


When n = 2:


a2 = 3a2–1 + 2


a2 = 3a1 + 2


a2 = 3(3) + 2


a2 = 9 + 2


a2 = 11


When n = 3:


a3 = 3a3–1 + 2


a3 = 3a2 + 2


a3 = 3(11) + 2


a3 = 33 + 2


a3 = 35


When n = 4:


a4 = 3a4–1 + 2


a4 = 3a3 + 2


a4 = 3(35) + 2


a4 = 105 + 2


a4 = 107


First four terms of sequence are 3, 11, 35, 107.


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