Please do not post the same question twice.

We have a_1 = 4 and a_(n+1) = -3a_n

Note that a_n is the term before a_(n+1)

So the first five terms are:

a_1 = 4

a_2 = a_(1 + 1) = -3a_1 = -3(4) = -12

a_3 = a_(2 + 1) = -3a_2 = -3(-12) = 36

a_4 = a_(3 + 1) = -3a_3 = -3(36) = -108

a_5 = a_(4 + 1) = -3a_4 = -3(-108) = 324

Note that 15 – 9 = 21 – 15 = 27 – 21 = 6. so we have a common difference between the terms. Thus we have an arithmetic sequence. The terms of an arithmetic sequence is given by:Find the 65th term of the sequence 9, 15, 21, 27,.....

a_n = a_1 + (n – 1)d, where a_n is the nth term, a_1 is the first term, n is the current number of the term, and d is the common difference. So for this sequence, we have:

a_n = 9 + (n – 1)6 = 9 – 6 + 6n = 3 + 6n

So the 65th term is a_65 = 3 + 6(65) = 393

For 15 + 5 + 5/3 + … I don’t think it should be a -15Find the sum of each infinate geometric series.

-15+5+5/3+... _________

Note that these terms come from a geometric sequence with common ratio r = 1/3 and the first term a = 15. since |r|<1, the infinite sum is given by:

S = a/(1 – r) = 15/(1 – 1/3) = 15/(2/3) = 45/2

E-4(3/4)n-1 ___________

For -4(3/4)^(n – 1), we have a geometric series with a = -4, r = ¾, since |r|<1, again the infinite sum is given by:

S = a/(1 – r) = -4/(1 – ¾) = -4/(1/4) = -16