1)Find a1 in a geometric series for which Sn=300,r=-3,and n=4

Question

A)15
B)15/2
C)-15
D)1/15
I chose A

5)Find the fifth term of the sequence in which a1=-3,and aN+1=3aN-n
A)-301
B)-99
C)-193
D)-341
I don't know

Answer 1

Jon, for #1 the equation is
300 = a((-3)^4 - 1)/(-3-1)
300 = a(80)/-4
a = -15

since there are only 4 terms, this is easy to check
-15 + 45 + (-135) + 405 = 300

for you #2, I will assume you meant
a₁ = -3
a_N+1 = 3_N - n

then:
a₁ = -3
a₂ = 3a₁ - 2 = -11
a₃ = 3a₂ - 3 = -36
a₄ = 3a₃ - 4 = -112
a₅ = 3a₄ - 5 = -341 which is choice D

Answer 2

find a1 for given geometric series.

Sn= 405, r=3, n=4