The answer to that question is B.
If a = first term and r = common ratio we have
a + ar + ar^2 = 13 and ar^2 / a = r^2 = 9
so r = 3
and a + 3a + 9a = 13
so a = 1
so they are 1,3 and 9
2.
in geometric series we have
4 , 4r ,4r^2 , 60
Arithmetic;
4, 4r , 4r + d , 4r + 2d
so we have the system of equations
4r + 2d = 60
4r^2 = 4r + d
From first equation
2r + d = 30
so d = 30 - 2r
Substitute for d in second equation:-
4r^2 - 4r - (30-2r) = 0
4r^2 - 2r - 30 =0
2r^2 - r - 15 = 0
(r - 3)(2r + 5) = 0
r = 3 or -2.5
r must be positive so its = 3
and d = 30 - 2(3) = 24
and the numbers are 4*3 = 12 , 4*3^2 = 36
first 3 are 4 , 12 and 36 ( in geometric)
and last 3 are 12, 36 and 60 ( in arithmetic)
The 2 numbers we ause are 12 and 36.
Answer:
Step-by-step explanation:
Answer: (5, -6) and 3
Step-by-step explanation:
