The sum of the 1st term and 2nd term of a geometric series is 30 and the sum of infinity is 54 determine the 1st term a, and common ration with r>0
Let the first term be "a" and the common ratio be "r".
Given that the sum of the 1st term and 2nd term is 30, we have:
a + ar = 30
a(1 + r) = 30
Given that the sum of the infinite series is 54, we have:
a / (1 - r) = 54
a = 54(1 - r)
Substitute this into the first equation:
54(1 - r)(1 + r) = 30
54(1 - r^2) = 30
54 - 54r^2 = 30
-54r^2 = -24
r^2 = 24 / 54
r^2 = 4 / 9
r = 2 / 3
Substitute r = 2/3 back into the equation a = 54(1 - r):
a = 54(1 - 2/3)
a = 54(1/3)
a = 18
Therefore, the 1st term is 18 and the common ratio is 2/3.