+60 What is the sum of the geometric series in which a1 = 5, r = 3, and an = 1,215? Hint: an = a1(r)n − 1, where a1 is the first term and r is the common ratio. Hint: cap s sub n equals start fraction a sub one left parenthesis one minus r to the power of n end power right parenthesis over one minus r end fraction comma r does not equal one comma where a1 is the first term and r is the common ratio. A.Sn= −1,820 B.Sn = −605 C.Sn = 1,820 D.Sn = 605
+23245 What is the sum of a finite geometric series whose first term is 5, common ratio is 3, and nth term is 1215?
The nth term of a geometric series can be found by using the formula: an = a1 · rn - 1.
For this problem, an = 1215, a1 = 5, and r = 3 ---> 1215 = 5 · 3n - 1
divide both sides by 5: 243 = 3n - 1
take the log of both sides: log(243) = log(3n - 1)
simplify: log(243) = (n - 1) · log(3)
divide both sides by log(3): log(243) / log(3) = n - 1
simplify: 5 = n - 1
add 1 to both sides: n = 6
To find the sum, use this formula: Sum = a(1 - rn) / (1 - r)
Sum = 5(1 - 36) / (1 - 3)
Sum = 5(-728) / (-2)
Sum = -3640 / -2
Sum = 1820
