Answer :
The sum of the geometric series with a first term of 3125, nth term of 1, and a common ratio of 1/5 is found to be approximately 1249.92 after determining the number of terms to be 6.
The student's question involves finding the sum of a geometric series with a given first term (a1), a given nth term (an), and a common ratio (r). We are given that a1 = 3125, an = 1, and r = 1/5. To find the sum, we can use the formula for the sum of a finite geometric series:
Sn = a1 * (1 - rn) / (1 - r)
First, we must determine the number of terms in the series (n). Since the terms of a geometric series are defined by a1 * rn-1, we have:
1 = 3125 * (1/5)n-1
Lets solve for n:
5n-1 = 3125
5n-1 = 55
n - 1 = 5
n = 6
Now, we can calculate the sum:
S6 = 3125 * (1 - (1/5)6) / (1 - 1/5)
S6 = 3125 * (1 - 1/15625) / (4/5)
S6 = 3125 * (15624/15625) / (4/5)
S6 = 3125 * (15624/15625) * (5/4)
S6 = (3125 * 15624 * 5) / (15625 * 4)
S6 = (78120000) / (62500)
S6 = 1249.92
Therefore, the sum of the geometric series is approximately 1249.92.
