High School

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Find the sum of a geometric series for which:

- First term, \( a_1 = 3125 \)
- Last term, \( a_n = 1 \)
- Common ratio, \( r = \frac{1}{5} \)

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.

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