Find the sum of the geometric series 1 + 0.8 + 0.8^2 +0.8^3 + ... + 0.8^{19}

Mathematics

1 answer:

Ainat [17]3 years ago

5 0

Answer:

S20 ≈ 4.942

Step-by-step explanation:

Sum of a geometric series is expressed as Sn = a(1-rⁿ)/1-r if r<1

a is the first term

r is the common ratio

n is the number of terms

Given the geometric series

1 + 0.8 + 0.8^2 +0.8^3 + ... + 0.8^{19}

Given a = 1,

r = 0.8/1 = 0.8²/0.8 = 0.8

n = 20 (The total number of terms in the series is 20)

Substituting this values in the formula above.

S20 = 1(1-0.8^20)/1-0.8

S20 = 1-0.01153/0.2

S20 = 0.9885/0.2

S20 ≈ 4.942

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Let the first and smaller number be x. Since the second number is 5 times the first, the second number is 5x. We also know both of these numbers must add to 50; thus, we can create the following equation.

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This is the value of the first number. The number you're looking for isn't specified in the question description, so I'll just provide the value of the second number. The second number has to be 5 times the first.

Thus, we just multiply 5 to x.

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