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Difficulty: MediumArithmetic and Geometric Sequences and Series

An infinite geometric series has a first term of 1212 and a sum of 88. What is the common ratio of this series?

  1. −12-\frac{1}{2}Answer
  2. B
    12\frac{1}{2}
  3. C
    13\frac{1}{3}
  4. D
    23\frac{2}{3}
  5. E
    52\frac{5}{2}

Answer

The correct common ratio is −12-\frac{1}{2}.
The correct answer is −12-\frac{1}{2}. The formula for the sum of an infinite geometric series is S=a11−rS = \frac{a_1}{1 - r}. Substituting the first term a1=12a_1 = 12 and the sum S=8S = 8 gives 8=121−r8 = \frac{12}{1 - r}. Multiplying both sides by 1−r1 - r yields 8(1−r)=128(1 - r) = 12, which simplifies to 8−8r=128 - 8r = 12. Subtracting 88 from both sides gives −8r=4-8r = 4. Dividing by −8-8 gives r=−12r = -\frac{1}{2}. Since ∣−12∣<1|-\frac{1}{2}| < 1, the series converges.

Step-by-Step Solution

1
Write the formula for the sum of an infinite geometric series.
S=a11−rS = \frac{a_1}{1 - r}
This formula relates the sum (SS), the first term (a1a_1), and the common ratio (rr) of an infinite geometric series where ∣r∣<1|r| < 1.
2
Substitute the given values into the formula.
8=121−r8 = \frac{12}{1 - r}
Replacing SS with 88 and a1a_1 with 1212 leaves the common ratio rr as the only unknown variable.
3
Solve for the common ratio rr.
8(1−r)=12⇒8−8r=12⇒−8r=4⇒r=−48=−128(1 - r) = 12 \Rightarrow 8 - 8r = 12 \Rightarrow -8r = 4 \Rightarrow r = -\frac{4}{8} = -\frac{1}{2}
Multiplying by the denominator and isolating rr yields the value of the common ratio.

Key Concept

Sum of an infinite geometric series

Alternative Method

Test the given choices by substituting each value of rr back into the sum formula S=121−rS = \frac{12}{1 - r} to see which one yields a sum of 88. For example, testing −12-\frac{1}{2} gives 121−(−0.5)=121.5=8\frac{12}{1 - (-0.5)} = \frac{12}{1.5} = 8, which matches the given sum.
Estimated Time:1m 0s
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Topics

Arithmetic and Geometric Sequences and Series

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