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Difficulty: MediumArithmetic and Geometric Sequences and Series

An arithmetic sequence has a first term of 22 and a common difference of dd. A geometric sequence has a first term of 44 and a common ratio of rr. The third term of the arithmetic sequence is equal to the third term of the geometric sequence. If d=rd = r and d≠1d \neq 1, what is the value of dd?

  1. A
    22
  2. B
    12\frac{1}{2}
  3. C
    −1-1
  4. −12-\frac{1}{2}Answer
  5. E
    13\frac{1}{3}

Answer

−12-\frac{1}{2}
The correct answer is −12-\frac{1}{2}. The third term of the arithmetic sequence is a3=a1+2d=2+2da_3 = a_1 + 2d = 2 + 2d. The third term of the geometric sequence is g3=g1r2=4r2g_3 = g_1 r^2 = 4r^2. Given that d=rd = r, we set the two terms equal: 2+2d=4d22 + 2d = 4d^2. Rearranging and dividing by 22 yields 2d2−d−1=02d^2 - d - 1 = 0, which factors as (2d+1)(d−1)=0(2d + 1)(d - 1) = 0. Since the problem specifies that d≠1d \neq 1, the only valid solution is d=−12d = -\frac{1}{2}.

Step-by-Step Solution

1
Write the formulas for the third term of both the arithmetic and geometric sequences.
For the arithmetic sequence: a3=a1+(3−1)d=2+2da_3 = a_1 + (3 - 1)d = 2 + 2d. For the geometric sequence: g3=g1⋅r3−1=4r2g_3 = g_1 \cdot r^{3 - 1} = 4r^2.
This establishes the algebraic expressions for the third terms using the given first terms.
2
Substitute dd for rr and set the two expressions equal to each other.
Since d=rd = r, we substitute dd into the geometric term to get g3=4d2g_3 = 4d^2. Setting them equal gives 2+2d=4d22 + 2d = 4d^2.
The problem states that the third terms are equal and that the common difference equals the common ratio.
3
Solve the quadratic equation 4d2−2d−2=04d^2 - 2d - 2 = 0 for dd.
Divide the equation by 22 to get 2d2−d−1=02d^2 - d - 1 = 0. Factoring this gives (2d+1)(d−1)=0(2d + 1)(d - 1) = 0. The roots are d=−12d = -\frac{1}{2} and d=1d = 1. Since the problem specifies d≠1d \neq 1, we have d=−12d = -\frac{1}{2}.
Solving the quadratic equation gives the possible values of the common difference, and the constraint rules out d=1d = 1.

Key Concept

Relating arithmetic and geometric sequence terms and solving the resulting quadratic equation.

Alternative Method

Instead of factoring, the quadratic formula can be used to solve 2d2−d−1=02d^2 - d - 1 = 0: d=−(−1)±(−1)2−4(2)(−1)2(2)=1±94=1±34d = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(2)(-1)}}{2(2)} = \frac{1 \pm \sqrt{9}}{4} = \frac{1 \pm 3}{4}. This yields d=1d = 1 and d=−12d = -\frac{1}{2}. Since the problem specifies d≠1d \neq 1, the only valid solution is d=−12d = -\frac{1}{2}.
Estimated Time:1m 30s
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Topics

Arithmetic and Geometric Sequences and Series

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