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Difficulty: HardArithmetic and Geometric Sequences and Series

An arithmetic sequence a1,a2,a3,…a_1, a_2, a_3, \dots with a positive common difference dd and a geometric sequence b1,b2,b3,…b_1, b_2, b_3, \dots with a positive common ratio rr both have the same positive first term (a1=b1>0a_1 = b_1 > 0). If the 3rd term of the arithmetic sequence equals the 3rd term of the geometric sequence (a3=b3a_3 = b_3), and the 7th term of the arithmetic sequence equals the 5th term of the geometric sequence (a7=b5a_7 = b_5), what is the value of rr?

  1. 2\sqrt{2}Answer
  2. B
    3\sqrt{3}
  3. C
    22
  4. D
    233\frac{2\sqrt{3}}{3}
  5. E
    222\sqrt{2}

Answer

2\sqrt{2}
Using the formulas for the nn-th terms, a3=a1+2da_3 = a_1 + 2d and b3=a1r2b_3 = a_1 r^2, giving 2d=a1(r2−1)2d = a_1(r^2 - 1). Substituting 6d=3a1(r2−1)6d = 3a_1(r^2 - 1) into a7=a1+6d=a1r4a_7 = a_1 + 6d = a_1 r^4 yields a1+3a1(r2−1)=a1r4a_1 + 3a_1(r^2 - 1) = a_1 r^4. Dividing by a1>0a_1 > 0 gives 1+3r2−3=r41 + 3r^2 - 3 = r^4, which simplifies to r4−3r2+2=0r^4 - 3r^2 + 2 = 0. Factoring gives (r2−1)(r2−2)=0(r^2 - 1)(r^2 - 2) = 0. Since d>0d > 0 and a1>0a_1 > 0, we must have r2>1r^2 > 1, eliminating r2=1r^2 = 1. Therefore, r2=2r^2 = 2, and since r>0r > 0, r=2r = \sqrt{2}.

Step-by-Step Solution

1
Express sequence terms in terms of first term a1a_1, common difference dd, and common ratio rr
a3=a1+2da_3 = a_1 + 2d, a7=a1+6da_7 = a_1 + 6d, b3=a1r2b_3 = a_1 r^2, and b5=a1r4b_5 = a_1 r^4
Apply the standard formulas for the nn-th term of arithmetic (an=a1+(n−1)da_n = a_1 + (n-1)d) and geometric (bn=b1rn−1b_n = b_1 r^{n-1}) sequences.
2
Set up equations based on given equality of terms
Equation 1: a1+2d=a1r2  ⟹  2d=a1(r2−1)a_1 + 2d = a_1 r^2 \implies 2d = a_1(r^2 - 1);
Equation 2: a1+6d=a1r4a_1 + 6d = a_1 r^4
Translate the given conditions a3=b3a_3 = b_3 and a7=b5a_7 = b_5 into algebraic relations.
3
Substitute 2d2d from Equation 1 into Equation 2
a1+3(2d)=a1+3a1(r2−1)=a1r4a_1 + 3(2d) = a_1 + 3a_1(r^2 - 1) = a_1 r^4
Express 6d6d as 3(2d)3(2d) to eliminate dd from the system.
4
Divide by a1a_1 (since a1>0a_1 > 0) and simplify to solve for rr
1+3r2−3=r4  ⟹  r4−3r2+2=0  ⟹  (r2−1)(r2−2)=01 + 3r^2 - 3 = r^4 \implies r^4 - 3r^2 + 2 = 0 \implies (r^2 - 1)(r^2 - 2) = 0
Reduce the equation to a quadratic in terms of r2r^2.
5
Determine the valid root for rr
r2=2  ⟹  r=2r^2 = 2 \implies r = \sqrt{2} (since r>0r > 0 and d>0d > 0 implies r2>1r^2 > 1)
Since d>0d > 0 and a1>0a_1 > 0, 2d=a1(r2−1)>02d = a_1(r^2 - 1) > 0, which requires r2>1r^2 > 1. Thus r2=1r^2 = 1 is rejected, leaving r2=2r^2 = 2.

Key Concept

Arithmetic and Geometric Sequences Alignment
Estimated Time:2m 0s
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Topics

Arithmetic and Geometric Sequences and Series

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