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Difficulty: Very hardArithmetic and Geometric Sequences and Series

An arithmetic sequence a1,a2,a3,…a_1, a_2, a_3, \dots has a first term a1=aa_1 = a and a non-zero common difference dd. A geometric sequence g1,g2,g3,…g_1, g_2, g_3, \dots has a first term g1=ag_1 = a and a common ratio r>1r > 1. The third term of the arithmetic sequence is equal to the second term of the geometric sequence (a3=g2a_3 = g_2), and the seventh term of the arithmetic sequence is equal to the third term of the geometric sequence (a7=g3a_7 = g_3). If the sum of the first five terms of the arithmetic sequence is 150, what is the value of the fifth term of the geometric sequence, g5g_5?

  1. 240Answer
  2. B
    480
  3. C
    800
  4. D
    96
  5. E
    810,000

Answer

The fifth term of the geometric sequence is 240.
The correct answer is 240. The term relationships a3=g2a_3 = g_2 and a7=g3a_7 = g_3 translate to a+2d=ara + 2d = ar and a+6d=ar2a + 6d = ar^2. Expressing the first equation as 2d=a(r−1)2d = a(r - 1) and substituting it into the second equation yields a+3a(r−1)=ar2a + 3a(r - 1) = ar^2. Dividing by aa gives the quadratic equation r2−3r+2=0r^2 - 3r + 2 = 0. Since r>1r > 1, we find r=2r = 2, which implies d=a/2d = a/2. The sum of the first five terms of the arithmetic sequence is 5(a+2d)=1505(a + 2d) = 150, which simplifies to a+2d=30a + 2d = 30. Substituting d=a/2d = a/2 gives 2a=302a = 30, so a=15a = 15. The fifth term of the geometric sequence is then g5=a⋅r4=15⋅24=240g_5 = a \cdot r^4 = 15 \cdot 2^4 = 240.

Step-by-Step Solution

1
Express the given sequence term relationships in terms of aa, dd, and rr.
a3=a+2da_3 = a + 2d, a7=a+6da_7 = a + 6d, g2=arg_2 = ar, and g3=ar2g_3 = ar^2.
To represent the equality of terms algebraically.
2
Solve the system of equations a+2d=ara + 2d = ar and a+6d=ar2a + 6d = ar^2 for the common ratio rr.
From the first equation, 2d=a(r−1)⇒d=a(r−1)22d = a(r - 1) \Rightarrow d = \frac{a(r - 1)}{2}. Substituting this into the second equation gives a+3a(r−1)=ar2a + 3a(r - 1) = ar^2. Dividing by aa (since a≠0a \neq 0) yields r2−3r+2=0r^2 - 3r + 2 = 0, which factors as (r−1)(r−2)=0(r-1)(r-2) = 0. Since r>1r > 1, we must have r=2r = 2. Thus, d=a/2d = a/2.
To find the relationship between the arithmetic common difference, the geometric common ratio, and the first term.
3
Use the sum of the first five terms of the arithmetic sequence to solve for aa.
S5=5(a+2d)=150⇒a+2d=30S_5 = 5(a + 2d) = 150 \Rightarrow a + 2d = 30. Substituting d=a/2d = a/2 gives a+a=30⇒2a=30⇒a=15a + a = 30 \Rightarrow 2a = 30 \Rightarrow a = 15.
To determine the numerical value of the first term.
4
Calculate the fifth term of the geometric sequence.
g5=a⋅r4=15⋅24=15⋅16=240g_5 = a \cdot r^4 = 15 \cdot 2^4 = 15 \cdot 16 = 240.
To find the requested term of the geometric sequence.

Key Concept

Solving systems of linear and exponential relationships using arithmetic and geometric sequence properties.
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Topics

Arithmetic and Geometric Sequences and Series

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