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Geometric Sequences

12 practice questions with full worked solutions. Free, no account needed.

Problems & worked solutions

Problem #0282 International

Problem 1 — Geometric Sequences

The numbers 4,12,36,108,…4, 12, 36, 108, \ldots form a geometric progression. The common ratio qq equals:

Show answer & worked solution
  1. A. 22
  2. B. 13\dfrac{1}{3}
  3. C. 33✓ correct
  4. D. 44

q=b2b1=124=3q = \dfrac{b_2}{b_1} = \dfrac{12}{4} = 3.

Problem #0281 International

Problem 2 — Geometric Sequences

In the geometric progression (bn)n≥1(b_n)_{n \ge 1}, b1=2b_1 = 2 and q=3q = 3. Determine b5b_5.

Show answer & worked solution
  1. A. 5454
  2. B. 8181
  3. C. 162162✓ correct
  4. D. 486486

b5=b1⋅q4=2⋅34=2⋅81=162b_5 = b_1 \cdot q^{4} = 2 \cdot 3^{4} = 2 \cdot 81 = 162.

Problem #0006 International

Problem 3 — Series

What is the sum of the infinite geometric series 1+12+14+⋯1 + \tfrac{1}{2} + \tfrac{1}{4} + \cdots?

Show answer & worked solution
  1. A. 32\dfrac{3}{2}
  2. B. 22✓ correct
  3. C. 52\dfrac{5}{2}
  4. D. Diverges

Geometric series with a=1a=1, r=12r=\tfrac{1}{2}. Sum =a1−r=11−1/2=2= \tfrac{a}{1-r} = \tfrac{1}{1-1/2} = 2.

Problem #0012 International

Problem 4 — Sum of Terms

Compute ∑k=052k\displaystyle\sum_{k=0}^{5} 2^k.

Show answer & worked solution
  1. A. 3131
  2. B. 6262
  3. C. 6363✓ correct
  4. D. 127127

With r=2r=2 and n=5n=5: 26−12−1=631=63\dfrac{2^6 - 1}{2-1} = \dfrac{63}{1} = 63. (Check: 1+2+4+8+16+32=631+2+4+8+16+32 = 63.)

Problem #0287 International

Problem 5 — Geometric Sequences

A bank deposit of $1,000\$1{,}000 earns annual compound interest at 10%10\%. After 33 years, the balance is closest to:

Show answer & worked solution
  1. A. $1,300\$1{,}300
  2. B. $1,320\$1{,}320
  3. C. $1,331\$1{,}331✓ correct
  4. D. $1,350\$1{,}350

A=1000⋅1.13=1000⋅1.331=$1,331A = 1000 \cdot 1.1^3 = 1000 \cdot 1.331 = \$1{,}331.

Problem #0285 International

Problem 6 — Geometric Sequences

In a geometric progression with b1=5b_1 = 5 and q=2q = 2, the term bn=320b_n = 320. Determine nn.

Show answer & worked solution
  1. A. 55
  2. B. 66
  3. C. 77✓ correct
  4. D. 88

5⋅2n−1=320⇒2n−1=64=26⇒n−1=6⇒n=75 \cdot 2^{n-1} = 320 \Rightarrow 2^{n-1} = 64 = 2^6 \Rightarrow n - 1 = 6 \Rightarrow n = 7.

Problem #0286 International

Problem 7 — Geometric Sequences

Compute the sum 1+12+14+18+⋯1 + \dfrac{1}{2} + \dfrac{1}{4} + \dfrac{1}{8} + \cdots.

Show answer & worked solution
  1. A. 32\dfrac{3}{2}
  2. B. 22✓ correct
  3. C. 52\dfrac{5}{2}
  4. D. ∞\infty

b1=1b_1 = 1, q=12q = \dfrac{1}{2}: S=11−1/2=2S = \dfrac{1}{1 - 1/2} = 2.

Problem #0283 International

Problem 8 — Geometric Sequences

Compute the sum of the first 1010 terms of the geometric progression 1,2,4,8,…1, 2, 4, 8, \ldots.

Show answer & worked solution
  1. A. 511511
  2. B. 512512
  3. C. 10231023✓ correct
  4. D. 10241024

S10=1⋅210−12−1=1023S_{10} = 1 \cdot \dfrac{2^{10} - 1}{2 - 1} = 1023.

Problem #0284 International

Problem 9 — Geometric Sequences

The numbers 33, xx, 2727 are in geometric progression (with positive ratio). Determine xx.

Show answer & worked solution
  1. A. 55
  2. B. 99✓ correct
  3. C. 1515
  4. D. 30\sqrt{30}

x2=3⋅27=81⇒x=9x^2 = 3 \cdot 27 = 81 \Rightarrow x = 9 (positive ratio).

Problem #0290 International

Problem 10 — Geometric Sequences

For the geometric progression 1,2,4,8,16,32,…1, 2, 4, 8, 16, 32, \ldots, the sum b2+b4+b6b_2 + b_4 + b_6 equals:

Show answer & worked solution
  1. A. 3636
  2. B. 4242✓ correct
  3. C. 4848
  4. D. 5656

b2=2b_2 = 2, b4=8b_4 = 8, b6=32b_6 = 32. The sum is 2+8+32=422 + 8 + 32 = 42. (These three form a GP with first term 22 and ratio 44.)

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