Geometric Sequences: Practice Problems with Step-by-Step Solutions

Question: Find the terms \(a_2\), \(a_3\), \(a_4\), and \(a_5\) of a geometric sequence if \(a_1 = 10\) and the common ratio \(r = -1\).

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Solution:

Use the recurrence definition \(a_n = a_{n-1} \times r\):

\[ a_2 = a_1 \times r = 10 \times (-1) = -10 \] \[ a_3 = a_2 \times r = -10 \times (-1) = 10 \] \[ a_4 = a_3 \times r = 10 \times (-1) = -10 \] \[ a_5 = a_4 \times r = -10 \times (-1) = 10 \]

Question: Find the 10th term of a geometric sequence if \(a_1 = 45\) and the common ratio \(r = 0.2\).

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Solution:

Use the formula \(a_n = a_1 \times r^{\,n-1}\):

\[ a_{10} = 45 \times (0.2)^{9} \] \[ a_{10} = 45 \times (0.2)^9 \approx 2.304 \times 10^{-5} \]

Question: Find \(a_{20}\) of a geometric sequence whose first terms are \(-\tfrac{1}{2}, \tfrac{1}{4}, -\tfrac{1}{8}, \tfrac{1}{16}, \dots\)

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Solution:

First, find the common ratio \(r\):

\[ r = \frac{a_2}{a_1} = \frac{\tfrac{1}{4}}{-\tfrac{1}{2}} = -\frac{1}{2} \]

Verify with the next terms:

\[ r = \frac{a_3}{a_2} = \frac{-\tfrac{1}{8}}{\tfrac{1}{4}} = -\frac{1}{2} \]

Now use \(a_n = a_1 r^{\,n-1}\):

\[ a_{20} = \left(-\frac{1}{2}\right) \times \left(-\frac{1}{2}\right)^{19} = \left(-\frac{1}{2}\right)^{20} = \frac{1}{2^{20}} \]

Question: Given \(a_{10} = \tfrac{3}{512}\) and \(a_{15} = \tfrac{3}{16384}\) of a geometric sequence, find the exact value of \(a_{30}\).

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Solution:

Write equations for both terms:

\[ a_{10} = a_1 r^{9} = \frac{3}{512} \] \[ a_{15} = a_1 r^{14} = \frac{3}{16384} \]

Divide \(a_{15}\) by \(a_{10}\) to eliminate \(a_1\):

\[ \frac{a_{15}}{a_{10}} = r^{5} = \frac{3/16384}{3/512} = \frac{512}{16384} = \frac{1}{32} \]

Taking the fifth root gives \(r = \tfrac{1}{2}\). Substitute back into the \(a_{10}\) equation:

\[ \frac{3}{512} = a_1 \left(\frac{1}{2}\right)^9 = \frac{a_1}{512} \implies a_1 = 3 \]

Finally, find \(a_{30}\):

\[ a_{30} = a_1 r^{29} = 3 \left(\frac{1}{2}\right)^{29} = \frac{3}{536870912} \]

Question: Find the sum \(S = \displaystyle\sum_{k=1}^{6} 3^{\,k - 1}\).

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Solution:

Method 1 (Direct expansion):

\[ S = 3^0 + 3^1 + 3^2 + 3^3 + 3^4 + 3^5 \] \[ S = 1 + 3 + 9 + 27 + 81 + 243 = 364 \]

Method 2 (Using the formula): The sequence has \(a_1 = 1\), \(r = 3\), \(n = 6\). Using \(S_n = a_1 \dfrac{1 - r^n}{1 - r}\):

\[ S_6 = 1 \cdot \frac{1 - 3^{6}}{1 - 3} = \frac{1 - 729}{-2} = \frac{-728}{-2} = 364 \]

Question: Find the sum \(S = \displaystyle\sum_{i=1}^{10} 8 \times \left(\tfrac{1}{4}\right)^{\,i - 1}\).

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Solution:

The terms form a geometric sequence:

\[ 8, \; 8\left(\tfrac{1}{4}\right)^1, \; 8\left(\tfrac{1}{4}\right)^2, \; \dots, \; 8\left(\tfrac{1}{4}\right)^9 \]

So \(a_1 = 8\), \(r = \tfrac{1}{4}\), \(n = 10\). Apply the finite sum formula:

\[ S_{10} = 8 \cdot \frac{1 - (1/4)^{10}}{1 - 1/4} = 8 \cdot \frac{1 - (1/4)^{10}}{3/4} = \frac{32}{3}\left(1 - (1/4)^{10}\right) \] \[ S_{10} \approx 10.67 \quad \text{(rounded to 2 decimal places)} \]

Question: Write the rational number \(5.31313131\dots\) as the ratio of two integers.

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Solution:

Express the repeating decimal as an infinite sum:

\[ 5.313131\dots = 5 + 0.31 + 0.0031 + 0.000031 + \dots \]

The fractional part \(0.31 + 0.0031 + 0.000031 + \dots\) is a geometric series with \(a_1 = 0.31\) and \(r = 0.01\). Apply \(S = \dfrac{a_1}{1 - r}\):

\[ S = \frac{0.31}{1 - 0.01} = \frac{0.31}{0.99} = \frac{31}{99} \]

Add the integer part back:

\[ 5.313131\dots = 5 + \frac{31}{99} = \frac{495}{99} + \frac{31}{99} = \frac{526}{99} \]

Try solving these exercises. Answers are provided below.

1. Find \(a_{20}\) given that \(a_3 = \tfrac{1}{2}\) and \(a_5 = 8\).
2. Find \(a_{30}\) given that the first terms of a geometric sequence are \(-2, 1, -\tfrac{1}{2}, \tfrac{1}{4}, \dots\)
3. Find \(r\) given that \(a_1 = 10\) and \(a_{20} = 10^{-18}\).
4. Write \(0.9717171\dots\) as a ratio of two positive integers.

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Answers:

• a) \(a_{20} = \pm 2^{33}\)
  (from \(r^2 = 16 \implies r = \pm 4\); substituting into \(a_{20} = a_3 r^{17}\) gives \(\pm 2^{33}\))
• b) \(a_{30} = \dfrac{1}{2^{28}}\)
• c) \(r = 0.1\)
• d) \(\dfrac{481}{495}\)