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Sequences & Series

Geometric Progression, Series & Sums

Introduction

A geometric sequence is a sequence such that any element after the first is obtained by multiplying the preceding element by a constant called the common ratio which is denoted by \(r\). The common ratio (\(r\)) is obtained by dividing any term by the preceding term, i.e.,

\[ r = \frac{a_2}{a_1} = \frac{a_3}{a_2} = \cdots = \frac{a_n}{a_{n-1}} \]
\(r\)common ratio
\(a_1\)first term
\(a_2\)second term
\(a_3\)third term
\(a_{n-1}\)the term before the \(n\)th term
\(a_n\)the \(n\)th term

The geometric sequence is sometimes called the geometric progression or GP, for short.

For example, the sequence 1, 3, 9, 27, 81 is a geometric sequence. Note that after the first term, the next term is obtained by multiplying the preceding element by 3.

The geometric sequence has its sequence formation: \(a_1,\ a_1 r,\ a_1 r^2, \ldots,\ a_1 r^{n-1},\ a_1 r^n\).

To find the \(n\)th term of a geometric sequence we use the formula:

\[ a_n = a_1 r^{n-1} \]
\(r\)common ratio
\(a_1\)first term
\(a_{n-1}\)the term before the \(n\)th term
\(n\)number of terms

Sum of Terms in a Geometric Progression

Finding the sum of terms in a geometric progression is easily obtained by applying the formulas:

  • \(n\)th partial sum of a geometric sequence

    \[ S_n = \frac{a_1(1 - r^n)}{1 - r}, \qquad r \neq 1 \]

  • Sum to infinity

    \[ S_\infty = \sum_{n=1}^{\infty} a r^{n-1} = \frac{a_1}{1-r}, \quad -1 \lt r \lt 1 \]
\(S_n\)sum of GP with \(n\) terms
\(S_\infty\)sum of GP with infinitely many terms
\(a_1\)the first term
\(r\)common ratio
\(n\)number of terms

Examples of Common Problems to Solve

Write down a specific term in a Geometric Progression

Question

Write down the 8th term in the Geometric Progression \(1,\ 3,\ 9,\ \ldots\)

Solution
  1. Step 1 — Write down key terms
    \[ a_1 = 1;\quad a_2 = 3;\quad a_3 = 9;\quad n = 8 \]
  2. Step 2 — Find the common ratio \(r\) using \(r = \frac{a_2}{a_1}\)
    \[ r = \frac{a_2}{a_1} = \frac{3}{1} = 3 \]
  3. Step 3 — Substitute \(n = 8\) into \(a_n = a_1 r^{n-1}\)
    \[ a_8 = a_1 r^{8-1} \]
  4. Step 4 — Substitute \(a_1 = 1\) and \(r = 3\)
    \[ = (1)(3)^7 \]
  5. Step 5 — Multiply \((1)\) and \((3)^7\)
    \[ = (3)^7 \]
  6. Step 6 — Simplify \((3)^7 = 2187\)
    \[ a_8 = 2187 \]

    That is, the 8th term of the progression is two thousand, one hundred and eighty-seven.

Finding the number of terms in a Geometric Progression

Question

Find the number of terms in the geometric progression \(6,\ 12,\ 24,\ \ldots,\ 1536\)

Solution
  1. Step 1 — Write down key terms
    \[ a_1 = 6;\quad a_2 = 12;\quad a_3 = 24;\quad a_n = 1536 \]
  2. Step 2 — Find \(r\) using \(r = \frac{a_2}{a_1}\)
    \[ r = \frac{a_2}{a_1} = \frac{12}{6} = 2 \]
  3. Step 3 — Substitute the values of \(a_1\), \(a_n\) and \(r\) into \(a_n = a_1 r^{n-1}\) to find \(n\)
    \[ 1536 = (6)(2)^{n-1} \]
  4. Step 4 — Divide both sides by 6
    \[ 256 = (2)^{n-1} \]
  5. Step 5 — Change 256 to its exponential form whose base \(= r\)
    \[ 2^8 = 2^{n-1} \]
  6. Step 6 — Equate the indices since they both have the same base
    \[ 8 = n - 1 \]
  7. Step 7 — Add 1 to both sides
    \[ 8 + 1 = n \]
  8. Step 8 — Add 8 and 1
    \[ 9 = n \]

    Hence, 1536 is the 9th term: one thousand, five hundred and thirty-six is the ninth term of the progression.

Finding the sum of a Geometric Series

Question

Find the sum of the geometric series \(-2,\ \dfrac{1}{2},\ -\dfrac{1}{8},\ \ldots,\ -\dfrac{1}{32768}\)

Solution
  1. Step 1 — Write down key terms
    \[ a_1 = -2;\quad a_2 = \frac{1}{2};\quad a_3 = -\frac{1}{8};\quad a_n = -\frac{1}{32768} \]
  2. Step 2 — Find \(r\) using \(r = \frac{a_2}{a_1}\)
    \[ r = \frac{a_2}{a_1} = \frac{\frac{1}{2}}{-2} = -\frac{1}{4} \]
  3. Step 3 — Substitute the values of \(a_1\), \(a_n\) and \(r\) into \(a_n = a_1 r^{n-1}\) to find \(n\)
    \[ -\frac{1}{32768} = (-2)\left(-\frac{1}{4}\right)^{n-1} \]
  4. Step 4 — Divide both sides by \(-2\)
    \[ \frac{1}{65536} = \left(-\frac{1}{4}\right)^{n-1} \]
  5. Step 5 — Change \(\frac{1}{65536}\) to its exponential form whose base \(= r\)
    \[ \left(-\frac{1}{4}\right)^{8} = \left(-\frac{1}{4}\right)^{n-1} \]
  6. Step 6 — Equate the indices since they both have the same base
    \[ 8 = n - 1 \]
  7. Step 7 — Add 1 to both sides
    \[ 8 + 1 = n \]
  8. Step 8 — Add 8 and 1
    \[ 9 = n \]
  9. Step 9 — Substitute the values of \(a_1\), \(r\) and \(n\) into \(S_n = \frac{a_1(1-r^n)}{1-r}\)
    \[ S_9 = \frac{-2\left(1 - \left(-\frac{1}{4}\right)^{9}\right)}{1 - \left(-\frac{1}{4}\right)} \]
  10. Step 10 — Evaluate \(\left(-\frac{1}{4}\right)^9\) then subtract \(-\frac{1}{4}\) from 1
    \[ = \frac{-2\left(1 - \left(-\frac{1}{262144}\right)\right)}{\frac{5}{4}} \]
  11. Step 11 — Multiply \(-2\) by \(\frac{262145}{262144}\)
    \[ = \frac{-\frac{262145}{131072}}{\frac{5}{4}} \]
  12. Step 12 — Divide \(-\frac{262145}{131072}\) by \(\frac{5}{4}\)
    \[ S_9 = -\frac{52429}{32768} \]

    That is, the sum of the first nine terms is minus fifty-two thousand, four hundred and twenty-nine over thirty-two thousand, seven hundred and sixty-eight.

Finding the sum of a Geometric Series to Infinity

Question

Work out the sum \(\sum_{r=1}^{\infty} \left(\dfrac{1}{3}\right)^{r}\)

Solution
  1. Step 1 — Expand the given sum
    \[ \sum_{r=1}^{\infty} \left(\frac{1}{3}\right)^{r} = \left(\frac{1}{3}\right)^{1} + \left(\frac{1}{3}\right)^{2} + \left(\frac{1}{3}\right)^{3} + \ldots + \left(\frac{1}{3}\right)^{n} + \ldots \]
    \[ = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \ldots + \left(\frac{1}{3}\right)^{n} + \ldots \]
  2. Step 2 — Solve for \(r\) using \(r = \frac{a_2}{a_1}\)
    \[ r = \frac{a_2}{a_1} = \frac{\frac{1}{9}}{\frac{1}{3}} = \frac{1}{3} \]
  3. Step 3 — Substitute \(a_1 = \frac{1}{3}\) and \(r = \frac{1}{3}\) into \(S_\infty = \frac{a_1}{1-r}\)
    \[ S_\infty = \frac{\frac{1}{3}}{1 - \frac{1}{3}} \]
  4. Step 4 — Subtract \(\frac{1}{3}\) from 1
    \[ = \frac{\frac{1}{3}}{\frac{2}{3}} \]
  5. Step 5 — Divide \(\frac{1}{3}\) by \(\frac{2}{3}\)
    \[ S_\infty = \frac{1}{2} \]

    That is, the sum to infinity of the series is one half.

Converting a Recurring Decimal to a Fraction

Decimals that occur in repetition infinitely or are repeated in period are called recurring decimals.

For example, 0.22222222... is a recurring decimal because the number 2 is repeated infinitely.

The recurring decimal 0.22222222... can be written as \(0.\dot{2}\).

Another example is 0.234523452345... is a recurring decimal because the number 2345 is repeated periodically.

Thus, it can be written as \(0.\dot{2}\dot{3}\dot{4}\dot{5}\) or it can also be expressed in fractions.

Question

Express \(0.\dot{7}\dot{2}\) as a fraction in its lowest terms

Solution
  1. Step 1 — Equivalent of the recurring decimal
    \[ 0.\dot{7}\dot{2} = 0.727272\ldots \]
  2. Step 2 — Let \(x\) stand for the recurring decimal
    \[ \text{Let } x = 0.727272\ldots \qquad [1] \]
  3. Step 3 — Multiply \(x\) and \(0.727272\ldots\) by 100
    \[ 100x = 72.7272\ldots \qquad [2] \]
  4. Step 4 — Subtract the equation [1] from [2]
    \[ 100x - x = 72.7272\ldots - 0.727272\ldots \]
  5. Step 5 — Evaluate both sides of the equation
    \[ 99x = 72 \]
  6. Step 6 — Divide both sides by 99
    \[ x = \frac{72}{99} \]
  7. Step 7 — Simplify \(\frac{72}{99}\)
    \[ x = \frac{8}{11} \]

    That is, as a fraction in its lowest terms, the recurring decimal is eight elevenths.

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