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Sequences & Series

Question 2 of 8

Geometric Progression, Series & Sums

Question

Find the number of terms in the Geometric Progression \(2,\ -1,\ \dfrac{1}{2},\ \ldots,\ \dfrac{1}{128}\)

Show solution Hide solution Fully worked — 8 steps
  1. Write down key terms
    \[ a_1 = 2;\quad a_2 = -1;\quad a_3 = \frac{1}{2};\quad a_n = \frac{1}{128} \]
  2. Find \(r\) using \(r = \frac{a_2}{a_1}\)
    \[ r = \frac{a_2}{a_1} = \frac{-1}{2} = -\frac{1}{2} \]
  3. Substitute the values of \(a_1\), \(a_n\) and \(r\) into \(a_n = a_1 r^{n-1}\) to find \(n\)
    \[ \frac{1}{128} = (2)\left(-\frac{1}{2}\right)^{n-1} \]
  4. Divide both sides by 2
    \[ \frac{1}{256} = \left(-\frac{1}{2}\right)^{n-1} \]
  5. Change \(\frac{1}{256}\) to its exponential form whose base \(= r\)
    \[ \left(-\frac{1}{2}\right)^{8} = \left(-\frac{1}{2}\right)^{n-1} \]
  6. Equate the indices since they both have the same base
    \[ 8 = n - 1 \]
  7. Add 1 to both sides
    \[ 8 + 1 = n \]
  8. Add 8 and 1
    \[ 9 = n \]

    Hence, \(\frac{1}{128}\) is the 9th term: one over one hundred and twenty-eight is the ninth term of the progression.