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Sequences & Series

Question 1 of 8

Geometric Progression, Series & Sums

Question

Write down the 11th term of the Geometric progression \(\dfrac{1}{3},\ \dfrac{1}{6},\ \dfrac{1}{12},\ \ldots\)

Show solution Hide solution Fully worked — 6 steps
  1. Write down key terms
    \[ a_1 = \frac{1}{3};\quad a_2 = \frac{1}{6};\quad a_3 = \frac{1}{12};\quad n = 11 \]
  2. Find the common ratio \(r\) using \(r = \frac{a_2}{a_1}\)
    \[ r = \frac{a_2}{a_1} = \frac{\frac{1}{6}}{\frac{1}{3}} = \frac{1}{2} \]
  3. Substitute \(n = 11\) into \(a_n = a_1 r^{n-1}\)
    \[ a_{11} = a_1 r^{11-1} \]
  4. Substitute \(a_1 = \frac{1}{3}\) and \(r = \frac{1}{2}\)
    \[ = \left(\frac{1}{3}\right)\left(\frac{1}{2}\right)^{10} \]
  5. Evaluate \(\left(\frac{1}{2}\right)^{10} = \frac{1}{1024}\)
    \[ = \left(\frac{1}{3}\right)\left(\frac{1}{1024}\right) \]
  6. Multiply \(\left(\frac{1}{3}\right)\) and \(\left(\frac{1}{1024}\right)\)
    \[ a_{11} = \frac{1}{3072} \]

    That is, the 11th term of the progression is one over three thousand and seventy-two.