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Sequences & Series

Question 3 of 8

Geometric Progression, Series & Sums

Question

The 5th term of a GP is \(-48\) and the 7th term is \(-12\). Find the values of the common ratio and the first term

Show solution Hide solution Fully worked — 11 steps
  1. Write down key terms
    \[ a_5 = -48;\quad a_7 = -12 \]
  2. Substitute \(a_5 = -48\) into \(a_5 = a_1 r^{5-1}\)
    \[ -48 = a_1 r^{5-1} \]
  3. Divide both sides by \(r^4\) to solve for \(a_1\)
    \[ -\frac{48}{r^4} = a_1 \qquad [1] \]
  4. Substitute \(a_7 = -12\) into \(a_7 = a_1 r^{7-1}\)
    \[ -12 = a_1 r^{7-1} \]
  5. Divide both sides by \(r^6\) to solve for \(a_1\)
    \[ -\frac{12}{r^6} = a_1 \qquad [2] \]
  6. Equate equations [1] and [2]
    \[ -\frac{48}{r^4} = -\frac{12}{r^6} \]
  7. Multiply both sides by \(-\frac{r^6}{48}\) to solve for \(r\)
    \[ r^2 = \frac{1}{4} \]
  8. Take the square root of both sides
    \[ r = \pm\frac{1}{2} \]
  9. Substitute the values of \(r\) into equation [1] to get \(a_1\)
    \[ -\frac{48}{\left(\frac{1}{2}\right)^{4}} = a_1 \quad\text{and}\quad -\frac{48}{\left(-\frac{1}{2}\right)^{4}} = a_1 \]
  10. Evaluate \(\left(-\frac{1}{2}\right)^{4} = \frac{1}{16}\)
    \[ -\frac{48}{\frac{1}{16}} = a_1 \quad\text{and}\quad -\frac{48}{\frac{1}{16}} = a_1 \]
  11. State the common ratio and the first term

    Dividing \(-48\) by \(\frac{1}{16}\) gives \(-768\) in both cases:

    \[ -768 = a_1 \quad\text{and}\quad -768 = a_1 \]

    Hence, the common ratio is \(\pm\frac{1}{2}\) and \(a_1 = -768\): the common ratio is plus or minus one half, and the first term is minus seven hundred and sixty-eight.