Standard High contrast

Sequences & Series

Question 4 of 8

Geometric Progression, Series & Sums

Question

The sum of the first three terms of a GP is \(-\dfrac{3}{2}\) and the sum to infinity of the GP is \(\dfrac{3}{52}\). If the GP has a positive common ratio \(r\), find \(r\) and the first term

Show solution Hide solution Fully worked — 14 steps

Note: solving the given conditions formally leads to \(r = 3\). But a geometric progression only has a sum to infinity when \(\lvert r \rvert \lt 1\), so no progression with \(r = 3\) has a sum to infinity at all — the conditions of this question are contradictory, and no geometric progression satisfies them. The working below shows the formal algebra.

  1. Write down key terms
    \[ S_3 = -\frac{3}{2} \quad\text{and}\quad S_\infty = \frac{3}{52} \]
  2. Substitute \(S_3 = -\frac{3}{2}\) into \(S_3 = \frac{a_1(1-r^3)}{1-r}\)
    \[ -\frac{3}{2} = \frac{a_1(1-r^3)}{1-r} \]
  3. Multiply both sides by \(\frac{(1-r)}{(1-r^3)}\) to find \(a_1\)
    \[ -\frac{3(1-r)}{2(1-r^3)} = a_1 \qquad [1] \]
  4. Substitute \(S_\infty = \frac{3}{52}\) into \(S_\infty = \frac{a_1}{1-r}\)
    \[ \frac{3}{52} = \frac{a_1}{1-r} \]
  5. Multiply both sides by \(1 - r\) to find \(a_1\)
    \[ \frac{3(1-r)}{52} = a_1 \qquad [2] \]
  6. Equate equations [1] and [2]
    \[ \frac{3(1-r)}{52} = -\frac{3(1-r)}{2(1-r^3)} \]
  7. Multiply both sides by \(-\frac{52(1-r^3)}{3(1-r)}\)
    \[ -(1 - r^3) = 26 \]
  8. Expand the bracket
    \[ -1 + r^3 = 26 \]
  9. Add 1 to both sides
    \[ r^3 = 26 + 1 \]
    \[ r^3 = 27 \]
  10. Take the cube root of both sides
    \[ r = 3 \]
  11. Substitute \(r = 3\) into equation [2]
    \[ \frac{3(1-3)}{52} = a_1 \]
  12. Evaluate \(3(1-3)\)
    \[ \frac{-6}{52} = a_1 \]
  13. Simplify \(\frac{-6}{52}\)
    \[ -\frac{3}{26} = a_1 \]
  14. State the answer

    Hence \(r = 3\) and \(a_1 = -\frac{3}{26}\): formally, the common ratio is three and the first term is minus three twenty-sixths — but see the note above: these values cannot belong to a progression with a sum to infinity.