The second term of a geometric series is -2 3 a... - JAMB Mathematics 2023 Question

Given \(T_2 = -\frac{2}{3}\) and \(S_\infty = \frac{3}{2}\) for a geometric series, let's find the common ratio \(r\).

The formula for the second term (\(T_2\)) in a geometric series is given by \(T_2 = ar\), where \(a\) is the first term and \(r\) is the common ratio.

We are given \(T_2 = -\frac{2}{3}\), so \(ar = -\frac{2}{3}\) ---equation (i).

The formula for the sum to infinity (\(S_\infty\)) in a geometric series is given by \(S_\infty = \frac{a}{1 - r}\).

We are given \(S_\infty = \frac{3}{2}\), so \(\frac{a}{1 - r} = \frac{3}{2}\) ---equation (ii).

Now, let's solve for \(a\) in terms of \(r\) using equation (i):

\[ar = -\frac{2}{3}\]

\[a = -\frac{2}{3r}\] ---equation (iii).

Substitute equation (iii) into equation (ii):

\[\frac{-\frac{2}{3r}}{1 - r} = \frac{3}{2}\]

Solve for \(r\):

\[-\frac{2}{3r(1 - r)} = \frac{3}{2}\]

Cross-multiply:

\[-4 = 9r(1 - r)\]

Expand and rearrange:

\[9r^2 - 9r - 4 = 0\]

Now, solve this quadratic equation. The solutions for \(r\) are:

\[r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]

where \(a = 9\), \(b = -9\), and \(c = -4\).

\[r = \frac{9 \pm \sqrt{81 + 144}}{18}\]

\[r = \frac{9 \pm \sqrt{225}}{18}\]

\[r = \frac{9 \pm 15}{18}\]

Two possible values for \(r\) are:

\[r_1 = \frac{24}{18} = \frac{4}{3}\]

\[r_2 = \frac{-6}{18} = -\frac{1}{3}\]

For a geometric series, the common ratio (\(|r|\)) must be less than 1 for the series to converge to a finite value. Therefore, \(r = -\frac{1}{3}\) is the valid solution.

Hence, the correct answer is \(-\frac{1}{3}\)