The second term of a geometric series is \(^{-2}/_3\) and its sum to infinity is...

The second term of a geometric series is \(^{-2}/_3\) and its sum to infinity is \(^3/_2\). Find its common ratio.

\(T_2 = \frac{-2}{3};S_\infty \frac {3}{2}\)

\(T_n = ar^n - 1\)

∴ \(T_2 = ar = \frac{-2}{3}\)---eqn.(i)

\(S_\infty = \frac{a}{1 - r} = \frac{3}{2}\)---eqn.(ii)

= 2a = 3(1 - r)

= 2a = 3 - 3r

∴ a = \(\frac{3 - 3r}{2}\)

Substitute \(\frac{3 - 3r}{2}\) for a in eqn.(i)

= \(\frac{3 - 3r}{2} \times r = \frac{-2}{3}\)

= \(\frac{3r - 3r^2}{2} = \frac{-2}{3}\)

= 3(3r - 3r\(^2\)) = -4

= 9r - 9r\(^2\) = -4

= 9r\(^2\) - 9r - 4 = 0

= 9r\(^2\) - 12r + 3r - 4 = 0

= 3r(3r - 4) + 1(3r - 4) = 0

= (3r - 4)(3r + 1) = 0

∴ r = \(\frac{4}{3} or - \frac{1}{3}\)

For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1.

∴ r = -\(^1/_3\) (since |-\(^1/_3\)| < 1)

There is an explanation video available below.

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