Series, Progressions and the Binomial Expansion: Question 3

Worked solution

Part (a): Common ratio and first term

Let the amplitudes form a geometric progression with first term $a$ and common ratio $r$. The second term is $u_2 = ar = 18$ and the third term is $u_3 = ar^2 = 12$. Dividing the later term by the earlier one:

$r = \frac{u_3}{u_2} = \frac{12}{18} = \frac{2}{3}.$

Then, from $ar = 18$:

$a = \frac{18}{r} = 18 \div \frac{2}{3} = 18\times\frac{3}{2} = 27.$

Check: with $a=27$ and $r=\frac{2}{3}$, $u_2 = 27\times\frac{2}{3}=18$ ✓ and $u_3 = 18\times\frac{2}{3}=12$ ✓. Both given values are reproduced, confirming $a=27$ and $r=\frac{2}{3}$.

Part (b): Sum of the first 10 oscillations

$S_n = \frac{a(1-r^n)}{1-r}$

$S_{10} = \frac{27\left(1-\left(\frac{2}{3}\right)^{10}\right)}{1-\frac{2}{3}} = \frac{27\left(1-\frac{1024}{59049}\right)}{\frac{1}{3}} = 81\left(1-\frac{1024}{59049}\right).$

Since $\left(\frac{2}{3}\right)^{10} = \frac{1024}{59049} \approx 0.017342$:

$S_{10} \approx 81\times(1-0.017342) = 81\times0.982658 \approx 79.595.$

Check by adding the terms directly: $27, 18, 12, 8, 5.333, 3.556, 2.370, 1.580, 1.053, 0.702$. Summing these ten values also gives $79.595$, confirming the formula result.

So $S_{10} \approx 79.6$ mm (to $1$ decimal place).

Part (c): Convergence and sum to infinity

Since $|r| = \frac{2}{3} < 1$, each term is smaller in size than the one before it, so as more and more terms are added, the amount being added shrinks towards zero and the running total approaches a fixed limit, the progression converges.

$S_\infty = \frac{a}{1-r} = \frac{27}{1-\frac{2}{3}} = \frac{27}{\frac{1}{3}} = 27\times3 = 81.$

Check: the partial sum found in part (b) was $S_{10}\approx79.595$, only $81-79.595=1.405$ short of this limit. The tail sum from the 11th term onward is itself a geometric series with first term $u_{11}=u_{10}\times r \approx0.702\times\frac{2}{3}\approx0.468$ and the same ratio $\frac{2}{3}$, so its sum is $\frac{0.468}{1-\frac{2}{3}}\approx1.405$, exactly matching the shortfall, so the two calculations agree.

Final answers

• (a) Common ratio $r = \boxed{\dfrac{2}{3}}$, first term $a = \boxed{27}$ mm.
• (b) $S_{10} \approx \boxed{79.6}$ mm.
• (c) The progression converges because $|r|=\frac{2}{3}<1$; sum to infinity $S_\infty = \boxed{81}$ mm.