Series, Progressions and the Binomial Expansion: Question 8

Worked solution

Part (a): Finding the common ratio

The sum to infinity of a geometric progression with first term $a$ and common ratio $r$ (where $|r|<1$) is

$S_\infty = \frac{a}{1-r}.$

Here $a=8$ and $S_\infty=20$, so

$20 = \frac{8}{1-r} \implies 1-r = \frac{8}{20} = 0.4 \implies r = 1-0.4 = 0.6.$

Check: $|r|=0.6<1$, so a sum to infinity genuinely exists, and substituting back, $\dfrac{8}{1-0.6}=\dfrac{8}{0.4}=20$ ✓, matching the given value.

Part (b): The 4th term

$u_4 = ar^3 = 8\times(0.6)^3 = 8\times0.216 = 1.728.$

Check: listing the terms directly, $u_1=8$, $u_2=8(0.6)=4.8$, $u_3=4.8(0.6)=2.88$, $u_4=2.88(0.6)=1.728$. Building up term by term gives the same value.

Part (c): Least $n$ for which $S_n > 19.9$

$S_n = \frac{a(1-r^n)}{1-r} = \frac{8\bigl(1-0.6^n\bigr)}{0.4} = 20\bigl(1-0.6^n\bigr).$

We need

$20\bigl(1-0.6^n\bigr) > 19.9 \implies 1-0.6^n > 0.995 \implies 0.6^n < 0.005.$

Taking natural logs of both sides (and reversing the inequality, since $\ln(0.6)<0$):

$n\ln(0.6) < \ln(0.005) \implies n > \frac{\ln(0.005)}{\ln(0.6)}.$

Using $\ln(0.005)\approx-5.2983$ and $\ln(0.6)\approx-0.5108$:

$n > \frac{-5.2983}{-0.5108} \approx 10.37.$

Since $n$ must be a positive integer, the least candidate is $n=11$; we verify this directly rather than relying on the rounded boundary:

$0.6^{10} \approx 0.0060466, \qquad S_{10} = 20(1-0.0060466) \approx 19.879,$

$0.6^{11} \approx 0.0036280, \qquad S_{11} = 20(1-0.0036280) \approx 19.927.$

Check: $S_{10}\approx19.879 < 19.9$, so $10$ terms are not enough, while $S_{11}\approx19.927 > 19.9$, so $11$ terms are enough. This confirms the least value of $n$ is $11$.

Final answers

• (a) Common ratio $r = \boxed{0.6}$.
• (b) $4$th term $u_4 = \boxed{1.728}$.
• (c) The least value of $n$ for which $S_n>19.9$ is $\boxed{11}$.