4.6 Geometric sequences and series, and the sum to infinity — Practise with help

1. $3 \times 2^3 = 3 \times 8 = 24$.
2. $u_2 = 0.5(8) = 4$, so $u_3 = 0.5(4) = 2$.
3. The sequence is arithmetic with $a = 2$ and $d = 4$, so $S_5 = \frac{5}{2}(2(2) + 4(4)) = \frac{5}{2}(20) = 50$.

If question 2 was hard, revisit recurrences in Section 4.3 before you continue. A geometric sequence is exactly a recurrence that multiplies by a fixed number.

Find the common ratio by dividing a term by the one before it. Then use $u_n = a r^{n-1}$.

Here $a = 5$ and $r = 15 \div 5 = 3$. Substitute $n = 6$, so the power of $3$ is $5$.

The 6th term is $1215$. The power is $5$, one less than the position.

Use $S_n = \dfrac{a(1 - r^n)}{1 - r}$ with $a = 2$ and $r = 3$.

Substitute $n = 8$: the numerator is $2(1 - 3^8)$ and the denominator is $1 - 3 = -2$. Work out $3^8 = 6561$ first.

The sum of the first 8 terms is $6560$. Dividing by a negative denominator and multiplying by a negative numerator is what turns the answer positive.

Find $r$ first, and check that $|r| < 1$ before you use the formula.

$r = \frac{6}{24} = \frac{1}{4}$, which is less than $1$ in size. Then use $S_\infty = \frac{a}{1 - r}$ with $a = 24$.

The series settles on $32$. Check the reason: dividing by $\frac{3}{4}$ is the same as multiplying by $\frac{4}{3}$, and the terms shrink quickly.

The third term is $a r^2$. Set it equal to $20$ and solve for $r$.

$5r^2 = 20$, so $r^2 = 4$. Take the positive square root, because the question says the ratio is positive.

The sequence is $5, 10, 20, \dots$, and the third term is indeed $20$. Without the positive condition, $r = -2$ would also fit.

Check the condition first: $|0.9| < 1$, so the series converges.

$S_\infty = \dfrac{100}{1 - 0.9} = \dfrac{100}{0.1}$.

The total can never reach $1000$ exactly, but it gets as close as you like. The large ratio is why the total is ten times the first term.

Split the decimal into $0.4 + 0.04 + 0.004 + \dots$ and read off the first term and the common ratio.

Here $a = 0.4$ and $r = 0.1$, so use $S_\infty = \frac{a}{1 - r}$.

Check: $4 \div 9 = 0.444\dots$, so the fraction is correct.

This is a finite geometric series, so use $S_n$ with $a = 600$, $r = 0.8$ and $n = 12$. Do not use the sum to infinity.

$S_{12} = \frac{600(1 - 0.8^{12})}{0.2} = 3000(1 - 0.8^{12})$. Work out $0.8^{12}$ first, then the bracket.

so

The appeal raises £2793.84 in the first year. The sum to infinity is £3000, so the year has collected almost everything the appeal can ever raise.