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

1. The term

Answer: Here $a = 5$ and $r = 3$. Then $u_6 = 5(3)^{6-1} = 5(243) = 1215$. The power is $5$, one less than the position, because the first term is the starting value.

2. The sum

Answer: Here $a = 2$ and $r = 3$. Then $S_8 = \frac{2(1 - 3^8)}{1 - 3} = \frac{2(1 - 6561)}{-2} = 6560$. Both the numerator and the denominator are negative, so the answer is positive.

3. The values of r

Answer: The third term is $ar^2 = 5r^2$, so $5r^2 = 20$ and $r^2 = 4$. Then $r = 2$ or $r = -2$. Both values give a third term of $20$, because the term depends on the square of the ratio. A condition such as "the terms are positive" selects one of them.

4. The sum to infinity

Answer: The common ratio is $r = \frac{6}{24} = \frac{1}{4}$, and $|r| < 1$, so the series converges. Then $S_\infty = \frac{24}{1 - \frac{1}{4}} = \frac{24}{\frac{3}{4}} = 32$. The total settles on $32$.

5. The sum to infinity

Answer: The condition $|0.9| < 1$ holds, so the sum to infinity exists: $S_\infty = \frac{100}{1 - 0.9} = \frac{100}{0.1} = 1000$. The total is ten times the first term, because the ratio is close to $1$. It approaches $1000$ but never reaches it.

6. The fraction

Answer: Split the decimal: $0.444\dots = 0.4 + 0.04 + 0.004 + \dots$, which is geometric with $a = 0.4$ and $r = 0.1$. Then $S_\infty = \frac{0.4}{1 - 0.1} = \frac{0.4}{0.9} = \frac{4}{9}$. The recurring decimal is exactly the fraction $\frac{4}{9}$.

7. The term

Answer: The common ratio is $r = \frac{20}{100} = \frac{1}{5}$. Then $u_{10} = 100\left(\frac{1}{5}\right)^{9} = \frac{100}{1953125} = \frac{4}{78125}$. The exact fraction is the answer; a decimal would be rounded and tiny.

8. The sum

Answer: Here $a = 4$ and $r = 3$. Then $S_{10} = \frac{4(1 - 3^{10})}{1 - 3} = \frac{4(1 - 59049)}{-2} = \frac{4(-59048)}{-2} = 118096$. The sum is close to twice the last term, $4(3^9) = 78732$, which is the pattern for a ratio of $3$.

9. The sum to infinity

Answer: The common ratio is $r = \frac{-4}{8} = -\frac{1}{2}$, and $\left|-\frac{1}{2}\right| < 1$, so the series converges even though the signs alternate. Then $S_\infty = \frac{8}{1 - \left(-\frac{1}{2}\right)} = \frac{8}{\frac{3}{2}} = \frac{16}{3}$. Subtracting a negative ratio makes the denominator larger, so the total is smaller than it would be for a positive ratio of the same size.

10. The ratio, the sum and the reason

Answer: The common ratio is $r = \frac{4}{2} = 2$, and the sum of the first 10 terms is $\frac{2(1 - 2^{10})}{1 - 2} = \frac{2(-1023)}{-1} = 2046$. The series has no sum to infinity because $|r| \ge 1$: each term is larger than the one before, so the running total grows without limit. The formula $S_\infty = \frac{a}{1 - r}$ is not available here.

11. The number of terms

Answer: The sum of the first $n$ terms is $S_n = \frac{100(1.1^n - 1)}{1.1 - 1} = 1000(1.1^n - 1)$. For $n = 7$, $S_7 = 1000(1.9487171 - 1) = 948.72$, which is below $1000$. For $n = 8$, $S_8 = 1000(2.14358881 - 1) = 1143.59$, which is above. So the smallest number of terms is $8$. A series with a ratio just above $1$ takes many terms to grow, which is why the answer is checked on both sides of the boundary.

12. The 12-month total and the sum to infinity

Answer: Here $a = 600$ and $r = 0.8$, so $S_{12} = \frac{600(1 - 0.8^{12})}{1 - 0.8} = 3000(1 - 0.8^{12}) = 2793.841569792$, which is £2793.84 to the nearest penny. The sum to infinity is $\frac{600}{1 - 0.8} = 3000$. After 12 months the appeal has raised £206.16 less than the £3000 it can never quite reach.