Sequences and Series: practice solutions, Grade 11 – download the PDF

1 Name that sequence ★★★

1. Differences: \(7, 7, 7\). Arithmetic with \(d = 7\).
2. Ratios: \(2, 2, 2\). Geometric with \(r = 2\).
3. Differences \(2, 3, 4\) and ratios \(2, 1.75, \dots\) are not constant. Neither.
4. Ratios: \(\dfrac{27}{81} = \dfrac13\), \(\dfrac{9}{27} = \dfrac13\), \(\dfrac{3}{9} = \dfrac13\). Geometric with \(r = \dfrac13\).

2 First terms of an arithmetic sequence ★★★

(a) \(9, 13, 17, 21, 25\).

(b) \(a_n = 9 + (n-1)\cdot 4 = 4n + 5\).

(c) \(a_{12} = 4\cdot 12 + 5 = 53\).

3 First terms of a geometric sequence ★★★

\(5, 15, 45, 135, 405\).

\(a_7 = 5\cdot 3^{6} = 5\cdot 729 = 3{,}645\).

4 Explicit to recursive ★★★

(a) \(5, 3, 1, -1\).

(b) The difference is \(-2\): \(a_1 = 5\) and \(a_{n+1} = a_n - 2\).

(c) \(a_{15} = 7 - 30 = -23\).

5 Your first sigma sum ★★★

The terms are \(5, 7, 9, 11\). The sum is \(5 + 7 + 9 + 11 = 32\).

Check: \(\dfrac{4\,(5 + 11)}{2} = 32\).

6 Even numbers up to 50 ★★★

The sequence is arithmetic with \(a_1 = 2\), \(d = 2\), \(a_n = 50\). Then \(2n = 50\) gives \(n = 25\) terms.

\(S_{25} = \dfrac{25\,(2 + 50)}{2} = 25\cdot 26 = 650\).

7 Converge or diverge? ★★★

(a) \(r = \dfrac13\), \(|r| < 1\), so \(S = \dfrac{6}{1 - \frac13} = \dfrac{6}{\frac23} = 9\).

(b) \(r = 2\), \(|r| \ge 1\): the series diverges and has no sum.

8 Which term is it? ★★★

The formula is \(a_n = 4 + (n-1)\cdot 5 = 5n - 1\).

(a) \(5n - 1 = 139\) gives \(5n = 140\), so \(n = 28\). It is the 28th term.

(b) \(5n - 1 = 200\) gives \(n = 40.2\), which is not a whole number. So 200 is not a term.

9 Finding the ratio from two terms ★★★

From \(a_2\) to \(a_5\) there are 3 steps, so \(r^3 = \dfrac{324}{12} = 27\) and \(r = 3\).

\(a_1 = \dfrac{12}{3} = 4\). The formula is \(a_n = 4\cdot 3^{n-1}\).

Check: \(a_5 = 4\cdot 81 = 324\).

10 Two terms of an arithmetic sequence ★★★

Five steps separate \(a_4\) and \(a_9\): \(5d = 42 - 17 = 25\), so \(d = 5\).

\(a_1 = 17 - 3\cdot 5 = 2\), and \(a_n = 2 + (n-1)\cdot 5 = 5n - 3\).

\(a_{20} = 97\), so \(S_{20} = \dfrac{20\,(2 + 97)}{2} = 10\cdot 99 = 990\).

11 A geometric sum ★★★

\(a_1 = 4\) and \(r = 3\).

\(S_7 = 4\cdot\dfrac{1 - 3^7}{1 - 3} = 4\cdot\dfrac{-2186}{-2} = 4\cdot 1093 = 4{,}372\).

12 Theater seats ★★★

The row counts form an arithmetic sequence with \(a_1 = 14\), \(d = 2\).

(a) \(a_{18} = 14 + 17\cdot 2 = 48\) seats.

(b) \(S_{18} = \dfrac{18\,(14 + 48)}{2} = 9\cdot 62 = 558\).

The theater has 558 seats.

13 A sigma sum with powers ★★★

The terms are \(6, 12, 24, 48, 96\), a geometric series with \(a_1 = 6\), \(r = 2\), 5 terms.

\(S_5 = 6\cdot\dfrac{1 - 2^5}{1 - 2} = 6\cdot 31 = 186\).

Check by adding: \(6 + 12 + 24 + 48 + 96 = 186\).

14 A repeating decimal as a fraction ★★★

\(0.363636\dots = 0.36 + 0.0036 + 0.000036 + \dots\), with \(a_1 = 0.36\) and \(r = 0.01\).

\(S = \dfrac{0.36}{1 - 0.01} = \dfrac{0.36}{0.99} = \dfrac{36}{99} = \dfrac{4}{11}\).

15 Saving for a bike ★★★

Deposits form an arithmetic sequence with \(a_1 = 50\), \(d = 15\). So \(S_n = \dfrac{n\,[100 + 15(n-1)]}{2} = \dfrac{15n^2 + 85n}{2}\).

We need \(15n^2 + 85n \ge 4000\), that is \(3n^2 + 17n - 800 \ge 0\). The positive root is \(n = \dfrac{-17 + \sqrt{9889}}{6} \approx 13.7\).

Test: \(S_{13} = \dfrac{13\cdot 280}{2} = 1{,}820 < 2{,}000\) and \(S_{14} = \dfrac{14\cdot 295}{2} = 2{,}065 \ge 2{,}000\).

She reaches her goal in month 14.

16 The bouncing ball ★★★

Down: 20 ft. Then each rebound height is traveled twice (up and down). The rebound heights are \(12, 7.2, 4.32, \dots\), geometric with \(a_1 = 12\), \(r = 0.6\).

Sum of rebounds: \(\dfrac{12}{1 - 0.6} = 30\) ft.

Total distance: \(20 + 2\cdot 30 = 80\) ft, about 24.4 m.

17 Yearly deposits ★★★

The first deposit grows for 10 years and the last one for 1 year. The total is \(500(1.04) + 500(1.04)^2 + \dots + 500(1.04)^{10}\), a geometric series with \(a_1 = 500\cdot 1.04 = 520\), \(r = 1.04\), \(n = 10\).

\(S_{10} = 520\cdot\dfrac{1.04^{10} - 1}{0.04} \approx 520\cdot 12.0061 \approx 6{,}243.18\) dollars.

18 Three numbers in an arithmetic sequence ★★★

Write them \(m - d,\ m,\ m + d\). The sum is \(3m = 27\), so \(m = 9\).

The product is \(9(81 - d^2) = 693\), so \(81 - d^2 = 77\) and \(d^2 = 4\), \(d = \pm 2\).

The numbers are \(7, 9, 11\) (or \(11, 9, 7\)). Check: \(7 + 9 + 11 = 27\) and \(7\cdot 9\cdot 11 = 693\).

19 Solving for x ★★★

(a) It is geometric with \(a_1 = 8\), \(r = x\). It converges exactly when \(|x| < 1\), that is \(-1 < x < 1\).

(b) \(\dfrac{8}{1 - x} = 20\), so \(1 - x = 0.4\) and \(x = 0.6\). This satisfies \(|x| < 1\).

20 Geometric means ★★★

We need \(a_1 = 5\) and \(a_5 = 80\), so \(5r^4 = 80\), \(r^4 = 16\), \(r = 2\) or \(r = -2\).

If \(r = 2\): \(5, 10, 20, 40, 80\). If \(r = -2\): \(5, -10, 20, -40, 80\).

21 An alternating series ★★★

(a) Terms: \(9, -4.5, 2.25, -1.125\). \(S_4 = 5.625\). Formula: \(9\cdot\dfrac{1 - (1/16)}{3/2} = 9\cdot\dfrac{15/16}{3/2} = 5.625\).

(b) \(r = -\tfrac12\), \(|r| < 1\): \(S = \dfrac{9}{1 + \frac12} = 6\).

(c) \(6 - 5.625 = 0.375\).