Practice Questions on Sequences & Series (IB Math AA SL Style)

Solution:

1. Common difference \(d\)
   From an arithmetic sequence, \(d = u_2 - u_1 = 17 - 20 = -3\).
2. Find \(u_6\)
   General term of an arithmetic sequence: \[ u_n = u_1 + (n-1)d. \] For \(n=6\): \[ u_6 = 20 + (6-1)(-3) = 20 + 5 \cdot (-3) = 20 - 15 = 5. \]
3. Find \(S_6\)
   The sum of the first \(n\) terms in an arithmetic sequence is: \[ S_n = \frac{n}{2} \bigl(u_1 + u_n\bigr). \] So, \[ S_6 = \frac{6}{2}\bigl(u_1 + u_6\bigr) = 3 \times (20 + 5) = 3 \times 25 = 75. \]

Solution:

1. Common difference \(d\)
   Since \(u_2 - u_1 = 3 - 7 = -4\), we have \(d = -4\).
2. Find \(u_7\)
   Using \(u_n = u_1 + (n-1)d\): \[ u_7 = 7 + (7-1)(-4) = 7 + 6 \cdot (-4) = 7 - 24 = -17. \]
3. Find \(S_7\)
   Recall: \[ S_n = \frac{n}{2}(u_1 + u_n). \] Here, \(n=7\): \[ S_7 = \frac{7}{2} \bigl(7 + (-17)\bigr) = \frac{7}{2} \times (-10) = 7 \times (-5) = -35. \]

Solution:

1. Common ratio \(r\)
   In a geometric sequence, \(r = \frac{u_2}{u_1} = \frac{12}{4} = 3.\)
2. Find \(u_5\)
   General term for a geometric sequence: \[ u_n = u_1 \, r^{\,n-1}. \] For \(n=5\): \[ u_5 = 4 \times 3^{\,5-1} = 4 \times 3^4 = 4 \times 81 = 324. \]
3. Find \(S_5\)
   Sum of the first \(n\) terms in a geometric sequence (for \(r \neq 1\)): \[ S_n = u_1 \frac{r^n - 1}{r - 1}. \] Here \(n=5\), \(u_1=4\), and \(r=3\): \[ S_5 = 4 \times \frac{3^5 - 1}{3 - 1} = 4 \times \frac{243 - 1}{2} = 4 \times \frac{242}{2} = 4 \times 121 = 484. \]

Solution:

1. Common ratio \(r\)
   Compute \(r = \frac{u_2}{u_1} = \frac{8}{16} = \tfrac{1}{2}.\)
2. Find \(u_6\)
   \[ u_n = u_1 \, r^{\,n-1}. \] For \(n=6\): \[ u_6 = 16 \left(\tfrac{1}{2}\right)^{5} = 16 \times \tfrac{1}{32} = \tfrac{16}{32} = \tfrac{1}{2}. \]
3. Find \(S_6\)
   Using \[ S_n = u_1 \frac{r^n - 1}{r - 1} \quad (\text{for }r \neq 1), \] we get \[ S_6 = 16 \times \frac{\left(\tfrac{1}{2}\right)^6 - 1}{\tfrac{1}{2} - 1} = 16 \times \frac{\tfrac{1}{64} - 1}{- \tfrac{1}{2}} = 16 \times \frac{- \tfrac{63}{64}}{- \tfrac{1}{2}}. \] Simplify step by step: \[ -\tfrac{63}{64} \div -\tfrac{1}{2} = \tfrac{63}{64} \times 2 = \tfrac{63}{32}. \] So \[ S_6 = 16 \times \tfrac{63}{32} = \tfrac{16 \times 63}{32} = \tfrac{16}{32} \times 63 = \tfrac{1}{2} \times 63 = 31.5. \] Or written as fraction: \[ S_6 = 31.5 \quad \text{or} \quad \tfrac{63}{2}. \]

Solution:

1. First 3 terms
   \(u_1 = 5.\)
   \(u_2 = 5 + 2 = 7.\)
   \(u_3 = 7 + 2 = 9.\)
2. Find \(u_{10}\)
   General term: \(u_n = u_1 + (n-1)d.\)
   For \(n=10\): \[ u_{10} = 5 + (10-1)\times 2 = 5 + 9\times 2 = 5 + 18 = 23. \]
3. Calculate \(S_{10}\)
   Sum formula: \[ S_n = \frac{n}{2} (u_1 + u_n). \] Here \(n=10\): \[ S_{10} = \frac{10}{2} (5 + 23) = 5 \times 28 = 140. \]

Solution:

(a) We see that \(u_1 = 5\) and \(u_2 = 8\). The common difference is
\(d = u_2 - u_1 = 8 - 5 = 3\).

(b) The nth term of an arithmetic sequence is \[ u_n = u_1 + (n - 1)d. \] For \(n = 10\): \[ u_{10} = 5 + (10 - 1)\times 3 = 5 + 9\times 3 = 5 + 27 = 32. \]

(c) The sum of the first n terms is \[ S_n = \frac{n}{2}\bigl(u_1 + u_n\bigr). \] Thus, for \(n = 10\): \[ S_{10} = \frac{10}{2} \bigl(5 + 32\bigr) = 5 \times 37 = 185. \]

Solution:

(a) We observe \(u_1 = 0\) and \(u_2 = -4\). Hence:
\(d = u_2 - u_1 = -4 - 0 = -4.\)

(b) Using \(\displaystyle u_n = u_1 + (n-1)d\):
\(u_{10} = 0 + (10-1)\times (-4) = 9 \times (-4) = -36.\)

(c) The sum of the first n terms formula: \(\displaystyle S_n = \frac{n}{2}\bigl(u_1 + u_n\bigr)\). For \(n=10\):
\(\displaystyle S_{10} = \frac{10}{2}\,\bigl(0 + (-36)\bigr) = 5 \times (-36) = -180.\)

Solution:

(a) Here \(u_1 = 12\) and \(u_2 = 9\). So
\(d = u_2 - u_1 = 9 - 12 = -3.\)

(b) Using \(u_n = u_1 + (n - 1)d\):
\(u_{10} = 12 + (10 - 1)\times (-3) = 12 + 9\times(-3) = 12 - 27 = -15.\)

(c) The sum formula: \(\displaystyle S_n = \frac{n}{2}(u_1 + u_n)\). So for \(n=10\):
\(\displaystyle S_{10} = \frac{10}{2}(12 + (-15)) = 5 \times (-3) = -15.\)

Solution:

(a) From \(u_1 = 1\) and \(u_2 = 3\), we get
\(d = 3 - 1 = 2.\)

(b) Using \(u_n = u_1 + (n-1)d\):
\(u_{10} = 1 + (10 - 1)\times 2 = 1 + 9\times 2 = 1 + 18 = 19.\)

(c) The sum of the first n terms: \(\displaystyle S_n = \frac{n}{2}(u_1 + u_n)\). Thus for \(n=10\):
\(\displaystyle S_{10} = \frac{10}{2}(1 + 19) = 5 \times 20 = 100.\)

Solution:

(a) Here \(u_1 = 4\) and \(u_2 = 10\). Hence
\(d = 10 - 4 = 6.\)

(b) Using the formula \(u_n = u_1 + (n-1)d\):
\(u_{10} = 4 + (10 - 1)\times 6 = 4 + 9\times 6 = 4 + 54 = 58.\)

(c) Recall the sum formula: \(\displaystyle S_n = \frac{n}{2}(u_1 + u_n)\). For \(n=10\):
\(\displaystyle S_{10} = \frac{10}{2}\,(4 + 58) = 5 \times 62 = 310.\)