These examples combine arithmetic progressions (AP) and geometric progressions (GP), similar to exam-style questions.
| Arithmetic Progression (AP) | Geometric Progression (GP) | |
|---|---|---|
| Definition | \(u_{n+1} - u_n = d\) (constant difference) | \(\dfrac{u_{n+1}}{u_n} = r\) (constant ratio) |
| nth term | \(u_n = a + (n - 1)d\) | \(u_n = ar^{\,n-1}\) |
| Sum of first \(n\) terms | \(S_n = \dfrac{n}{2}\bigl(2a + (n - 1)d\bigr) = \dfrac{n}{2}(u_1 + u_n)\) | \(S_n = a\,\dfrac{1 - r^{n}}{1 - r}\) for \(r \ne 1\) |
| Sum to infinity | Not defined (terms do not tend to 0) | \(S_{\infty} = \dfrac{a}{1 - r}\) if \(|r| < 1\) |
Question
A sequence \(\{u_n\}\) is defined as follows:
Solution
(i) First six terms
Odd terms (AP): \(a_{\text{AP}} = 4\), \(d = 3\).
\[
u_1 = 4,\quad
u_3 = 4 + 2\cdot 3 = 10,\quad
u_5 = 4 + 4\cdot 3 = 16.
\]
Even terms (GP): \(a_{\text{GP}} = 2\), \(r = \dfrac{1}{2}\).
\[
u_2 = 2,\quad
u_4 = 2\left(\frac{1}{2}\right)^{2-1} = 1,\quad
u_6 = 2\left(\frac{1}{2}\right)^{3-1} = 2\left(\frac{1}{2}\right)^{2} = \frac{1}{2}.
\]
So the first six terms are: \[ 4,\; 2,\; 10,\; 1,\; 16,\; \frac{1}{2},\dots \]
(ii) Formula for \(u_{2k-1}\)
The odd terms form an AP: \[ u_1 = 4,\quad u_3 = 10,\quad u_5 = 16,\dots \] These correspond to \(k = 1,2,3,\dots\) with common difference 6 between successive odd indices, so \[ u_{2k-1} = 4 + (k - 1)\cdot 6 = 6k - 2,\quad k = 1,2,3,\dots \]
(iii) Sum of the first 10 terms
We have 5 odd terms \(u_1, u_3, u_5, u_7, u_9\) and 5 even terms \(u_2, u_4, u_6, u_8, u_{10}\).
Odd terms (AP part)
The odd terms (in terms of \(k\)) are \[ u_{2k-1} = 6k - 2,\quad k = 1,2,3,4,5. \] Sum of these 5 terms: \[ \begin{aligned} S_{\text{odd}} &= \sum_{k=1}^{5} (6k - 2) \\ &= 6\sum_{k=1}^{5} k - 2\cdot 5 \\ &= 6\cdot \frac{5\cdot 6}{2} - 10 \\ &= 6\cdot 15 - 10 \\ &= 90 - 10 \\ &= 80. \end{aligned} \]
Even terms (GP part)
Even terms form a GP: \[ u_{2}, u_{4}, u_{6}, u_{8}, u_{10} = 2,\, 1,\, \tfrac{1}{2},\, \tfrac{1}{4},\, \tfrac{1}{8}. \] This is a GP with first term \(2\), common ratio \(\dfrac{1}{2}\) and 5 terms: \[ \begin{aligned} S_{\text{even}} &= 2\cdot \frac{1 - \left(\frac{1}{2}\right)^{5}}{1 - \frac{1}{2}} \\ &= 2\cdot \frac{1 - \frac{1}{32}}{\frac{1}{2}} \\ &= 2\cdot \frac{\frac{31}{32}}{\frac{1}{2}} \\ &= 2\cdot \frac{31}{32}\cdot 2 \\ &= \frac{124}{32} = \frac{31}{8}. \end{aligned} \]
Total sum of first 10 terms: \[ S_{10} = S_{\text{odd}} + S_{\text{even}} = 80 + \frac{31}{8} = \frac{640}{8} + \frac{31}{8} = \frac{671}{8}. \]
So \[ \boxed{S_{10} = \dfrac{671}{8}}. \]
Question
An arithmetic progression has first term \(5\) and common difference \(3\). A geometric progression has first term \(5\) and common ratio \(2\).
Solution
1(a) AP: 15th term
For the AP: \(a = 5\), \(d = 3\). \[ u_n = a + (n - 1)d. \] So \[ u_{15} = 5 + 14\cdot 3 = 5 + 42 = 47. \]
1(b) AP: sum of first 15 terms
\[ \begin{aligned} S_{15} &= \frac{15}{2}\bigl(2a + (15 - 1)d\bigr) \\ &= \frac{15}{2}\bigl(2\cdot 5 + 14\cdot 3\bigr) \\ &= \frac{15}{2}(10 + 42) \\ &= \frac{15}{2}\cdot 52 \\ &= 15 \cdot 26 \\ &= 390. \end{aligned} \] So \(S_{15} = \boxed{390}\).
2(a) GP: 8th term
For the GP: \(a = 5\), \(r = 2\). \[ u_n = ar^{n-1} = 5\cdot 2^{n-1}. \] So \[ u_8 = 5\cdot 2^{7} = 5\cdot 128 = 640. \]
2(b) GP: smallest \(n\) with \(u_n > 500\)
We want \[ u_n = 5\cdot 2^{n-1} > 500. \] Divide both sides by 5: \[ 2^{n-1} > 100. \] Check powers of 2: \[ 2^{6} = 64,\quad 2^{7} = 128. \] So we need \(2^{n-1} \ge 128\), i.e. \(n - 1 \ge 7 \Rightarrow n \ge 8\).
Therefore the smallest integer \(n\) such that \(u_n > 500\) is \[ \boxed{n = 8}. \]
3. Comment on growth
The AP increases by a fixed amount \(d = 3\) each time, so its terms increase steadily. The GP doubles each time, so its terms grow much faster; by the 8th term the GP already exceeds 600, while the AP’s 15th term is only 47. This shows how geometric growth quickly overtakes arithmetic growth.
Question
The first three terms of an arithmetic progression are \(3, 7, 11\). The first term of a geometric progression is 18 and its common ratio is \(r\), where \(|r| < 1\).
Solution
1. Sum of first 6 terms of the AP
For the AP: \(a = 3\), \(d = 4\). Check: \(3, 7, 11,\dots\) difference \(= 4\).
\[ \begin{aligned} S_6 &= \frac{6}{2}\bigl(2a + (6 - 1)d\bigr) \\ &= 3(2\cdot 3 + 5\cdot 4) \\ &= 3(6 + 20) \\ &= 3 \cdot 26 \\ &= 78. \end{aligned} \]
So the sum of the first 6 terms of the AP is \(\boxed{78}\).
2. Sum to infinity of the GP and value of \(r\)
For the GP: \(a_{\text{GP}} = 18\), common ratio \(r\), \(|r| < 1\). Sum to infinity: \[ S_{\infty} = \frac{18}{1 - r}. \] We are told: \[ S_{\infty} = S_6(\text{AP}) = 78. \] So \[ \begin{aligned} \frac{18}{1 - r} &= 78, \\ 1 - r &= \frac{18}{78} = \frac{3}{13}, \\ r &= 1 - \frac{3}{13} = \frac{10}{13}. \end{aligned} \]
Since \(\left|\dfrac{10}{13}\right| < 1\), the sum to infinity formula is valid. Hence \[ \boxed{r = \dfrac{10}{13}}. \]
3. Sum of the first 4 terms of the GP
Now \(a = 18\), \(r = \dfrac{10}{13}\), \(n = 4\). \[ \begin{aligned} S_4 &= 18\,\frac{1 - r^{4}}{1 - r} = 18\,\frac{1 - \left(\frac{10}{13}\right)^{4}}{1 - \frac{10}{13}} \\ &= 18\,\frac{1 - \left(\frac{10}{13}\right)^{4}}{\frac{3}{13}} = 18 \cdot \frac{13}{3}\left(1 - \left(\frac{10}{13}\right)^{4}\right). \end{aligned} \] Simplify the constant factor: \[ 18 \cdot \frac{13}{3} = 6 \cdot 13 = 78. \] So \[ S_4 = 78\left(1 - \left(\frac{10}{13}\right)^{4}\right). \]
Therefore, the sum of the first four terms of the geometric progression is \[ \boxed{S_4 = 78\left(1 - \left(\dfrac{10}{13}\right)^{4}\right)}. \] (You can leave it in this exact form, or evaluate numerically if needed.)