Question

A sequence \(\{u_n\}\) is defined as follows:

• The odd-numbered terms \(u_1, u_3, u_5, \dots\) form an arithmetic progression with first term 4 and common difference 3.
• The even-numbered terms \(u_2, u_4, u_6, \dots\) form a geometric progression with first term 2 and common ratio \(\dfrac{1}{2}\).

1. Write down the first six terms of the sequence.
2. Find an expression for \(u_{2k-1}\), the \((2k-1)\)th term (odd) of the sequence.
3. Find the sum of the first 10 terms of the sequence.

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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 \(2\).

1. For the arithmetic progression, find:
   • (a) the 15th term,
   • (b) the sum of the first 15 terms.
2. For the geometric progression, find:
   • (a) the 8th term,
   • (b) the smallest value of \(n\) such that \(u_n > 500\).
3. Comment briefly on the difference in growth of the two sequences.

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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 \(|r| < 1\).

1. Find the sum of the first 6 terms of the arithmetic progression.
2. Given that the sum to infinity of the geometric progression is equal to the sum of the first 6 terms of the arithmetic progression, find the value of \(r\).
3. For this value of \(r\), find the sum of the first 4 terms of the geometric progression.

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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.)