QUESTION 8 8.1 $$ x ; 4 ; y $$ form an arithmetic sequence while $$ x ; \sqrt{15} ; y $$ form a geometric sequence. Calculate $$ x $$ and $$ y $$. 8.2 Consider an arithmetic series: $$ S_{n}=n^{2}+3 n $$ : 8.2.1 Calculate the second term of this series. 8.2.2 The first two terms of this sequence form a geometric sequence. Calculate the sum of the first 6 terms of this geometric sequence.

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**8.1**

We are given that $$x$$, $$4$$, and $$y$$ form an arithmetic sequence. This means the difference between consecutive terms is constant:
$$
4 - x = y - 4.
$$
Solving for $$y$$, we get:
$$
y = 8 - x.
$$

We are also given that $$x$$, $$\sqrt{15}$$, and $$y$$ form a geometric sequence. In a geometric sequence the ratio between consecutive terms is constant. Thus:
$$
\frac{\sqrt{15}}{x} = \frac{y}{\sqrt{15}}.
$$
Cross multiplying gives:
$$
\sqrt{15} \cdot \sqrt{15} = x y \quad \Rightarrow \quad 15 = x y.
$$
Substitute $$y = 8 - x$$ into the equation:
$$
x(8 - x) = 15.
$$
This simplifies to:
$$
-x^2 + 8x - 15 = 0 \quad \Rightarrow \quad x^2 - 8x + 15 = 0.
$$
Factor the quadratic:
$$
(x - 3)(x - 5) = 0.
$$
Thus, the solutions are:
$$
x = 3 \quad \text{or} \quad x = 5.
$$
Using $$y = 8 - x$$:
- If $$x = 3$$, then $$y = 8 - 3 = 5$$.
- If $$x = 5$$, then $$y = 8 - 5 = 3$$.

**8.2**

We are given an arithmetic series with sum
$$
S_n = n^2 + 3n.
$$

*8.2.1 Calculate the second term of this series.*

The $$n$$th term $$a_n$$ is given by:
$$
a_n = S_n - S_{n-1}.
$$
For $$n = 2$$:
$$
S_2 = 2^2 + 3 \times 2 = 4 + 6 = 10,
$$
$$
S_1 = 1^2 + 3 \times 1 = 1 + 3 = 4.
$$
Thus:
$$
a_2 = S_2 - S_1 = 10 - 4 = 6.
$$

*8.2.2 The first two terms of this sequence form a geometric sequence. Calculate the sum of the first 6 terms of this geometric sequence.*

The first two terms of the arithmetic series are:
$$
a_1 = S_1 = 4 \quad \text{and} \quad a_2 = 6.
$$
Since these form a geometric sequence, the common ratio $$r$$ is:
$$
r = \frac{a_2}{a_1} = \frac{6}{4} = \frac{3}{2}.
$$
The sum $$S_6$$ of the first 6 terms of a geometric sequence with first term $$a_1 = 4$$ is given by:
$$
S_6 = a_1 \frac{r^6 - 1}{r - 1}.
$$
Substitute the values:
$$
S_6 = 4 \frac{\left(\frac{3}{2}\right)^6 - 1}{\frac{3}{2} - 1}.
$$
Calculate the denominator:
$$
\frac{3}{2} - 1 = \frac{1}{2}.
$$
Thus:
$$
S_6 = 4 \cdot \frac{\left(\frac{3}{2}\right)^6 - 1}{\frac{1}{2}} = 8\left[\left(\frac{3}{2}\right)^6 - 1\right].
$$
Now calculate $$\left(\frac{3}{2}\right)^6$$:
$$
\left(\frac{3}{2}\right)^6 = \frac{3^6}{2^6} = \frac{729}{64}.
$$
Then:
$$
S_6 = 8\left(\frac{729}{64} - 1\right) = 8\left(\frac{729 - 64}{64}\right) = 8\left(\frac{665}{64}\right) = \frac{5320}{64} = \frac{665}{8}.
$$

**Summary of Answers:**

- For 8.1:
  - Either $$x = 3$$ and $$y = 5$$ or $$x = 5$$ and $$y = 3$$.

- For 8.2:
  - 8.2.1: The second term is $$6$$.
  - 8.2.2: The sum of the first 6 terms of the geometric sequence is $$\frac{665}{8}$$.
**8.1 Solutions:** - $$x = 3$$ and $$y = 5$$ - $$x = 5$$ and $$y = 3$$ **8.2 Solutions:** - 8.2.1: The second term is $$6$$. - 8.2.2: The sum of the first 6 terms of the geometric sequence is $$\frac{665}{8}$$.

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