2) In an arithmetic sequence $$ T_{1}=(1-x)^{2} $$ and $$ T_{2}=1+x^{2}: $$ a) Write down an expression for the common difference in terms of $$ x $$. b) Determine the value(s) of $$ x $$ if $$ S_{10}=310 $$. 3) Using your knowledge of series: a) Prove that the sum of the first $$ n $$, odd natural numbers is $$ n^{2} $$. b) Prove that the sum of the first $$ n $$, even natural numbers is $$ n^{2}+n $$. 4) The sum of the first 5 terms of an arithmetic sequence is 0 and the fiffth term is 8 . Determine $$ \mathbf{t} $$ common difference and the first term. 5) In a geometric series the first term is 36 , the last term is 16 and the sum of the series is 76 . Determine the common ratio and the number of terms in the sequence.

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### Problem 2
Given the arithmetic sequence where $$ T_{1}=(1-x)^{2} $$ and $$ T_{2}=1+x^{2} $$:

#### a) Write down an expression for the common difference in terms of $$ x $$.

The common difference $$ d $$ in an arithmetic sequence is given by:

$$
d = T_{2} - T_{1}
$$

Substituting the expressions for $$ T_{1} $$ and $$ T_{2} $$:

$$
d = (1 + x^{2}) - (1 - x)^{2}
$$

Now, let's simplify $$ (1 - x)^{2} $$:

$$
(1 - x)^{2} = 1 - 2x + x^{2}
$$

Thus, we have:

$$
d = (1 + x^{2}) - (1 - 2x + x^{2}) = 1 + x^{2} - 1 + 2x - x^{2} = 2x
$$

So, the expression for the common difference is:

$$
d = 2x
$$

#### b) Determine the value(s) of $$ x $$ if $$ S_{10}=310 $$.

The sum of the first $$ n $$ terms of an arithmetic sequence is given by:

$$
S_{n} = \frac{n}{2} (2a + (n-1)d)
$$

For $$ n = 10 $$ and $$ S_{10} = 310 $$:

$$
310 = \frac{10}{2} (2T_{1} + 9d)
$$

This simplifies to:

$$
310 = 5(2T_{1} + 9d)
$$

Dividing both sides by 5:

$$
62 = 2T_{1} + 9d
$$

Substituting $$ T_{1} = (1-x)^{2} $$ and $$ d = 2x $$:

$$
62 = 2(1-x)^{2} + 9(2x)
$$

Expanding this:

$$
62 = 2(1 - 2x + x^{2}) + 18x
$$
$$
62 = 2 - 4x + 2x^{2} + 18x
$$
$$
62 = 2x^{2} + 14x + 2
$$

Rearranging gives:

$$
2x^{2} + 14x + 2 - 62 = 0
$$
$$
2x^{2} + 14x - 60 = 0
$$

Dividing the entire equation by 2:

$$
x^{2} + 7x - 30 = 0
$$

Now, we can solve this quadratic equation using the quadratic formula:

$$
x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}
$$

Here, $$ a = 1, b = 7, c = -30 $$:

$$
x = \frac{-7 \pm \sqrt{7^{2} - 4 \cdot 1 \cdot (-30)}}{2 \cdot 1}
$$
$$
x = \frac{-7 \pm \sqrt{49 + 120}}{2}
$$
$$
x = \frac{-7 \pm \sqrt{169}}{2}
$$
$$
x = \frac{-7 \pm 13}{2}
$$

Calculating the two possible values for $$ x $$:

1. $$ x = \frac{6}{2} = 3 $$
2. $$ x = \frac{-20}{2} = -10 $$

Thus, the values of $$ x $$ are $$ 3 $$ and $$ -10 $$.

### Problem 3
#### a) Prove that the sum of the first $$ n $$, odd natural numbers is $$ n^{2} $$.

The first $$ n $$ odd natural numbers are $$ 1, 3, 5, \ldots, (2n-1) $$. The sum can be expressed as:

$$
S = 1 + 3 + 5 + \ldots + (2n-1)
$$

This can be rewritten as:

$$
S = \sum_{k=1}^{n} (2k - 1)
$$

Using the formula for the sum of an arithmetic series:

$$
S = n \cdot \frac{(1 + (2n - 1))}{2} = n \cdot \frac{(2n)}{2} = n^2
$$

Thus, the sum of the first $$ n $$ odd natural numbers is $$ n^{2} $$.

#### b) Prove that the sum of the first $$ n $$, even natural numbers is $$ n^{2}+n $$.

The first $$ n $$ even natural numbers are $$ 2, 4, 6, \ldots, 2n $$. The sum can be expressed as:

$$
S = 2 + 4 + 6 + \ldots + 2n
$$

Factoring out 2 gives:

$$
S = 2(1 + 2 + 3 + \ldots + n) = 2 \cdot \frac{n(n + 1)}{2} = n(n + 1)
$$

Thus, the sum of the first $$ n $$ even natural numbers is $$ n(n + 1) = n^{2} + n $$.

### Problem 4
The sum of the first 5 terms of an arithmetic sequence is 0, and the fifth term is 8.

Let the first term be $$ a $$ and the common difference be $$ d $$.

The sum of the first 5 terms is:

$$
S_5 = \frac{5}{2} (2a + 4d) = 0
$$

This simplifies to:

$$
2a + 4d = 0 \quad \text{(1)}
$$

The fifth term is given by:

$$
T_5 = a + 4d = 8 \quad \text{(2)}
$$

Now we can solve these two equations simultaneously. From equation (1):

$$
2a = -4d \implies a = -2d
$$

Substituting $$ a = -2d $$ into equation (2):

$$
-2d + 4d = 8
$$
$$
2d = 8 \implies d = 4
$$

Now substituting $$ d = 4 $$ back into $$ a = -2d $$:

$$
a = -2(4) = -8
$$

Thus, the first term $$ a $$ is $$ -8 $$ and the common difference $$ d $$ is $$ 4 $$.

### Problem 5

**Problem 2:** a) The common difference $$ d $$ is $$ 2x $$. b) The values of $$ x $$ are $$ 3 $$ and $$ -10 $$. **Problem 3:** a) The sum of the first $$ n $$ odd natural numbers is $$ n^{2} $$. b) The sum of the first $$ n $$ even natural numbers is $$ n^{2} + n $$. **Problem 4:** - First term $$ a = -8 $$ - Common difference $$ d = 4 $$ **Problem 5:** Cannot determine the common ratio and the number of terms with the given information.

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