Let's break down and solve each part of the question in the image you provided.

(a) The first, second, and last terms in an arithmetic progression are 56, 53, and -22, respectively.

We need to find the sum of all the terms in the arithmetic progression.

Let the first term be $a_1 = 56$, the second term be $a_2 = 53$, and the last term be $a_n = -22$.

For an arithmetic progression, the common difference $d$ between consecutive terms can be calculated by:

$d = a_2 - a_1 = 53 - 56 = -3$

To find the number of terms $n$, we use the formula for the $n$-th term of an arithmetic progression:

$a_n = a_1 + (n-1)d$

Substituting the known values:

$-22 = 56 + (n-1)(-3)$

Simplifying:

$-22 = 56 - 3(n-1)$ $-22 - 56 = -3(n-1)$ $-78 = -3(n-1)$ $n-1 = \frac{-78}{-3} = 26$ $n = 27$

So, the progression has 27 terms.

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

$S_n = \frac{n}{2} (a_1 + a_n)$

Substituting the values:

$S_{27} = \frac{27}{2} (56 + (-22)) = \frac{27}{2} \times 34 = 27 \times 17 = 459$

Thus, the sum of all the terms in the progression is 459.

(b) The first, second, and third terms of a geometric progression are $2k + 6$, $2k$, and $k + 2$ respectively.

We need to:

(i) Find the value of $k$.

In a geometric progression, the ratio between consecutive terms is constant. So, we set the ratio between the first and second terms equal to the ratio between the second and third terms:

$\frac{2k}{2k + 6} = \frac{k+2}{2k}$

Cross-multiplying:

$2k \times 2k = (2k + 6)(k + 2)$

Expanding both sides:

$4k^2 = (2k \cdot k) + (2k \cdot 2) + (6 \cdot k) + (6 \cdot 2)$ $4k^2 = 2k^2 + 4k + 6k + 12$ $4k^2 = 2k^2 + 10k + 12$

Bringing all terms to one side:

$4k^2 - 2k^2 - 10k - 12 = 0$ $2k^2 - 10k - 12 = 0$

Dividing the equation by 2:

$k^2 - 5k - 6 = 0$

Factoring the quadratic equation:

$(k - 6)(k + 1) = 0$

Thus, $k = 6$ or $k = -1$.

Since the problem states that $k$ is a positive constant, we take $k = 6$.

(ii) Find the sum of the first three terms.

Now that we know $k = 6$, we can substitute this into the expressions for the first, second, and third terms of the geometric progression:

• First term: $2k + 6 = 2(6) + 6 = 18$
• Second term: $2k = 2(6) = 12$
• Third term: $k + 2 = 6 + 2 = 8$

The sum of the first three terms is:

$18 + 12 + 8 = 38$

Thus, the sum of the first three terms is 38.

Let me know if you want more details or have any questions! Here are 5 related questions to explore further:

1. How do you find the common difference for any arithmetic progression if only two terms are given?
2. What is the general formula for the $n$-th term of a geometric progression?
3. How can you use factoring to solve any quadratic equation?
4. Can an arithmetic progression have a common difference of zero? What would it look like?
5. How do you calculate the product of terms in a geometric progression?

Tip: For quadratic equations, always check if it can be factored before using the quadratic formula—it might save time!