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

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**(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**.

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**(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 \( k = 6 \).

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**(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!

The first, second, and last terms in an arithmetic progression are 56, 53, and -22 respectively. Find the sum of all the terms in the progression. The first, second, and third terms of a geometric progression are 2k + 6, 2k, and k + 2 respectively. Find the value of k and the sum of the first three terms.

In this problem, learn how to calculate the sum of an arithmetic progression and solve for k in a geometric progression. Includes detailed steps on finding common differences, sum formulas, and solving quadratic equations. Suitable for students in Grades 10-12.

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