The given sequence is $4, 7, 10, 13, \dots$, which is an arithmetic progression (AP) with:

• First term $a = 4$
• Common difference $d = 3$

We are asked to find how many terms $n$ of this arithmetic progression must be taken so that their sum equals 69.

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

$S_n = \frac{n}{2} \times \left(2a + (n-1) \cdot d\right)$

Substitute the known values $a = 4$, $d = 3$, and $S_n = 69$ into the formula:

$69 = \frac{n}{2} \times \left(2(4) + (n-1) \cdot 3\right)$

Simplifying:

$69 = \frac{n}{2} \times \left(8 + 3n - 3\right)$ $69 = \frac{n}{2} \times (5 + 3n)$ $138 = n \times (5 + 3n)$ $138 = 5n + 3n^2$ $3n^2 + 5n - 138 = 0$

This is a quadratic equation. We can solve it using the quadratic formula:

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

For $3n^2 + 5n - 138 = 0$, we have:

• $a = 3$
• $b = 5$
• $c = -138$

Let's solve this equation.The solutions to the quadratic equation are:

$n_1 = 6 \quad \text{and} \quad n_2 = -7.67$

Since the number of terms $n$ must be a positive integer, we discard the negative solution. Therefore, $n = 6$.

Thus, 6 terms of the progression must be taken for their sum to be 69.

Would you like more details, or do you have any other questions?

Here are five related questions:

1. How do we find the sum of an arithmetic progression?
2. What happens if the common difference in the progression changes?
3. Can the quadratic formula always solve arithmetic progression problems?
4. How do we verify that the sum of the first 6 terms is indeed 69?
5. How would the solution change if the required sum were different?

Tip: When solving quadratic equations, always check for negative or non-real solutions and ensure the answer makes sense in the context of the problem.