Let's tackle the two problems one by one.

Problem 5:

The problem gives us the 4th term ($a_4 = 7$) and the 30th term ($a_{30} = 85$) of an arithmetic sequence. We are to find the first term $a$ and the common difference $d$, and then use them to find the 7th term ($a_7$).

Solution:

The general term of an arithmetic sequence is given by: $a_n = a + (n-1) \cdot d$

1. For the 4th term ($a_4 = 7$): $a + 3d = 7$

2. For the 30th term ($a_{30} = 85$): $a + 29d = 85$

Now, we can solve these two equations simultaneously to find $a$ and $d$.

Step 1: Subtract the first equation from the second.

$(a + 29d) - (a + 3d) = 85 - 7$ $26d = 78 \quad \Rightarrow \quad d = 3$

Step 2: Substitute $d = 3$ into the first equation.

$a + 3(3) = 7 \quad \Rightarrow \quad a + 9 = 7 \quad \Rightarrow \quad a = -2$

Now we have $a = -2$ and $d = 3$.

Step 3: Find the 7th term ($a_7$).

$a_7 = a + 6d = -2 + 6(3) = -2 + 18 = 16$

Final Answer for Problem 5:

• $a = -2$
• $d = 3$
• 7th term = 16

Problem 6:

We are given that the 3rd term ($a_3 = 18$) and the 6th term ($a_6 = 486$) of an arithmetic sequence. We need to find the general rule for the sequence, i.e., the first term $a$ and the common difference $d$.

Solution:

Using the general formula for the $n$-th term: $a_n = a + (n-1) \cdot d$

1. For the 3rd term ($a_3 = 18$): $a + 2d = 18$

2. For the 6th term ($a_6 = 486$): $a + 5d = 486$

We now solve these two equations simultaneously.

Step 1: Subtract the first equation from the second.

$(a + 5d) - (a + 2d) = 486 - 18$ $3d = 468 \quad \Rightarrow \quad d = 156$

Step 2: Substitute $d = 156$ into the first equation.

$a + 2(156) = 18 \quad \Rightarrow \quad a + 312 = 18 \quad \Rightarrow \quad a = 18 - 312 = -294$

Thus, the general term of the sequence is: $a_n = -294 + (n-1) \cdot 156$

Final Answer for Problem 6:

• $a = -294$
• $d = 156$
• General rule: $a_n = -294 + (n-1) \cdot 156$

Would you like more details or further clarifications on any of these steps?

Here are five related questions to explore further:

1. How can we derive the formula for the sum of an arithmetic sequence?
2. What happens to an arithmetic sequence if the common difference is zero?
3. How can we find the $n$-th term of an arithmetic sequence if the first term and the sum of the sequence are known?
4. How do geometric sequences differ from arithmetic sequences in their structure?
5. Can we find the number of terms in an arithmetic sequence if we know the sum and the common difference?

Tip: When solving simultaneous equations, subtraction is often the easiest way to eliminate one of the variables, making it simpler to find a solution.