Let's go through the questions in the image one by one and solve them.

1. Find the 16th term of an arithmetic sequence whose first three terms are 7, –4, –15, ...

In an arithmetic sequence, the general form is:

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

Where $a_n$ is the nth term, $a_1$ is the first term, $n$ is the term number, and $d$ is the common difference.

• First term $a_1 = 7$
• Common difference $d = -4 - 7 = -11$

Thus, the 16th term ($a_{16}$) is:

$a_{16} = 7 + (16 - 1)(-11) = 7 + 15(-11) = 7 - 165 = -158$

The 16th term is –158.

2. Find the first term of an arithmetic sequence whose 6th term is –4 and the 10th term is –28.

We have the formula for an arithmetic sequence:

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

We know:

• $a_6 = -4$ → $-4 = a_1 + 5d$
• $a_{10} = -28$ → $-28 = a_1 + 9d$

Subtracting the first equation from the second:

$(-28) - (-4) = (a_1 + 9d) - (a_1 + 5d)$ $-24 = 4d \quad \Rightarrow \quad d = -6$

Now substitute $d = -6$ into the first equation:

$-4 = a_1 + 5(-6)$ $-4 = a_1 - 30 \quad \Rightarrow \quad a_1 = 26$

The first term is 26.

3. Find the 10th term of the geometric sequence 3, –6, 12, ...

The formula for the nth term of a geometric sequence is:

$a_n = a_1 \cdot r^{n-1}$

Where $r$ is the common ratio. To find $r$:

$r = \frac{-6}{3} = -2$

Thus, the 10th term is:

$a_{10} = 3 \cdot (-2)^{9} = 3 \cdot (-512) = -1536$

The 10th term is –1536.

4. Find the 11th term of the harmonic sequence $\frac{1}{8}, \frac{1}{11}, \frac{1}{14}, \frac{1}{17}, \dots$

For harmonic sequences, the reciprocals form an arithmetic sequence. The terms are $8, 11, 14, 17, \dots$, with a common difference $d = 3$.

To find the 11th term of the arithmetic sequence:

$a_n = a_1 + (n - 1)d$ $a_{11} = 8 + (11 - 1)(3) = 8 + 30 = 38$

So the 11th term of the harmonic sequence is:

$\frac{1}{38}$

5. Which term is 132 in the sequence whose $a_1 = 13$ and $d = 7$?

The formula for the nth term is:

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

We know $a_n = 132$, $a_1 = 13$, and $d = 7$:

$132 = 13 + (n - 1)(7)$ $132 - 13 = (n - 1)(7) \quad \Rightarrow \quad 119 = (n - 1)(7)$ $n - 1 = \frac{119}{7} = 17 \quad \Rightarrow \quad n = 18$

The 18th term is 132.

Would you like me to continue solving the remaining questions?

Related Questions:

1. How do you calculate the common difference in an arithmetic sequence?
2. What is the general formula for the nth term in a geometric sequence?
3. How do harmonic sequences differ from arithmetic sequences?
4. Can you find the sum of terms in an arithmetic progression?
5. How do you calculate the nth term in a Fibonacci sequence?

Tip:

Always check whether the problem involves arithmetic, geometric, or harmonic sequences before deciding the appropriate formula to apply.