Question

find the indicated geometric means for each pair of nonconsecutive terms.

1. 4 and 256; 2 means
2. 256 and 81; 3 means
3. \\(\frac{4}{7}\\) and 7; 1 mean
4. -2 and 54; 2 means
5. 1 and 27; 2 means
6. 48 and -750; 2 means
7. \\(i\\) and -1; 4 means
8. \\(t^8\\) and \\(t^{-7}\\); 4 means

Explanation:

Identify the geometric sequence formula

To find the geometric means between two nonconsecutive terms \(a_1\) and \(a_n\), we treat them as terms of a Geometric Sequence. If there are \(k\) geometric means between \(a_1\) and \(a_n\), then \(a_n\) is the \((k+2)\)-th term of the sequence:

$$a_n = a_1 \cdot r^{k+1}$$

where \(r\) is the Common Ratio.

Solve for the common ratio in each problem

We solve for \(r\) using \(r = \sqrt[k+1]{\frac{a_n}{a_1}}\).

• Question 32: 4 and 256; 2 means. \(a_1 = 4\), \(a_4 = 256\).

$$r^3 = \frac{256}{4} = 64 \implies r = 4$$

Means: \(4 \cdot 4 = 16\), \(16 \cdot 4 = 64\).

• Question 33: 256 and 81; 3 means. \(a_1 = 256\), \(a_5 = 81\).

$$r^4 = \frac{81}{256} \implies r = \pm \frac{3}{4}$$

Using positive \(r = \frac{3}{4}\): \(192, 144, 108\).
Using negative \(r = -\frac{3}{4}\): \(-192, 144, -108\).

• Question 34: \(\frac{4}{7}\) and 7; 1 mean. \(a_1 = \frac{4}{7}\), \(a_3 = 7\).

$$r^2 = \frac{7}{4/7} = \frac{49}{4} \implies r = \pm \frac{7}{2}$$

Means: \(\pm 2\).

• Question 35: \(-2\) and 54; 2 means. \(a_1 = -2\), \(a_4 = 54\).

$$r^3 = \frac{54}{-2} = -27 \implies r = -3$$

Means: \(6, -18\).

Solve remaining problems

• Question 36: 1 and 27; 2 means. \(a_1 = 1\), \(a_4 = 27\).

$$r^3 = \frac{27}{1} = 27 \implies r = 3$$

Means: \(3, 9\).

• Question 37: 48 and \(-750\); 2 means. \(a_1 = 48\), \(a_4 = -750\).

$$r^3 = \frac{-750}{48} = -\frac{125}{8} \implies r = -\frac{5}{2}$$

Means: \(-120, 300\).

• Question 38: \(t\) and \(-1\); 4 means. \(a_1 = t\), \(a_6 = -1\).

$$r^5 = \frac{-1}{t} = -t^{-1} \implies r = -t^{-1/5}$$

Means: \(-t^{4/5}, t^{3/5}, -t^{2/5}, t^{1/5}\).

• Question 39: \(t^8\) and \(t^{-7}\); 4 means. \(a_1 = t^8\), \(a_6 = t^{-7}\).

$$r^5 = \frac{t^{-7}}{t^8} = t^{-15} \implies r = t^{-3}$$

Means: \(t^5, t^2, t^{-1}, t^{-4}\).

Answer:

No.	Answer
33	\(\pm 192, 144, \pm 108\)
34	\(\pm 2\)
35	\(6, -18\)
36	\(3, 9\)
37	\(-120, 300\)
38	\(-t^{4/5}, t^{3/5}, -t^{2/5}, t^{1/5}\)
39	\(t^5, t^2, t^{-1}, t^{-4}\)