Find the indicated number of missing terms (means) in each given sequence:
points each)
a. arithmetic means between and

Ask by Gibson Chan. in the United States

Mar 14,2025

Question

Find the indicated number of missing terms (means) in each given sequence:
points each)
a. arithmetic means between and

Ask by Gibson Chan. in the United States

Mar 14,2025

UpStudy AI · Accepted Answer

We are given an arithmetic progression with first term
$$
t_0=\log_7(144)
$$
and sixth term
$$
t_6=\log_7\left(\frac{9}{4}\right)
$$
with 5 arithmetic means inserted between them. In an arithmetic progression the common difference $$ d $$ is given by
$$
d=\frac{t_6-t_0}{6}.
$$

**Step 1. Compute the common difference $$ d $$:**

Calculate the difference:
$$
t_6 - t_0 = \log_7\left(\frac{9}{4}\right)-\log_7(144).
$$
Using the logarithm property $$\log_b(x)-\log_b(y)=\log_b\left(\frac{x}{y}\right)$$, we have
$$
t_6-t_0 = \log_7\left(\frac{\frac{9}{4}}{144}\right)=\log_7\left(\frac{9}{576}\right).
$$
Notice that $$576= 144\times 4$$ and simplifying $$\frac{9}{576}$$ we get:
$$
\frac{9}{576}=\frac{1}{64}.
$$
Thus,
$$
t_6-t_0 = \log_7\left(\frac{1}{64}\right).
$$

Now divide by 6:
$$
d=\frac{\log_7\left(\frac{1}{64}\right)}{6}.
$$
We recognize that dividing by 6 inside the logarithm is equivalent to taking a 6th root:
$$
d = \log_7\left(\left(\frac{1}{64}\right)^{\frac{1}{6}}\right).
$$
Since $$64=2^6$$, we have:
$$
\left(\frac{1}{64}\right)^{\frac{1}{6}}=\frac{1}{2}.
$$
Therefore,
$$
d=\log_7\left(\frac{1}{2}\right).
$$

**Step 2. Find the arithmetic means:**

The $$n^\text{th}$$ term of an arithmetic progression is given by
$$
t_n=t_0+n\cdot d.
$$
We need to find $$t_1, t_2, t_3, t_4,$$ and $$t_5$$.

- For $$n=1$$:
  $$
  t_1=\log_7(144)+\log_7\left(\frac{1}{2}\right)=\log_7\left(144\cdot\frac{1}{2}\right)=\log_7(72).
  $$

- For $$n=2$$:
  $$
  t_2=\log_7(144)+2\log_7\left(\frac{1}{2}\right)=\log_7\left(144\cdot\left(\frac{1}{2}\right)^2\right)=\log_7(36).
  $$

- For $$n=3$$:
  $$
  t_3=\log_7(144)+3\log_7\left(\frac{1}{2}\right)=\log_7\left(144\cdot\left(\frac{1}{2}\right)^3\right)=\log_7(18).
  $$

- For $$n=4$$:
  $$
  t_4=\log_7(144)+4\log_7\left(\frac{1}{2}\right)=\log_7\left(144\cdot\left(\frac{1}{2}\right)^4\right)=\log_7(9).
  $$

- For $$n=5$$:
  $$
  t_5=\log_7(144)+5\log_7\left(\frac{1}{2}\right)=\log_7\left(144\cdot\left(\frac{1}{2}\right)^5\right)=\log_7\left(\frac{144}{32}\right)=\log_7\left(\frac{9}{2}\right),
  $$
  since $$\frac{144}{32}=\frac{9}{2}$$.

Thus, the 5 arithmetic means are:
$$
\log_7(72),\quad \log_7(36),\quad \log_7(18),\quad \log_7(9),\quad \log_7\left(\frac{9}{2}\right).
$$
The 5 arithmetic means between $$ \log_7(144) $$ and $$ \log_7\left(\frac{9}{4}\right) $$ are: $$ \log_7(72),\quad \log_7(36),\quad \log_7(18),\quad \log_7(9),\quad \log_7\left(\frac{9}{2}\right). $$

Find the indicated number of missing terms (means) in each given sequence:
points each)
a. arithmetic means between and

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Find the indicated number of missing terms (means) in each given sequence: points each) a. arithmetic means between and

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