Find the indicated number of missing terms (means) in each given sequence: \( (5 \) points each) a. \( \quad 5 \) arithmetic means between \( \log _{7}(144) \) and \( \log _{7}\left(\frac{9}{4}\right) \)

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: \[ \log_7(72),\quad \log_7(36),\quad \log_7(18),\quad \log_7(9),\quad \log_7\left(\frac{9}{2}\right). \]