91Ó°ÊÓ

Recall the formula for the nth term of a geometric progression: \[a_n = a_1 \times r^{(n-1)}\] where \(a_n\) is the nth term, \(a_1\) is the first term, r is the common ratio, and n is the number of terms. In this case, we have \(a_1 = 27\), \(a_n = \frac{32}{9}\).

For a geometric progression with a non-zero common ratio, the formula for the sum of the first n terms is: \[S_n = \frac{a_1 \times (1 - r^n)}{1 - r}\] In this case, we have \(S_n = \frac{665}{9}\).

Now, we can substitute \(a_1\) and \(a_n\) into the equation for the nth term to obtain: \[\frac{32}{9} = 27 \times r^{(n-1)}\] We can also substitute \(a_1\) and \(S_n\) into the equation for the sum of n terms to obtain: \[\frac{665}{9} = \frac{27 \times (1 - r^n)}{1 - r}\] This gives us a system of two equations with two variables, n and r: \[\begin{cases} \frac{32}{9} = 27 \times r^{(n-1)}\\ \frac{665}{9} = \frac{27 \times (1 - r^n)}{1 - r} \end{cases}\]

Divide the first equation by \(27\) to isolate the term \(r^{(n-1)}\): \[r^{(n-1)} = \frac{32}{243}\] Now, substitute this expression into the second equation: \[\frac{665}{9} = \frac{27 \times (1 - r^n)}{1 - r}\cdot\frac{243}{32}\] Solving for r we get: \[r = \frac{3}{4}\] Now, substitute the value of r back into the expression \(r^{(n-1)}\): \[\left(\frac{3}{4}\right)^{(n-1)} = \frac{32}{243}\] Applying the logarithm to both sides: \[(n-1)\log\left(\frac{3}{4}\right) = \log\left(\frac{32}{243}\right)\] Divide by \(\log\left(\frac{3}{4}\right)\) and solve for n: \[n-1 = \frac{\log\left(\frac{32}{243}\right)}{\log\left(\frac{3}{4}\right)}\] \[n = \frac{\log\left(\frac{32}{243}\right)}{\log\left(\frac{3}{4}\right)} + 1\] Using a calculator, we can find: \[n \approx 5\] So, the values of n and r are 5 and \(\frac{3}{4}\), respectively.