Given the problem, we have the fourth term $u_4 = 135$ and the fifth term $u_5 = 81$ of a geometric sequence. We'll proceed to solve each part.

(a) Find the common ratio $r$ of the sequence.

In a geometric sequence, each term is obtained by multiplying the previous term by the common ratio $r$. Therefore: $u_5 = u_4 \cdot r$ Substituting the given values: $81 = 135 \cdot r$ Solving for $r$: $r = \frac{81}{135} = \frac{3}{5} = 0.6$ So, the common ratio $r = 0.6$.

(b) Find $u_1$, the first term of the sequence.

Using the formula for the $n$-th term of a geometric sequence: $u_n = u_1 \cdot r^{n-1}$ For $u_4$, we have: $135 = u_1 \cdot r^{4-1} = u_1 \cdot r^3$ Substituting $r = 0.6$: $135 = u_1 \cdot (0.6)^3$ Calculating $(0.6)^3$: $0.6^3 = 0.216$ Thus: $135 = u_1 \cdot 0.216$ Solving for $u_1$: $u_1 = \frac{135}{0.216} = 625$ So, the first term $u_1 = 625$.

(c) Calculate the sum of the first 20 terms of the sequence.

The sum of the first $n$ terms of a geometric sequence is given by: $S_n = u_1 \cdot \frac{1 - r^n}{1 - r}$ For $n = 20$, $u_1 = 625$, and $r = 0.6$: $S_{20} = 625 \cdot \frac{1 - (0.6)^{20}}{1 - 0.6}$ Calculating $(0.6)^{20}$: $0.6^{20} \approx 0.000366$ Substituting: $S_{20} = 625 \cdot \frac{1 - 0.000366}{0.4}$ $= 625 \cdot \frac{0.999634}{0.4}$ $\approx 625 \cdot 2.499085 = 1561.928125$ So, the sum of the first 20 terms, $S_{20} \approx 1561.93$.

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Related Questions

1. How would you find the common ratio if given two non-consecutive terms?
2. What is the formula for the sum of an infinite geometric series, and when does it apply?
3. Can the common ratio be greater than 1 in a converging sequence?
4. What changes in the formula if the sequence is decreasing rather than increasing?
5. How does the sum formula change if the common ratio is negative?

Tip: For any geometric sequence, you can confirm your answer by calculating intermediate terms to ensure consistency with the initial values given.