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The first three terms of a geometric sequence are $3m - 4$, 6, and $m + 1$, where $m$ is an integer. Prove that $m$ satisfies the equation $3m^2 - m - 40 = 0$.
In a geometric sequence, the third term, $t_3$, is 64, and the seventh term, $t_7$, is 1024.
1. What is the common ratio of this sequence?
2. Determine the first term, $t_1$, of the sequence.
3. Find the total sum of the first 10 terms of the sequence.
Given that $a$, $3a$, and $a + 40$ are consecutive terms of a sequence:
(i) What is the value of $a$ if this sequence is geometric?
(ii) List the three terms of the sequence and identify the common ratio.
Consider the sequence $$S_n = \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \frac{1}{81} + \dots + \frac{1}{3^n}$$.
(a) Calculate $S_1, S_2, S_3, S_4,$ and $S_5$ in fractional form.
(b) Based on your calculations, hypothesize a formula for $S_n$.
(c) Determine $S_n$ using the formula $$S_n = \frac{u_1(1 - r^n)}{1 - r}$$.
(d) Discuss the behavior of $S_n$ as $n$ becomes very large.