A geometric sequence starts at 243, and every new term is found by dividing by 3. Which statement correctly gives 𝑇5 and describes what happens if the sequence continues? A. 𝑇5 =19683; multiplying by 3 instead produces geometric growth that moves increasingly far from 0. B. 𝑇5 =1; division occurs 5 times, and the sequence reaches 0 immediately afterwards. C. 𝑇5 =231; subtracting 3 each time creates a constant-difference sequence that eventually reaches 0. D. 𝑇5 =3; later positive terms approach 0 but do not reach 0. Handwritten evidence: Write an explicit or repeated-division rule, and calculate 𝑇5, and explain the difference between approaching 0 and reaching 0. Question 15 A geometric sequence starts at 3 and has a common ratio of 2. What is term 5? A. 𝑇5 =11, treating the ratio as a constant difference B. 𝑇5 =96, using the term number as the exponent C. 𝑇5 =48, using first 𝑥 ratio^(term-1)[𝑥𝑟𝑛−1] D. 𝑇5 =30, multiplying the first term, index and ratio once Handwritten evidence: Write the repeated multiplication or the explicit rule used. Question 16 Thandi writes 𝑇𝑛 = 4𝑛 + 5 for the sequence 5, 9, 13, 17, ... because the difference is 4 and the first term is 5. Which feedback is most useful? A. Keep the rule because it contains both important numbers, and a rule does not need to work for the first term. B. Change the coefficient to 5, because the first visible number must always multiply 𝑛 in an explicit rule. C. The coefficient 4 is suitable, but +5 gives the wrong first term; compare 5 with 4(1) to obtain � �𝑛 = 4𝑛+1. D. Replace it only with 𝑇𝑛 = 𝑇𝑛−1 + 4, because an arithmetic sequence cannot have an explicit rule, even when a rule in terms of 𝑛 is requested. Handwritten evidence: Test the learner's rule for two terms, derive the rule, and state a feedback question. (3)

The first term is $a_1 = 243$.
Dividing by 3 means the common ratio is $r = \frac{1}{3}$.

The repeated-division rule is $T_n = T_{n-1} \div 3$.

The explicit rule for a geometric sequence is $T_n = a_1 \cdot r^{n-1}$.

Using the explicit rule:

Alternatively, by repeated division:
$T_1 = 243$
$T_2 = 243 \div 3 = 81$
$T_3 = 81 \div 3 = 27$
$T_4 = 27 \div 3 = 9$
$T_5 = 9 \div 3 = 3$

Approaching 0 means that the terms of the sequence get progressively closer to 0 without ever actually becoming 0. In this sequence, each term is obtained by dividing the previous term by 3. Since the starting term (243) is positive, all subsequent terms will also be positive. Dividing a positive number by 3 always results in a positive number, never 0. Therefore, the terms will become smaller and smaller, getting arbitrarily close to 0, but they will never reach 0. Reaching 0 would imply that one of the terms in the sequence is exactly 0.

The final answer is $\boxed{\text{D. } T_5 =3\text{; later positive terms approach 0 but do not reach 0.}}$

The first term is $a_1 = 3$.
The common ratio is $r = 2$.

The explicit rule for a geometric sequence is $T_n = a_1 \cdot r^{n-1}$.

$T_1 = 3$
$T_2 = 3 \cdot 2 = 6$
$T_3 = 6 \cdot 2 = 12$
$T_4 = 12 \cdot 2 = 24$
$T_5 = 24 \cdot 2 = 48$

The final answer is $\boxed{\text{C. } T_5 =48\text{, using first } x \text{ ratio}^{\text{(term-1)}} [xr^{n-1}]}$.

Thandi's rule: $T_n = 4n + 5$
For $n=1$: $T_1 = 4(1) + 5 = 4 + 5 = 9$.
The actual first term of the sequence is 5. So, Thandi's rule gives an incorrect first term.
For $n=2$: $T_2 = 4(2) + 5 = 8 + 5 = 13$.
The actual second term of the sequence is 9. So, Thandi's rule also gives an incorrect second term.

The given sequence is 5, 9, 13, 17, ...
This is an arithmetic sequence.
The first term is $a_1 = 5$.
The common difference $d = 9 - 5 = 4$.
The explicit rule for an arithmetic sequence is $T_n = a_1 + (n-1)d$.
Substitute $a_1 = 5$ and $d = 4$:

"Your rule $T_n = 4n + 5$ correctly identifies the common difference as 4. However, when you substitute $n=1$ into your rule, you get $T_1 = 4(1) + 5 = 9$. The actual first term of the sequence is 5. How can you adjust the constant term in your rule (the '+5' part) so that it correctly gives the first term of the sequence when $n=1$?"

The final answer is $\boxed{\text{C. The coefficient 4 is suitable, but +5 gives the wrong first term; compare 5 with 4(1) to obtain } T_n = 4n+1.}$