The first sequence is $$2 \cdot 3^{n-1}$$ (geometric with ratio $$3$$). The second sequence is $$5^{n-1}$$ (geometric with ratio $$5$$). The sum sequence is $$2 \cdot 3^{n-1} + 5^{n-1}$$: $$3, 11, 43, 179, 787, \ldots$$. Differences: $$8, 32, 136, 608, \ldots$$ (not constant). Ratios: $$\frac{11}{3} \approx 3.67$$, $$\frac{43}{11} \approx 3.91$$, $$\frac{179}{43} \approx 4.16$$ (not constant). The sequence is neither arithmetic nor geometric. Choices A and D assume it's arithmetic. Choice B assumes it's geometric.

When you encounter a sequence problem, you need to identify the pattern by examining the differences between consecutive terms. This sequence tests your ability to recognize when a quadratic expression generates the terms.

Let's find the pattern by calculating the first and second differences:

First differences: $$7-2=5$$, $$17-7=10$$, $$32-17=15$$, $$52-32=20$$ Second differences: $$10-5=5$$, $$15-10=5$$, $$20-15=5$$

Since the second differences are constant (5), this confirms the sequence follows a quadratic pattern of the form $$an^2 + bn + c$$.

To verify the correct answer, test choice D with the given terms:

• For $$n=1$$: $$2.5(1)^2 - 2.5(1) + 2 = 2.5 - 2.5 + 2 = 2$$ ✓
• For $$n=2$$: $$2.5(4) - 2.5(2) + 2 = 10 - 5 + 2 = 7$$ ✓
• For $$n=3$$: $$2.5(9) - 2.5(3) + 2 = 22.5 - 7.5 + 2 = 17$$ ✓

Choice A gives incorrect values: for $$n=1$$, $$5(1) - 13(1) + 10 = 2$$, but for $$n=2$$, $$5(4) - 13(2) + 10 = 16 \neq 7$$.

Choice B similarly fails: for $$n=2$$, $$4(4) - 11(2) + 9 = 11 \neq 7$$.

Choice C also produces wrong values: for $$n=2$$, $$3(4) - 8(2) + 7 = 7$$, but for $$n=3$$, $$3(9) - 8(3) + 7 = 10 \neq 17$$.

Strategy tip: When you see constant second differences, you know you're dealing with a quadratic sequence. Always test your answer choice with multiple terms from the sequence to confirm it works consistently.

Computing the sequence: $$a_1 = 2$$, $$a_2 = 5$$, $$a_3 = 3(5) - 2(2) = 11$$, $$a_4 = 3(11) - 2(5) = 23$$, $$a_5 = 3(23) - 2(11) = 47$$, $$a_6 = 3(47) - 2(23) = 95$$, $$a_7 = 3(95) - 2(47) = 191$$. The 7th term is the first to exceed $$100$$. Choice A gives $$a_6 = 95 < 100$$. Choices C and D are later terms that also exceed $$100$$ but are not the first.

The sequence follows the pattern: $$a_n = 2a_{n-1} - 1$$ with $$a_1 = 3$$. Computing: $$a_1 = 3$$, $$a_2 = 2(3) - 1 = 5$$, $$a_3 = 2(5) - 1 = 9$$, $$a_4 = 2(9) - 1 = 17$$, $$a_5 = 2(17) - 1 = 33$$. The difference is $$a_5 - a_4 = 33 - 17 = 23$$. Choice A comes from incorrectly computing $$a_4 = 16$$ and $$a_5 = 31$$, giving difference $$22$$. Choice C is $$a_5 + 12 = 45$$. Choice D is $$a_4 + 30 = 47$$.

When you encounter geometric sequence problems, remember that each term is found by multiplying the previous term by a constant ratio $$r$$. The general formula is $$a_n = a_1 \cdot r^{n-1}$$, where $$a_1$$ is the first term.

Given that the 3rd term is $$18$$ and the 6th term is $$486$$, you can write:

• $$a_3 = a_1 \cdot r^2 = 18$$
• $$a_6 = a_1 \cdot r^5 = 486$$

To find the common ratio, divide the second equation by the first: $$\frac{a_6}{a_3} = \frac{a_1 \cdot r^5}{a_1 \cdot r^2} = r^3 = \frac{486}{18} = 27$$

Taking the cube root: $$r = 3$$

Now you can find $$a_1$$: $$18 = a_1 \cdot 3^2 = 9a_1$$, so $$a_1 = 2$$

Therefore, $$a_4 = a_1 \cdot r^3 = 2 \cdot 3^3 = 2 \cdot 27 = 54$$. The answer is A.

Looking at the wrong choices: B) $$36$$ would result from incorrectly calculating $$r = 2$$ instead of $$3$$. C) $$72$$ might come from miscalculating $$a_1 = 8$$ instead of $$2$$. D) $$81$$ would result from thinking $$a_4 = r^4 = 3^4$$, forgetting to include the first term coefficient.

Strategy tip: In geometric sequence problems, always use the ratio between given terms to find $$r$$ first. When terms aren't consecutive, remember that $$\frac{a_m}{a_n} = r^{m-n}$$. This approach avoids having to solve for $$a_1$$ until the end.