1. Problem: Find the value of the infinite series starting with $3 - 2 + \frac{4}{3} - \frac{8}{9} + ...$ corresponding to the given multiple-choice answers. Step 1: Observe the pattern in the series: $3, -2, \frac{4}{3}, -\frac{8}{9}, ...$. Step 2: Notice the terms alternate in sign and the absolute value follows a pattern. Step 3: Express terms generally: $$a_n = 3 \times \left(-\frac{2}{3}\right)^{n-1}$$ Step 4: Because $\left| -\frac{2}{3} \right| < 1$, the series is geometric with first term $a = 3$ and ratio $r = -\frac{2}{3}$. Step 5: Sum of infinite geometric series is $$S = \frac{a}{1-r} = \frac{3}{1 - (-\frac{2}{3})} = \frac{3}{1 + \frac{2}{3}} = \frac{3}{\frac{5}{3}} = \frac{3 \times 3}{5} = \frac{9}{5}$$. Final answer for first problem is $\boxed{\frac{9}{5}}$, which corresponds to option D. --- 2. Problem: Given arithmetic sequence with $a_5=4$ and $a_8=10$, find the sum of the first 10 terms. Step 1: Let the first term be $a$ and common difference $d$. Step 2: Use given terms: $$a + 4d = 4$$ $$a + 7d = 10$$ Step 3: Subtract equations: $$(a + 7d) - (a + 4d) = 10 - 4 \Rightarrow 3d = 6 \Rightarrow d = 2$$ Step 4: Substitute $d=2$ into first equation: $$a + 4 \times 2 = 4 \Rightarrow a + 8 = 4 \Rightarrow a = -4$$ Step 5: Sum of first 10 terms: $$S_{10} = \frac{10}{2} (2a + 9d) = 5 (2\times (-4) + 9 \times 2) = 5(-8 + 18) = 5 \times 10 = 50$$ Final answer for second problem is $\boxed{50}$ which corresponds to option A. --- 3. Problem: Five children have ages forming an arithmetic sequence. Youngest is 7, oldest is 19. Find sum of their ages 5 years later. Step 1: The 5 children's ages are arithmetic sequence with 5 terms. Step 2: Let first term $a=7$ and last term $l=19$. Step 3: Sum now: $$S_5 = \frac{5}{2} (a + l) = \frac{5}{2} (7 + 19) = \frac{5}{2} \times 26 = 65$$ Step 4: After 5 years, each child is 5 years older, so total addition: $$5 \times 5 = 25$$ Step 5: Sum after 5 years: $$65 + 25 = 90$$ Final answer for third problem is $\boxed{90}$, which corresponds to option C.