1. Problem 2: In an arithmetic sequence, given $u_6 = -5$ and $u_9 = -20$, find $S_{20}$ (sum of first 20 terms). 2. Formula for $n$th term of arithmetic sequence: $$u_n = a + (n-1)d$$ where $a$ is first term, $d$ common difference. 3. Sum of first $n$ terms: $$S_n = \frac{n}{2} (2a + (n-1)d)$$ 4. Use given terms to find $a$ and $d$: $$u_6 = a + 5d = -5$$ $$u_9 = a + 8d = -20$$ 5. Subtract equations: $$ (a + 8d) - (a + 5d) = -20 - (-5) \Rightarrow 3d = -15 \Rightarrow d = -5$$ 6. Substitute $d$ back: $$a + 5(-5) = -5 \Rightarrow a - 25 = -5 \Rightarrow a = 20$$ 7. Find $S_{20}$: $$S_{20} = \frac{20}{2} (2(20) + 19(-5)) = 10 (40 - 95) = 10 (-55) = -550$$ --- 1. Problem 4: Geometric series $0.5 - 0.1 + 0.02 + \dots$ with sum $S_n = 0.416$. Find number of terms $n$. 2. Formula for sum of geometric series: $$S_n = a \frac{1-r^n}{1-r}$$ where $a$ is first term, $r$ common ratio. 3. Identify $a=0.5$, $r = \frac{-0.1}{0.5} = -0.2$. 4. Substitute values: $$0.416 = 0.5 \frac{1 - (-0.2)^n}{1 - (-0.2)} = 0.5 \frac{1 - (-0.2)^n}{1.2} = \frac{0.5}{1.2} (1 - (-0.2)^n)$$ 5. Simplify: $$0.416 = 0.4167 (1 - (-0.2)^n)$$ 6. Divide both sides: $$\frac{0.416}{0.4167} = 1 - (-0.2)^n \Rightarrow 0.9983 = 1 - (-0.2)^n$$ 7. Rearrange: $$(-0.2)^n = 1 - 0.9983 = 0.0017$$ 8. Since $|r|<1$, solve: $$|(-0.2)^n| = 0.0017 \Rightarrow 0.2^n = 0.0017$$ 9. Take log base 10: $$n \log 0.2 = \log 0.0017 \Rightarrow n = \frac{\log 0.0017}{\log 0.2} \approx \frac{-2.7696}{-0.69897} \approx 3.96$$ 10. Since $n$ must be integer, $n=4$ terms. --- 1. Problem 5: Geometric progression with $u_3=4.5$ and $u_7=22.78125$. Find common ratio $r$ and first term $a$. 2. Formula for $n$th term: $$u_n = a r^{n-1}$$ 3. Write equations: $$u_3 = a r^2 = 4.5$$ $$u_7 = a r^6 = 22.78125$$ 4. Divide second by first: $$\frac{a r^6}{a r^2} = r^4 = \frac{22.78125}{4.5} = 5.0625$$ 5. Solve for $r$: $$r = \sqrt[4]{5.0625} = 1.5$$ 6. Substitute $r$ back: $$a (1.5)^2 = 4.5 \Rightarrow a \times 2.25 = 4.5 \Rightarrow a = 2$$ --- 1. Problem 7: Find number of terms in arithmetic sequence $4,7,10,\dots,61$. 2. Use formula for $n$th term: $$u_n = a + (n-1)d$$ 3. Here $a=4$, $d=3$, last term $u_n=61$. 4. Solve: $$61 = 4 + (n-1)3 \Rightarrow 61 - 4 = 3(n-1) \Rightarrow 57 = 3(n-1)$$ 5. Divide: $$\cancel{3}(n-1) = \frac{57}{\cancel{3}} = 19 \Rightarrow n-1=19 \Rightarrow n=20$$ --- 1. Problem 9: Three consecutive terms of geometric sequence are $x-3$, $6$, $x+2$. Find $x$. 2. Property: middle term squared equals product of neighbors: $$6^2 = (x-3)(x+2)$$ 3. Calculate: $$36 = x^2 - 3x + 2x - 6 = x^2 - x - 6$$ 4. Rearrange: $$x^2 - x - 6 - 36 = 0 \Rightarrow x^2 - x - 42 = 0$$ 5. Solve quadratic: $$x = \frac{1 \pm \sqrt{1 + 168}}{2} = \frac{1 \pm \sqrt{169}}{2} = \frac{1 \pm 13}{2}$$ 6. Solutions: $$x = 7 \text{ or } x = -6$$ --- 1. Problem 10a: Tank with 55 liters, water flows out at 7% per minute. Write sequence for volume after $n$ minutes. 2. Volume after 1 minute: $$55 \times (1 - 0.07) = 55 \times 0.93$$ 3. Sequence: $$V_n = 55 \times 0.93^n$$ for $n=1,2,3,\dots$ --- 1. Problem 10b: Type of sequence? 2. Since each term is previous term multiplied by constant $0.93$, it is a geometric sequence. --- 1. Problem 10c: Volume after 10 minutes: 2. Calculate: $$V_{10} = 55 \times 0.93^{10} \approx 55 \times 0.484 = 26.62$$ liters --- 1. Problem 10d: Water drained after 15 minutes: 2. Initial volume: 55 liters 3. Volume left after 15 minutes: $$V_{15} = 55 \times 0.93^{15} \approx 55 \times 0.335 = 18.43$$ liters 4. Water drained: $$55 - 18.43 = 36.57$$ liters --- 1. Problem 10e: Time to drain tank (volume approaches 0). 2. Since volume decreases by 7% each minute, it never reaches zero but approaches it. 3. To find time $t$ when volume is less than 1 liter: $$55 \times 0.93^t < 1$$ 4. Divide: $$0.93^t < \frac{1}{55} = 0.01818$$ 5. Take log: $$t \log 0.93 < \log 0.01818 \Rightarrow t > \frac{\log 0.01818}{\log 0.93} \approx \frac{-1.740}{-0.031} = 56.13$$ 6. So after about 57 minutes volume is less than 1 liter. --- 1. Problem 12: Loan €20987, total paid €22960 after 4 years, compounded monthly. Find annual interest rate $r$. 2. Formula for compound interest: $$A = P \left(1 + \frac{r}{m}\right)^{mt}$$ 3. Given: $$A=22960, P=20987, t=4, m=12$$ 4. Substitute: $$22960 = 20987 \left(1 + \frac{r}{12}\right)^{48}$$ 5. Divide: $$\frac{22960}{20987} = \left(1 + \frac{r}{12}\right)^{48} \Rightarrow 1.0949 = \left(1 + \frac{r}{12}\right)^{48}$$ 6. Take 48th root: $$1 + \frac{r}{12} = 1.0949^{1/48} \approx 1.0019$$ 7. Solve for $r$: $$\frac{r}{12} = 0.0019 \Rightarrow r = 0.0228 = 2.28\%$$ annual interest rate. --- 1. Problem 14: Expand $(3x - y)^6$ using binomial theorem. 2. Binomial theorem: $$(a - b)^n = \sum_{k=0}^n \binom{n}{k} a^{n-k} (-b)^k$$ 3. Here $a=3x$, $b=y$, $n=6$. 4. Expansion: $$\sum_{k=0}^6 \binom{6}{k} (3x)^{6-k} (-y)^k = \sum_{k=0}^6 \binom{6}{k} 3^{6-k} x^{6-k} (-1)^k y^k$$ 5. Write terms explicitly: $$729x^6 - 1458x^5 y + 1215x^4 y^2 - 540x^3 y^3 + 135x^2 y^4 - 18x y^5 + y^6$$ --- 1. Problem 17a: Find term in $x^5$ in expansion of $(x - 3)^9$. 2. General term: $$T_{k+1} = \binom{9}{k} x^{9-k} (-3)^k$$ 3. We want power of $x$ to be 5: $$9 - k = 5 \Rightarrow k = 4$$ 4. Term: $$T_5 = \binom{9}{4} x^5 (-3)^4 = 126 x^5 81 = 10206 x^5$$ --- 1. Problem 17b: Find term in $x^6$ in expansion of $-2x(x - 3)^9$. 2. Multiply $-2x$ by term in $(x - 3)^9$ with $x^5$ (from 17a) to get $x^6$ term. 3. Term: $$-2x \times 10206 x^5 = -20412 x^6$$ --- 1. Problem 22a: Arithmetic series with $u_{15} = 43$, $u_{31} = 183$. Find $a$ and $d$. 2. Use formula: $$u_n = a + (n-1)d$$ 3. Write equations: $$a + 14d = 43$$ $$a + 30d = 183$$ 4. Subtract: $$16d = 140 \Rightarrow d = 8.75$$ 5. Substitute back: $$a + 14(8.75) = 43 \Rightarrow a + 122.5 = 43 \Rightarrow a = -79.5$$ --- 1. Problem 22b: Find $u_{100}$: 2. Calculate: $$u_{100} = a + 99d = -79.5 + 99(8.75) = -79.5 + 866.25 = 786.75$$ --- 1. Problem 23a: Angelina deposits 3000, compound interest 1.5% per year, find interest after 10 years. 2. Formula: $$A = P(1 + r)^t$$ 3. Calculate: $$A = 3000 (1 + 0.015)^{10} = 3000 (1.015)^{10} \approx 3000 \times 1.1609 = 3482.7$$ 4. Interest earned: $$3482.7 - 3000 = 483$$ --- 1. Problem 23b: Additional $1200$ deposited annually from 2020 to 2029 (10 deposits), find total at start 2030. 2. Calculate future value of initial deposit: $$FV_1 = 3000 (1.015)^{11} \approx 3000 \times 1.178 = 3534$$ 3. Future value of annuity deposits: $$FV_2 = 1200 \times \frac{(1.015)^{10} - 1}{0.015} \approx 1200 \times 11.06 = 13272$$ 4. Total amount: $$3534 + 13272 = 16806$$ --- 1. Problem 24a: Brad deposits 5500, 2.75% compound interest per year, find value after 4 years. 2. Calculate: $$A = 5500 (1 + 0.0275)^4 = 5500 (1.0275)^4 \approx 5500 \times 1.1145 = 6130$$ --- 1. Problem 24b: Find time $t$ for investment to reach 12000. 2. Solve: $$12000 = 5500 (1.0275)^t \Rightarrow (1.0275)^t = \frac{12000}{5500} = 2.1818$$ 3. Take log: $$t \log 1.0275 = \log 2.1818 \Rightarrow t = \frac{0.339}{0.012} = 28.25$$ 4. Rounded to nearest year: 28 years. --- 1. Problem 29: Geometric series sum to infinity 120, common ratio 0.2, find 6th term. 2. Sum to infinity: $$S_\infty = \frac{a}{1-r} = 120 \Rightarrow a = 120 (1-0.2) = 120 \times 0.8 = 96$$ 3. 6th term: $$u_6 = a r^{5} = 96 \times 0.2^{5} = 96 \times 0.00032 = 0.03072$$ --- 1. Problem 30: Geometric series with $u_2=180$, $u_6=\frac{20}{9}$, find sum to infinity. 2. Use formula: $$u_n = a r^{n-1}$$ 3. Write equations: $$u_2 = a r = 180$$ $$u_6 = a r^{5} = \frac{20}{9}$$ 4. Divide: $$\frac{a r^{5}}{a r} = r^{4} = \frac{20/9}{180} = \frac{20}{9 \times 180} = \frac{20}{1620} = \frac{1}{81}$$ 5. Solve for $r$: $$r = \sqrt[4]{\frac{1}{81}} = \frac{1}{3}$$ 6. Find $a$: $$a \times \frac{1}{3} = 180 \Rightarrow a = 540$$ 7. Sum to infinity: $$S_\infty = \frac{a}{1-r} = \frac{540}{1 - \frac{1}{3}} = \frac{540}{\frac{2}{3}} = 540 \times \frac{3}{2} = 810$$