1. Find the first four terms of the sequence. a) Since the sequence is unclear due to symbols, please clarify the exact form. b) Similarly, please provide a clearer form for sequence b. c) Same for sequence c. 2. Find the formula for the nth term of the sequence. a) Given sequence 1, -1, 1, -1, 1, ... This alternates sign, so the nth term is $$a_n = (-1)^{n+1}$$. b) Given sequence 0, 3, 8, 15, 24, ... Observe differences: 3, 5, 7, 9 increasing by 2, so quadratic sequence. Assume formula $$a_n = n^2 - 1$$. Verify: For n=1, $$1^2 - 1 = 0$$. For n=2, $$4 - 1 = 3$$. Matches sequence. c) Given sequence 1, 5, 9, 13, 17, ... It's arithmetic with common difference 4. Formula: $$a_n = 1 + (n-1)*4 = 4n - 3$$. 3. Which sequences converge or diverge? a) Sequence depends on clarity. b) Sequence $$rac{n}{n} = 1$$ always converges to 1. c) Sequence unclear, need exact formula. 4. Find 20th term and sum of first 20 terms of arithmetic progressions. a) Sequence 2, 6, 10, 14, ... with common difference $$d=4$$ and first term $$a_1=2$$. 20th term: $$a_{20} = a_1 + (20-1)*d = 2 + 19*4 = 2 + 76 = 78$$. Sum of first 20 terms: $$S_{20} = \frac{20}{2} (a_1 + a_{20}) = 10(2 +78) = 10*80 = 800$$. b) Sequence -5, -3.5, -2, -0.5, ... Common difference is $$d=1.5$$. 20th term: $$a_{20} = -5 + 19*1.5 = -5 + 28.5 = 23.5$$. Sum of first 20 terms: $$S_{20} = 10(-5 + 23.5) = 10*18.5 = 185$$. 5. nth term and sum of arithmetic progressions. a) Sequence 4, 6, 8, 10, ... with $$a_1=4$$ and $$d=2$$. $$a_n = 4 + (n-1)*2 = 2n + 2$$. Sum: $$S_n = \frac{n}{2}(2a_1 + (n-1)d) = \frac{n}{2}(8 + 2(n-1)) = \frac{n}{2}(2n + 6) = n(n + 3)$$. b) Sequence 17, 14, 11, 8, ... with $$a_1=17$$ and $$d=-3$$. $$a_n = 17 - 3(n-1) = 20 - 3n$$. Sum: $$S_n = \frac{n}{2}(2*17 + (n-1)(-3)) = \frac{n}{2}(34 - 3n + 3) = \frac{n}{2}(37 - 3n)$$. c) Incomplete data, please clarify. 6. General formulas for arithmetic progression: - nth term: $$u_n = u_1 + (n-1)d$$. - Sum of n terms: $$S_n = \frac{n}{2} (2u_1 + (n-1)d)$$ or $$S_n = \frac{n}{2} (u_1 + u_n)$$. 7. Given $$u_4 = 40$$ and $$u_{12} = 42$$. Use formula: $$u_n = a + (n-1)d$$. Then: $$a + 3d = 40$$ $$a + 11d = 42$$. Subtract equations: $$8d = 2 ightarrow d = \frac{1}{4}$$. Then: $$a = 40 - 3(\frac{1}{4}) = 40 - 0.75 = 39.25$$. The AP is $$u_n = 39.25 + (n-1)*0.25$$. 8. The sum of 8th and 14th term is 50 and 5th term is 13. From $$u_n = a + (n-1)d$$, $$u_8 + u_{14} = 2a + (8-1)d + (14-1)d = 2a + 20d = 50$$ $$u_5 = a + 4d = 13$$. Solve system: From $$u_5$$: $$a = 13 - 4d$$. Substitute into first: $$2(13 - 4d) + 20d = 50$$ $$26 - 8d + 20d = 50$$ $$26 + 12d = 50 o 12d=24 o d=2$$. Then: $$a = 13 - 4*2 = 13 - 8 = 5$$. So AP is $$u_n = 5 + (n-1)*2$$. 9. Insert 8 arithmetic means between -2 and 14. Total terms: 10. Common difference: $$d = \frac{14 - (-2)}{9} = \frac{16}{9}$$. The sequence is: $$-2, -2 + d, -2 + 2d, ..., 14$$. 10. Find x consecutive integers starting at 8 with sum $$kx$$. Sum of first x integers starting at 8 is: $$S = \frac{x}{2} [2*8 + (x-1)*1] = \frac{x}{2} (16 + x - 1) = \frac{x}{2} (x + 15)$$. Set equal to given sum $$kx$$, solve for x if $$k$$ known. 11. Sum of 3 consecutive terms in AP is 33 and product is 1287. Let terms be: $$a - d, a, a + d$$. Sum: $$3a = 33 o a = 11$$. Product: $$(a - d) a (a + d ) = 1287$$ Substitute $$a=11$$: $$(11 - d) * 11 * (11 + d) = 1287$$ $$11 (121 - d^2) = 1287$$ $$121 - d^2 = \frac{1287}{11} = 117$$ $$d^2 = 121 -117 = 4 o d=\pm 2$$. Terms: 9, 11, 13 or 13, 11, 9. 12. Find nth term of harmonic sequence with first two terms 6, 3. Harmonic sequence terms are reciprocals of arithmetic sequence. Let $$a_1 = \frac{1}{6}, a_2 = \frac{1}{3}$$. Common difference: $$d = a_2 - a_1 = \frac{1}{3} - \frac{1}{6} = \frac{1}{6}$$. Arithmetic sequence term: $$a_n = \frac{1}{6} + (n-1) \frac{1}{6} = \frac{n}{6}$$. Harmonic sequence term: $$h_n = \frac{1}{a_n} = \frac{6}{n}$$. 13 to 36: Please clarify if you want further solutions for these problems.