Sucesiones repaso Klasse 12 [86 marks]


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1 Sucesiones repaso Klasse [8 marks] a. Consider an infinite geometric sequence with u = 40 and r =. (i) Find. u 4 (ii) Find the sum of the infinite sequence. (i) correct approach () e.g. u 4 = (40) u 4 = 5 N, listing terms (ii) correct substitution into formula for infinite sum e.g. =, S = 80 N 40 S 0.5 (4 ) 40 S = 0.5 () Consider an arithmetic sequence with n terms, with first term ( ) and eighth term ( 8). b. (i) Find the common difference. (ii) Show that S n = n 8n. [5 marks] (i) attempt to set up expression for e.g. + (8 )d u 8 e.g. 8 = + (8 )d, d = 4 N (ii) correct substitution into formula for sum e.g. S n e.g. = (4n 7), n + S n [5 marks] n AG N0 8 ( ) 7 = (( ) + (n )4) n S n = n 8n n n () c. The sum of the infinite geometric sequence is equal to twice the sum of the arithmetic sequence. Find n. [5 marks]
2 multiplying (AP) by or dividing S (infinite GP) by S e.g. S, n, 40 evidence of substituting into e.g. n 8n = 40, 4n 7n 80 ( = 0) attempt to solve their quadratic (equation) e.g. intersection of graphs, formula n = 0 [5 marks] S n A N S n = S In an arithmetic sequence, u = and = 8. u a. Find d. [ marks] attempt to find d u u e.g., 8 = + d d = N [ marks] b. Find u 0. [ marks] correct substitution () e.g. u 0 = + (0 ), = 0 u 0 = 59 N [ marks] u 0 c. Find S 0. [ marks] correct substitution () e.g. = ( + 59), S 0 0 S 0 = 0 [ marks] N 0 S 0 = ( + 9 )
3 An arithmetic sequence is given by 5, 8,,. (a) a. Write down the value of d. (b) Find (i) ; u 00 (ii) S 00. (c) Given that = 50, find the value of n. u n [7 marks] (a) d = N [ mark] (b) (i) correct substitution into term formula () e.g. u 00 = 5 + (99), 5 + (00 ) u 00 = 0 N (ii) correct substitution into sum formula () = ((5) + 99()), S S 00 = 550 N 00 S 00 = (5 + 0) (c) correct substitution into term formula () 50 = 5 + (n ), 50 = n + n = 500 N [ marks] Total [7 marks] Write down the value of d. b. [ mark] d = N [ mark] c. Find (i) ; u 00 (ii). S 00
4 (i) correct substitution into term formula () e.g. u 00 = 5 + (99), 5 + (00 ) u 00 = 0 N (ii) correct substitution into sum formula () = ((5) + 99()), S S 00 = 550 N 00 S 00 = (5 + 0) d. Given that u n = 50, find the value of n. [ marks] correct substitution into term formula () 50 = 5 + (n ), 50 = n + n = 500 N [ marks] Total [7 marks] 4. The sum of the first three terms of a geometric sequence is.755, and the sum of the infinite sequence is 440. Find the common ratio. [ marks] correct substitution into sum of a geometric sequence.755 = ( ), u + u r + u r =.755 correct substitution into sum to infinity u r = 440 attempt to eliminate one variable substituting r u r u = 440( r) correct equation in one variable r r.755 = 440( r)( ), () 440( r)( + r + r ) =.755 evidence of attempting to solve the equation in a single variable sketch, setting equation equal to zero, r = 0.95 = 9 0 [ marks] N4.755 = 440( r ) In an arithmetic sequence, the third term is 0 and the fifth term is. Find the common difference. 5a. [ marks]
5 5a. attempt to find d 0, 0 d = 4d, d =, d = d = N [ marks] Find the first term. 5b. [ marks] correct approach u = 4 [ marks] N () 0 = u +, 0 Find the sum of the first 0 terms of the sequence. 5c. correct substitution into sum or term formula correct simplification S 0 0 ( ), = , 4 + = 50 N u 0 () () The sides of a square are cm in length. The midpoints of the sides of this square are joined to form a new square and four triangles (diagram ). The process is repeated twice, as shown in diagrams and. Let Let x n A n denote the length of one of the equal sides of each new triangle. denote the area of each new triangle. a. The following table gives the values of x n and A n, for n. Copy and complete the table. (Do not write on this page.) n x n A n 8 4
6 valid method for finding side length = c, side ratios, 8, =, + = s x x 8 for area () 4 4 n x n 8 4 A n NN 8 The process described above is repeated. Find A. b. METHOD recognize geometric progression for r = () () A u n = u r n ( ) ; 4,,,,, 5 = N 4 A n (R) METHOD attempt to find () x N 8 ( ),,,,, x = A 5 ) = () c. Consider an initial square of side length k cm. The process described above is repeated indefinitely. The total area of the shaded [7 marks] rions is k cm. Find the value of k.
7 METHOD recognize infinite geometric series area of first triangle in terms of k (R) () attempt to substitute into sum of infinite geometric series (must have k) correct equation () valid attempt to solve their quadratic METHOD N recognizing that there are four sets of infinitely shaded rions with equal area area of original square is k so total shaded area is 4 k correct equation = k 4 k = 4k () () () valid attempt to solve their quadratic S n ( ) [7 marks] a r =, r < k ( k ) ( k ) k k = 4 k = 4, = 4k k = k, k = N k 8 k(k 4), k = 4 or k = 0 k k(k 4), k = 4 or k = 0 R The sums of the terms of a sequence follow the pattern S = + k, S = 5 + k, S = + 7k, S 4 = + 5k,, where k Z. 7a. Given that u = + k, find u, u and u 4. valid method u = S S, + k + u = 5 + k u = 4 + k, u = 7 + 4k, u 4 = 0 + 8k N4 7b. Find a general expression for u n.
8 correct AP or GP () finding common difference is, common ratio is valid approach using arithmetic and geometric formulas + (n ) and r n k u n = n + n k N4 Note: Award for n, for n k. The first three terms of a infinite geometric sequence are m,, m + 4, where m Z. 8a. Write down an expression for the common ratio, r. [ marks] correct expression for r N r =, m [ marks] m+4 8b. Hence, show that m satisfies the equation m + m 40 = 0. [ marks] correct equation m m [ marks] m+4 m+4 =, = () (m + 4)(m ) = AG m m m + 4m 4 =, m + m 4 = + m 40 = 0 N0 8c. Find the two possible values of m. valid attempt to solve (m + 8)(m 5) = 0, m = m = 8, m = 5 N ± d. Find the possible values of r.
9 attempt to substitute any value of m to find r r = 8,, r = 5+4 N 8e. The sequence has a finite sum. State which value of r leads to this sum and justify your answer. r = (may be seen in justification) valid reason R N0 r <, < < Notes: Award R for r < only if awarded. [ marks] 8f. The sequence has a finite sum. Calculate the sum of the sequence. finding the first term of the sequence which has r < () u = 9 (may be seen in formula) () u correct substitution of u and their r into, as long as r < 8, S 9 =, ( ) = (= 5.4) S N r International Baccalaureate Organization 0 International Baccalaureate  Baccalauréat International  Bachillerato Internacional Printed for Colio Aleman de Barranquilla
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