Cambridge A Level Mathematics 9709 — 2021 Oct/Nov Paper 1 · Variant 3

4 The first term of an arithmetic progression is 84 and the common difference is −3. (a) Find the smallest value of n for which the nth term is negative. [2] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ It is given that the sum of the first 2k terms of this progression is equal to the sum of the first k terms. (b) Find the value of k. [3] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................

Mark scheme: 4(a) ( ) 84 3 1 0 n − − = M1 OE, SOI. Allow either = 0 or < 0 (to -3). Smallest n is 30 A1 SC B2 for answer only n = 30 WWW. 2 4(b) ( )( ) ( )( ) 2 168 2 1 3 168 1 3 2 2 k k k k      + − − =  + − −              M1 A1 M1 for forming an equation using correct formula. A1 for at least one side correct. 19 k = A1 3

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5 B Y 9 cm 11 cm X C A In the diagram, X and Y are points on the line AB such that BX = 9 cm and AY = 11 cm. Arc BC is part of a circle with centre X and radius 9 cm, where CX is perpendicular to AB. Arc AC is part of a circle with centre Y and radius 11 cm. (a) Show that angle XYC = 0.9582 radians, correct to 4 significant figures. 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(b) Find the perimeter of ABC. 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Mark scheme: 5(a) Angle XYC = 1 9 sin 11 −      = 0.9582 or 9 sin leading to 0.9582 11 XYC XYC = = B1 AG. OE using cosine rule. 1 5(b) 2 2 11 9 40 XY = − = or using 0.9582 and trigonometry *M1 A1 9 11 AB theirXY = + − B1 FT OE e.g. 20 2 10 −√ , 2 9 2 10 11 2 10 + − + −√ Arc AC = 11 × 0.9582 M1 Arc BC = 9× π 2 M1 Perimeter = [13.6(8) + 10.5(4) +14.1(4) =] 38.4 A1 AWRT. Answer must be evaluated as a single decimal. 6

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