A-LevelMathematicsSeries9709/12 · Feb/Mar 2021 · Q99 Marks
The first term of a progression is cos θ, where 0 < θ < ½π.
- (a)(i)For the case where the progression is geometric, the sum to infinity is 1 . cos θ Show that the second term is cos θ sin² θ.[3]
- (a)(ii)Find the sum of the first 12 terms when θ = ⅓π, giving your answer correct to 4 significant figures.[2]
- (b)For the case where the progression is arithmetic, the first two terms are again cos θ and cos θ sin² θ respectively. Find the 85th term when θ = ⅓π.[4]
Mark scheme
Marking points from the official mark scheme (Feb/Mar 2021). The mark or mark type is shown in brackets.
- [1] 1/cosθ
- [1] 1-r = cos²θ leading to r=1-cos²θ (Eliminate fractions)
- [1] r = sin²θ leading to 2nd term = cos θ sin²θ (AG)
- [1] S12 = cos(π/3) [1-(sin²(π/3))¹²] / [1-sin²(π/3)] = 0.5[1-(0.75)¹²] / 1-0.75 (Evidence of correct substitution, use of S_n formula and attempt to evaluate)
- [1] 1.937
- [1] [d=] cosθsin²θ-cosθ (Use of d=u_2-u_1)
- [1] -1/8
- [1] [85th term =] 1/2 + 84 x (-1/8) (Use of a + 84d with a calculated value of d)
- [1] -10
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