O-LevelAdditional MathematicsSeries4037/12 · May/June 2021 · Q910 Marks

The first three terms of an arithmetic progression are –4, 8, 20. Find the smallest number of terms for which the sum of this arithmetic progression is greater than 2000.

(a)The first three terms of an arithmetic progression are –4, 8, 20. Find the smallest number of terms for which the sum of this arithmetic progression is greater than 2000.[4]
(b)(i)The 7th and 9th terms of a geometric progression are 27 and 243 respectively. Given that the geometric progression has a positive common ratio, find this common ratio,[2]
(b)(ii)the 30th term, giving your answer as a power of 3.[2]
(c)Explain why the geometric progression 1, sinθ, sin²θ, ... for -π/2 < θ < π/2, where θ is in radians, has a sum to infinity.[2]

Mark scheme

Marking points from the official mark scheme (May/June 2021). The mark or mark type is shown in brackets.

[1] B1 d=12
[1] M1 For use of sum formula to obtain a three term quadratic inequality or equation
[1] M1 Dep for attempt at critical value(s) using their quadratic, may be using a calculator, so may be implied by a correct answer of 20.
[1] A1 n = 20
[1] M1 For ar⁶ = 27 and ar⁸ = 243 with an attempt to eliminate a to obtain r². Allow other valid methods.
[1] A1 r = 3
[1] B1 for a = 1/27 or 3⁻³ nfww
[1] B1 3²⁶
[1] B1 Common ratio or r = sinθ
[1] B1 Dep on previous B1 -1<sinθ<1 or |sinθ|<1 or -1<r<1 or |r|<1 with no incorrect statements seen.

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