A-LevelMathematicsSeries9709/12 · May/June 2017 · Q78 Marks

The first two terms of an arithmetic progression are 16 and 24. Find the least number of terms of the progression which must be taken for their sum to exceed 20000.

(a)The first two terms of an arithmetic progression are 16 and 24. Find the least number of terms of the progression which must be taken for their sum to exceed 20000.[4]
(b)A geometric progression has a first term of 6 and a sum to infinity of 18. A new geometric progression is formed by squaring each of the terms of the original progression. Find the sum to infinity of the new progression.[4]

Mark scheme

Marking points from the official mark scheme (May/June 2017). The mark or mark type is shown in brackets.

(S_n =) n/2 [32+(n-1)8] and 20000
[M1 M1] correct formula used with d from 16+ d = 24
→ n² + 3n - 5000 (<,=,> 0)
[A1] A1 for correct expression linked to 20000.
→ (n = 69.2) → 70 terms needed.
[DM1] Simplification to a three term quadratic.
[A1] Condone use of 20001 throughout. Correct answer from trial and improvement gets 4/4.
a = 6, a/(1-r) = 18 → r = 2/3
[M1 A1] Correct S∞ formula used to find r.
New progression a = 36, r = 4/9 oe
[M1] Obtain new values for a and r by any valid method.
New S∞ = 36 / (1 - 4/9) → 64.8 or 324/5 oe
[A1] (Be aware that r = -2/3 leads to 64.8 but can only score M marks)

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