The first three terms of an arithmetic progression are (2x+1), 4(2x+1) and 7(2x+1), where x ≠ -1/2. Show that the sum to n terms can be written in the form (n/2)(2x+1)(An+B), where A and B are integers to be found.

Mark scheme

Marking points from the official mark scheme (May/June 2023). The mark or mark type is shown in brackets.

1. [B1] for n/2 (2(2x+1)+3(n-1)(2x+1)) or n/2 (2(2x+1)+(6x+3)(n-1)) oe
2. [A1] 3n-1
3. [M1] For equating their answer to (a) to (54n+37)(2x+1) and attempt to solve a 3-term quadratic equation in n to obtain n=...
4. [A1] 37 only
5. [M1] For attempt to solve to obtain a value for x. n must be a positive integer
6. [A1] x = -4
7. [B1] Award when a correct statement is first seen (2y+1)(3(2y+1))^(n-1) = 4(2y+1)(3(2y+1))^(n+1) Or (3(2y+1))^(n-1) = 4(3(2y+1))^(n+1) Or (2y+1)r^(n-1) = 4(2y+1)r^(n+1) Or ar^(n-1) = 4ar^(n+1)
8. [M1] Either M1 for an equation of the form (2y+1)² = k or (6y+3)² = m where k and m are numerical and not zero (may be expanded) with no terms in n Or M1 for r² = p, where p is numerical and not zero
9. [A1] for two answers (2y+1)² = 1/36 or r² = 1/36 => 2y+1 = ±1/6 or r = ±1/6 => 6y+3 = ±1/6)
10. [A1] For both 5/12 and -7/12 and no others
11. [B1] Allow 2sin²θ < 1 or -1 < 2sin²θ < 1 soi
12. [B1] May be implied by θ < π/4
13. [B1] 0 < θ < π/4 Or 0 < θ < 0.785 or better