The first three terms of an arithmetic progression are lg 3, 3 lg 3, 5 lg 3. Given that the sum to n terms of this progression can be written as 256 lg 81, find the value of n.
- (a)The first three terms of an arithmetic progression are lg 3, 3 lg 3, 5 lg 3. Given that the sum to n terms of this progression can be written as 256 lg 81, find the value of n.[5]
- (b)DO NOT USE A CALCULATOR IN THIS PART OF THE QUESTION. The first three terms of a geometric progression are ln 256, ln 16, ln 4. Find the sum to infinity of this progression, giving your answer in the form p ln 2.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.
- [5] B1 Must be exact (for Common difference = 2 lg 3)
- M1 For use of the sum formula (for n/2 (2 lg 3 + (n - 1)2 lg 3) = 256 lg 81 or n/2 (lg 9 + (n - 1)lg 9) = 512 lg 9)
- B1 Allow when working with decimal (for lg 81 = 4 lg 3 soi or lg 81 = 2 lg 9 soi)
- M1 Dep on first M mark, for attempt to simplify the sum equation by dividing through by lg 3 oe to obtain an equation in n only (for n² = 1024 oe)
- A1 Must have exact working through out (for n = 32 cao)
- [4] M1 For use of power rule to obtain the common ratio (for ln 256 = 4 ln 4, ln 16 = 2 ln 4 oe)
- A1 (for Common ratio = 0.5)
- M1 Allow ln 256 for first term and their r provided it is positive and < 1 (for S_∞ = (4 ln 4) / (1-their r) oe)
- A1 (for 16 ln 2)
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