A-LevelMathematicsSeries9709/12 · May/June 2019 · Q1010 Marks
(a) In an arithmetic progression, the sum of the first ten terms is equal to the sum of the next five terms. The first term is a. (b) The sum to infinity of a geometric progression is 9 times the sum of the first four terms. Given that the first term is 12, find the value of the fifth term.
- (a)(i)Show that the common difference of the progression is (1/3)a.[4]
- (a)(ii)Given that the tenth term is 36 more than the fourth term, find the value of a.[2]
- (b)The sum to infinity of a geometric progression is 9 times the sum of the first four terms. Given that the first term is 12, find the value of the fifth term.[4]
Mark scheme
Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.
- [M1] S10 = S15-S10 or S10 = S(11 to 15)
- [B1] 5(2a + 9d) oe
- [A1] 7.5(2a + 14d) – 5(2a + 9d) or 5/2 [(a + 10d) + (a+14d)] oe
- [A1] d = a/3 AG
- [M1] (a+9d)=36+(a+3d)
- [A1] a = 18
- [B1] S_inf = 9 × S_4; a/(1-r) = 9 a(1-r^4)/(1-r) or 9(a + ar+ ar²+ar³)
- [M1] 9(1 - r^4) = 1 where n = 3,4 or 5
- [A1] r^4 = 8/9 oe
- [A1] (5th term =) 10 2/3 or 10.7
Write your answer and get it marked
On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.
Sign up free →