2024 Y6 Timed Prac Rev Paper 3 (Soln)

RAFFLES INSTITUTION H2 Mathematics (9758) 2024 Year 6     2024 Year 6 Timed Practice Revision Practice Paper 3 Solution Source: 2021 Year 6 Term 3 Common Test Section A: PURE MATHS (52 Marks) 1 An arithmetic seri es has first term a and common difference d, where a and d are non-zero. If the first, fifth and eleventh terms of the arithmetic seri es are equal to the first, second and third terms of a geometric series respectively, find the exact sum of the first ten terms of the geometric series, in the form ka , where k is a simplified rational number. [4] Solution Comments [4] The first 3 terms of the GP are ,4 , 1 0aa da d    2 22 2 2 41 0 4 41 0 81 6 1 0 16 2 0 (Since 0)8 ad a d aa d ad a a d aa dd a a d da d add           Let the common ratio of the GP be r. 443 8 2 aaadr aa     . Sum of the first 10 terms = 10(1 ) 1 ar r   10 10 3 12 3 12 321 2 58025 512 a a a         The question asked for a simplified rational number, which is a number of the form p q , where p and q are integers, ie, express k as a fraction in lowest terms.

Alternatively, to find the required sum of the geometric series, we just need to find r. We can do so by eliminating d from the equations: 2 4 ( 1 ) 10 (2) ad a r ad a r     (1) (1 ) 4 ard  and thus substituting into (2): we have 2 2 2 (1 )10 4 4 10( 1) 4 since 0 25 3 0 31 or 2 araa r rra rr rr           We reject r = 1, because if r = 1 then (1) (or (2)  d = 0 The required sum of the first 10 terms then follows the same working as in the first solution. Note that the justification that r cannot be 1 has to be clearly stated as it is not given in the question.

2 (i) By writing   2 21 23rr in partial fractions, find an expression for  2 2 21 23 n r rr  in terms of n. [3] (ii) Hence find the exact value of 21 21 21 19 21 21 23 23 25  . [2] Solution Comments (i) [3] Let   2 21 23 21 23 AB rr r r      22 3 2 1Ar Br  When 12,121 3rA    When 32,123 1rB      21 1 21 23 21 23rr r r    22 21 1 21 23 21 23 11 57 11 + 79 11 + 91 1 11 + 21 21 11 + 21 23 11 52 3 nn rr rr r r nn nn n                         Check your partial fractions by either 1) re-combining the 2 fractions into a single one OR 2) using GC to check both sides of the expression give the same graph (both graphs should overlap completely when sketched) Do remember to write down the first 2 cancellations as well as the last one.

(ii) [2]  9 21 21 21 19 21 21 23 23 25 2 21 23r rr      