Question 9:
--- 9(a) ---
9(a) |  a  16
ar 16 , 100 leading to a and a1001r
 
 1r  r | B1 | Rearranging two algebraic statements to give a .
These can be implied by a correct equation in one variable.
1001rr 16 leading to 100r2 100r16 0 | *M1 | Using their two expressions and rearranging to get a
3-term quadratic expression with all of the terms on one
side. Condone sign errors only.
5r45r10
OR 4 1
leading to r or
 
25 252 4.25.4 5 5
2.25 | DM1 | Condone 5r45r1 following 100r2 100r16.
a = 20, a = 80 | A1 | SC: if DM0 scored SCB1 is available for sight of 20 and
80.
Alternative Method for Question 9(a)
 a  16 100a
ar 16, 100 leading to r and r 
 
 1r  a 100 | B1 | Rearranging two algebraic statements to give r.
These can be implied by a correct equation in one variable.
1600100a a2 leading to a2 100a16000 | *M1 | Using their two expressions and rearranging to get a
3-term quadratic expression with all of the terms on one
side. Condone sign errors and 160 instead of 1600 only.
100 1002 4.1600
a20a800 OR
2 | DM1
a = 20, a = 80 | A1 | SC: if DM0 scored SCB1 is available for sight of 20 and
80.
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Question | Answer | Marks | Guidance
--- 9(b) ---
9(b) | 4 1
r ,
5 5 | B1 | OE SOI
n1 n1
4 1
[u ] their 20their v  their80their
n  5   n  5   | B1FT | 2 expressions for the nth term FT their values from part (a)
if r less than 1.
Method 1 for final 2 marks
n1
1
20 4n1

5 | M1 | Correctly separating the numerator and denominator of
n1
4
their or one correct step towards the solution eg
 
5
4n2
u 80 .
n 5n1
n1 n1
1 1 1
u n  4 80   5  4n1 4n280 5   4n2v n | A1 | AG Given result clearly shown
Method 2 for final 2 marks
200.8n1 1
 4n1
800.2n1 4 | M1 | Dividing two nth terms of the correct format and
simplifying their terms in r.
=414n1 4n2 | A1 | AG
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Question | Answer | Marks | Guidance