Question 6:
--- 6(a) ---
6(a) | u = a r n − 1  a r + a r 2 = 6 and a r 3 = 8
n | M1
a r + a3 r 2 6 3
 =  1 + r = r 2
a r 8 4 | M1
 3 r 2 − 4 r − 4 = 0 * | A1*
(3)
(a)
Way 2 | 8 8
u = 8  u = , u =
4 3 r 2 r 2 | M1
8 8
+ =6
r r2 | M1
 3 r 2 − 4 r − 4 = 0 * | A1*
(3)
(b) | 2
r =−
3 | B1
8
a r 3 = 8  a = = . . .
3
 2 
−
3 | M1
u =−27
1 | A1ft
(3)
(c) | − 2 7
S = = . . .
  2 
1 − −
3 | M1
81
= −
5 | A1
(2)
(8 marks)
Notes:
(a) Ignore labelling and mark (a), (b) and (c) together
M1: Uses the correct n-th term formula for a GP to set up two equations in a and r
May also be in terms of u and r or u and r or u and r e.g.
1 2 3
u
u r + u r 2 = 6 , u r 3 = 8 o r u + u r = 6 , u r 2 = 8 o r 3 + u = 6 , r u = 8
1 1 1 2 2 2 r 3 3
Must be using the correct term formula so e.g. a r 2 + a r 3 = 6 and ar4 =8 is M0
1−r3
Alternatively may use the correct sum formula for the first equation, a −a=6 oe.
1−r
M1: Attempts to solve their two equations to get an equation in r. Look for an attempt to divide the
two equations, or an attempt to find a in terms of r from one and substitute into the other.
Allow slips but the algebra should essentially be correct so do not allow use of e.g. ar3 = (ar)3
Alternatively, attempts to eliminate r from the equation, r =
3
2
a

2
3
a
a
+
4
3
a
a 2
= 6
2 1
2a3 +4a3 =6. Award when they reach a quadratic in a
13
in this case.
A1*: CSO Note that as we are marking (a), (b) and (c) together, allow the printed answer to appear
anywhere as long as it follows correct work but the “=0” must be seen not implied.
Way 2:
M1: Uses u = 8 to write u and u in terms of r
4 2 3
M1: Uses u + u = 6 to get an equation in r
2 3
A1*: CSO
(b)
B1: Correct value of r seen or used in their working even if subsequently rejected and ignore any
other value offered e.g. r = 2
M1: Uses a value for r from solving the equation given in (a) where | r |  1 in one of their equations
from (a) to find a value for u or a. Allow slips but the algebra should essentially be correct
1
A1ft: Correct value and no other values. Follow through on
(
t h e
8
i r r
) 3
for a value of r with | r |  1
(c)
M1: Uses the correct sum formula with their a (u ) obtained from |r|1 and r where
1
| r |  1
PMT
, to find the sum
to infinity.
A1: Correct answer and no other values. Allow equivalents e.g. −16.2
Question | Scheme | Marks