Mathematics · Oct / Nov 2025 · Paper 21 · Question 8
YMA01
As printed in the paper
Show the answer and mark scheme
8(i)[3]
nS
=
a
+
ar
+
ar
2
+
.......
ar
n
−
1
=
+
+
+
−
and
=
+
+
+
nS
a
ar
ar
2
.......
ar
n
1
rS
ar
ar
2
ar
3
.......
ar
n
Finds
S and
rS and subtracts. E.g.
−
=
S
rS
....
n
n
−
n
a
(1
r
)
Completes proof
*
S
(1
−
r
)
=
a
(1
−
r
n
)
S
=
−
n
n
(1
r
)
8(ii)(a)[2]
Attempts
U =
×
with
n =
150 0.92
n
5or 6
(
)
(
)
6
litres) *
U
=
150
×
0.92
=
90.95 ( 91
≈
6
−
a
(1
r
n
)
8(ii)(b)[3]
Attempts
S
=
with
n =
39 / 40/ 41
a =
150/138
and
r =
0.92
(1
−
r
)
n
−
−
−
138(1 0.92 )
40
0.92(1 0.92 )
40
150(1
0.92 )
41
S
=
,
S
=
150
×
OR
S
=
−
150
(1 0.92)
−
(1 0.92)
−
(1
−
0.92)
1664 litres
Do not be too concerned with labelling here. It should be obvious which part of the question
candidates are attempting to answer
(i) Condone
S
S
throughout.
n
A proof via the infinite sum formula gains no credit unless you see a correct expression for
nS .
B1: Scored for a correct expression for
nS . For example,
=
+
+
+
−
nS
a
ar
ar
2
.......
ar
n
1
including the ''
with a minimum of 3 terms and must include the first and the nth term.
nS
''
would be acceptable but
is not.
nS
=
a
+
ar
+
.......
ar
n
−
1
nS
=
a
+
.......
ar
n
−
1
It cannot be scored if some terms are incorrect
nS
=
a
+
.......
+
ar
n
+
ar
n
−
1
Or they are not in the correct order
=
+
−
+
−
nS
a
.......
ar
n
1
ar
n
2
M1: States
=
+
+
+
(minimum of 3 correct terms including first and last
nS
a
ar
ar
2
.......
ar
n
−
1
terms)
and
=
+
+
(minimum of 3 correct terms including first and last terms )
rS
ar
ar
2
.......
ar
n
n
followed by an attempt to subtract either way around
A1*: Fully correct proof. Look for the following 3 lines (or their equivalents) following B1, M1
S
−
rS
=
a
−
ar
n
n
n
(
)
(
)
−
=
−
nS
1
r
a
1
r
n
−
a
(1
r
n
)
S
=
(Note that this is a given answer)
(1
−
r
)
n
Example
nS
:
a
ar
ar
2
.......
ar
n
−
1
rS
:
ar
ar
2
ar
3
.......
ar
n
n
rS
−
S
=
ar
n
−
a
n
n
(
)
(
)
−
−
a r
n
1
a
1
r
n
(
)
(
)
r
−
1
S
=
a r
n
−
1
S
=
S
=
(
)
(
)
n
r
−
1
n
1
−
r
n
Scores SC B0, M1, A1
(Note that this candidate subtracts the other way around, but the given answer must be
produced and follow correct working)
(ii) (a)
M1: Attempts
with
.
U =
150
×
0.92
n
n =
5or 6
n
You may see variations on this such as
(
)
5
150
×
92%
Allow step by step solutions such as
×
0.92
×
0.92
×
0.92
×
0.92
×
0.92
×
0.92
150
→
138
→
126.96
→
116.8032
→
107.458944
→
98.86222848
→
90.9532502
(with accuracy at least 1dp for the M1)
A1*: Awrt 90.95 but accept the truncated value 90.9. You should expect to see an answer
that isn’t exactly 91
ISW after sight of a correct answer
Note that candidates who just write
then award M1, A0 as this is
150 0.92
×
6
=
91
incorrect
If attempted via step by step solutions (listing) then intermediate answers must be
correct to 2dp rounded or truncated.
(b)
−
a
(1
r
n
)
M1: Attempts
S
=
with
n =
,
a =
150/138
and
r =
0.92
o.e. such as 92%
39 / 40/ 41
(1
−
r
)
n
A1: A correct expression. You may see variations of the following
138(1 0.92 )
−
40
92%(1
−
92% )
40
150(1
−
0.92 )
41
=
,
=
×
OR
=
−
S
S
150
S
150
(1 0.92)
−
(1
−
92%)
(1
−
0.92)
A1: cao 1664 litres
(
)
2
4
x
+
5
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