Mathematics · May / June 2025 · Paper 21 · Question 7

YMA01

As printed in the paper

The question as text

7. In this question you must show all stages of your working. Solutions relying entirely on calculator technology are not acceptable.

Read automatically from the paper; the printed page above is exact.

Show the answer and mark scheme

7(i)(a)[2]
Attempts
S
∞=
with
a
=
10,
r
=
1
−
r
10
S∞=
50
(b) Sets
k
−
1
k
−
1
10 0.8
×
<
0.0005
0.8
<
0.00005
e.g.
log(0.00005)
k
−>
1
or e.g.
k
−>
1
log
0.00005
(see notes)
log(0.8)
0.8
(
)
k
−>
1
44.38
k
=
46
7(i)(b)[3]
10 8 6.4 ...
+ +
+
8
a
Attempts
S
∞=
with
a
=
10,
r
=
1
−
r
10
S∞=
50
(b) Sets
k
−
1
k
−
1
10 0.8
×
<
0.0005
0.8
<
0.00005
e.g.
log(0.00005)
k
−>
1
or e.g.
k
−>
1
log
0.00005
(see notes)
log(0.8)
0.8
(
)
k
−>
1
44.38
k
=
46
Sets 850
+
(
−
1)
×−=
7
0
=
122.4
n
n
n =
122
States or implies that
122 2 850 121
S =
×
+
×−
7
=
52033
2
(i) (a)
a
8
M1: Attempts
∞=
with
=
10,
=
but condone slips on an incorrectly evaluated ''r'' provided the
S
a
r
1
−
10
r
method to finding it is correct and 1
−<
<. The expression with the values substituted in is sufficient.
1
r
A1: (
) 50
S∞=
(b)
Note that candidates may misread the number of 0s in 0.0005 but can still score M1M1A0
M1: Sets
k−
1
and proceeds to
k−<
1
(or they may manipulate correctly to
10 "0.8"
×
<
0.0005
"0.8"
0.00005
k <
"0.8"
0.00005 0.8
×
). If they take logs before this stage then they will need to proceed to e.g.
k−<
1
log"0.8"
log0.0005
−
log10
Condone other signs being used e.g. = or >
k
k
−
1...
M1: Uses a correct method to solve an equation or inequality of the type
...
q or
p
p
q
It cannot be for just stating the value for k. They only need to proceed as far as
k − on one side of
1
an equation or inequality. Evidence of using logs is required which cannot be e.g. 44.38… otherwise
M0A0
Do not be concerned with the direction of inequalities, if used, in their solution for this mark.
A1: (
) 46
k =
from correct working.
If inequalities are used in their method then the direction must be correct.
When using either formula condone missing brackets around (
)
(ii)
−
7
122 2 850
(
)
e.g.
850
+
(
n
−
1)
−
7
or e.g.
S
=
×
+
122 1
−
−
7
2
n
M1: Attempts to find the value of n that maximises
S
n
(
)
May set
850
+
(
−
1)
×−=
7
0
−
7
+
857
=
0
=
...
or show trial calculations of
n
n
n
= −. May be implied by 122.
4
u
123
n
Alternatively sets
2 850
(
1
)
7
and either
S
=
×
+
n
−
×−
2
n
1707
7
1707
differentiates and sets = 0 :
=
−
2
−
7
=
0
=
...
S
n
n
n
n
n
2
2
2
1707
7
1707
7
uses roots:
S
=
n
−
n
2
=
0
n
−
n
=
0
n
=
0,
n
=
...
and then finds the midpoint
n
2
2
2
2
of the two roots
1707
7
7
1707
7
1707
completes the square:
S
=
n
−
n
2
=
0
−
n
2
−
n
−
n
−
n
2
2
2
7
2
14
to find the position of the turning point (approximately 121.9)
In all cases condone the use of n instead of
n−
1
n =
122
A1: States or implies that
(if there are multiple attempts using different values of n then the others
must be rejected for this mark to be scored)
dM1: Attempts
S using a correct formula (they may attempt to find the last term and use
S
n
It is dependent upon the previous M.
Condone the incorrect value of n being selected but it must be a positive integer.
May see multiple attempts using different values of n which is acceptable.
A1: 52033only (i.e. if they have multiple attempts using different values then this answer must be clearly
identified or the others all rejected)
π
π
3
3tan 2
3
2
arctan
7(ii)[4]
Sets 850
+
(
−
1)
×−=
7
0
=
122.4
n
n
n =
122
States or implies that
122 2 850 121
S =
×
+
×−
7
=
52033
2
(i) (a)
a
8
M1: Attempts
∞=
with
=
10,
=
but condone slips on an incorrectly evaluated ''r'' provided the
S
a
r
1
−
10
r
method to finding it is correct and 1
−<
<. The expression with the values substituted in is sufficient.
1
r
A1: (
) 50
S∞=
(b)
Note that candidates may misread the number of 0s in 0.0005 but can still score M1M1A0
M1: Sets
k−
1
and proceeds to
k−<
1
(or they may manipulate correctly to
10 "0.8"
×
<
0.0005
"0.8"
0.00005
k <
"0.8"
0.00005 0.8
×
). If they take logs before this stage then they will need to proceed to e.g.
k−<
1
log"0.8"
log0.0005
−
log10
Condone other signs being used e.g. = or >
k
k
−
1...
M1: Uses a correct method to solve an equation or inequality of the type
...
q or
p
p
q
It cannot be for just stating the value for k. They only need to proceed as far as
k − on one side of
1
an equation or inequality. Evidence of using logs is required which cannot be e.g. 44.38… otherwise
M0A0
Do not be concerned with the direction of inequalities, if used, in their solution for this mark.
A1: (
) 46
k =
from correct working.
If inequalities are used in their method then the direction must be correct.
When using either formula condone missing brackets around (
)
(ii)
−
7
122 2 850
(
)
e.g.
850
+
(
n
−
1)
−
7
or e.g.
S
=
×
+
122 1
−
−
7
2
n
M1: Attempts to find the value of n that maximises
S
n
(
)
May set
850
+
(
−
1)
×−=
7
0
−
7
+
857
=
0
=
...
or show trial calculations of
n
n
n
= −. May be implied by 122.
4
u
123
n
Alternatively sets
2 850
(
1
)
7
and either
S
=
×
+
n
−
×−
2
n
1707
7
1707
differentiates and sets = 0 :
=
−
2
−
7
=
0
=
...
S
n
n
n
n
n
2
2
2
1707
7
1707
7
uses roots:
S
=
n
−
n
2
=
0
n
−
n
=
0
n
=
0,
n
=
...
and then finds the midpoint
n
2
2
2
2
of the two roots
1707
7
7
1707
7
1707
completes the square:
S
=
n
−
n
2
=
0
−
n
2
−
n
−
n
−
n
2
2
2
7
2
14
to find the position of the turning point (approximately 121.9)
In all cases condone the use of n instead of
n−
1
n =
122
A1: States or implies that
(if there are multiple attempts using different values of n then the others
must be rejected for this mark to be scored)
dM1: Attempts
S using a correct formula (they may attempt to find the last term and use
S
n
It is dependent upon the previous M.
Condone the incorrect value of n being selected but it must be a positive integer.
May see multiple attempts using different values of n which is acceptable.
A1: 52033only (i.e. if they have multiple attempts using different values then this answer must be clearly
identified or the others all rejected)
π
π
3
3tan 2
3
2
arctan

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