Mathematics · Oct / Nov 2021 · Paper 21 · Question 7
YMA01
As printed in the paper
Show the answer and mark scheme
7(i)[3]
n−
1
100
States or implies that
4 6
×
>
10
Takes logs correctly to produce an equation without powers
n
−
1
100
(
)
4 6
×
>
10
⇒
log 4
+
n
−
1 log6
>
100log10
n =
129
a
(ii) (a) States
ar = − and
6
r =
25
1
−
−
6
Combines to form an equation in r
⇒
=
25
(
)
r
1
−
r
(
)
2
⇒−=
6
25
r
1
−
r
⇒
25
r
−
25
r
−
6
=
0
*
6
1
7(ii)(a)[3]
n−
1
100
States or implies that
4 6
×
>
10
Takes logs correctly to produce an equation without powers
n
−
1
100
(
)
4 6
×
>
10
⇒
log 4
+
n
−
1 log6
>
100log10
n =
129
a
(ii) (a) States
ar = − and
6
r =
25
1
−
−
6
Combines to form an equation in r
⇒
=
25
(
)
r
1
−
r
(
)
2
⇒−=
6
25
r
1
−
r
⇒
25
r
−
25
r
−
6
=
0
*
6
1
r =
,
−
5
5
1
r = −
as
r <
1
(for S∞ to exist)
5
(
)
n
a
1
−
r
6
(d) Attempts
−
S
=
with
n
=
4,
r
=
their( )
c
and
a
=
1
−
r
their( )
c
4
4
1
30 1
−−
5
S
=
= 24.96 o.e.
4
1
1
−−
5
(i)
n−
1
100
B1:
States or implies the solution can be found by solving
4 6
×
...10
where ... can be = or
any inequality
N
100
M1: Shows a correct method of solving an equation of the form
4 6 ...10
×
by correctly
taking logs to produce an equation without powers. The log and index work must be
correct, but allow slips in rearranging terms.
A1:
n =
129
(ii)(a)
a
B1:
States or implies the correct two equations in a and r.
ar = − and
6
r =
25
1
−
a
M1: Combines
ar = − and
6
r =
25
to form a single equation in r
1
−
A1*: Proceeds to
25
r
2
−
25
r
−
6
=
0
showing at least one correct simplified intermediate line
and no errors
(ii)(b)
6
1
B1:
r =
,
−
or exact equivalent – award when seen even if not in part (b).
5
5
(ii)(c)
1
r = −
r <
1
(for S∞ to exist). Requires a minimal reason, but accept reasons that
B1:
as
5
reject 6
since it would mean all the terms are negative so cannot give a positive sum. Do
5
not accept just “as the GS is convergent”.
(
)
n
a
1
−
r
−
6
4,
their( )
M1: Attempts
S
=
with
n
=
r
=
c
and
a
=
. If there was no
1
−
r
their( )
c
4
attempt to answer (c) accept with either value from (b) for r.
If the correct formula is quoted then you can allow slips in substitution, but if the correct
formula is not quoted then the equation should be correct for their r and a.
Alternatively, they may find and add the first 4 terms.
24.96 or equivalent such as 624
A1:
25
4
d
y
7(ii)(b)[1]
r =
,
−
5
5
1
7(ii)(c)[1]
r = −
as
r <
1
(for S∞ to exist)
5
(
)
n
a
1
−
r
6
(d) Attempts
−
S
=
with
n
=
4,
r
=
their( )
c
and
a
=
1
−
r
their( )
c
4
4
1
30 1
−−
5
S
=
= 24.96 o.e.
4
1
1
−−
5
(i)
n−
1
100
B1:
States or implies the solution can be found by solving
4 6
×
...10
where ... can be = or
any inequality
N
100
M1: Shows a correct method of solving an equation of the form
4 6 ...10
×
by correctly
taking logs to produce an equation without powers. The log and index work must be
correct, but allow slips in rearranging terms.
A1:
n =
129
(ii)(a)
a
B1:
States or implies the correct two equations in a and r.
ar = − and
6
r =
25
1
−
a
M1: Combines
ar = − and
6
r =
25
to form a single equation in r
1
−
A1*: Proceeds to
25
r
2
−
25
r
−
6
=
0
showing at least one correct simplified intermediate line
and no errors
(ii)(b)
6
1
B1:
r =
,
−
or exact equivalent – award when seen even if not in part (b).
5
5
(ii)(c)
1
r = −
r <
1
(for S∞ to exist). Requires a minimal reason, but accept reasons that
B1:
as
5
reject 6
since it would mean all the terms are negative so cannot give a positive sum. Do
5
not accept just “as the GS is convergent”.
7(ii)(d)[2]
(
)
n
a
1
−
r
−
6
4,
their( )
M1: Attempts
S
=
with
n
=
r
=
c
and
a
=
. If there was no
1
−
r
their( )
c
4
attempt to answer (c) accept with either value from (b) for r.
If the correct formula is quoted then you can allow slips in substitution, but if the correct
formula is not quoted then the equation should be correct for their r and a.
Alternatively, they may find and add the first 4 terms.
24.96 or equivalent such as 624
A1:
25
4
d
y
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