Mathematics · Oct / Nov 2020 · Paper 21 · Question 8
YMA01
As printed in the paper
The question as text
8. A geometric series has first term a and common ratio r.
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[3]
S
=
a
+
ar
+
...
+
ar −
n
1
n
rS
=
ar
+
ar
2
+
...
+
ar
n
n
Writes down at least 3 correct terms of a geometric series and multiplies their sequence by r.
There may be extra incorrect terms but allow this mark if there are 3 correct terms in both
sequences and at least one “+” in both sequences but see special case below
S
−
rS
=
a
−
ar
n
or
rS
−
S
=
ar
n
−
a
n
n
n
n
Obtains either equation where both Sn and rSn had the correct first and last terms and at least
one other correct term but no incorrect terms. Both sides must be seen unfactorised.
(
)
a
1
−
r
n
(
)
(
)
1
−
r S
=
a
1
−
r
n
⇒
S
=
*
n
n
1
−
r
Factorises both sides and divides by 1 – r to obtain the printed answer
(
)
(
)
a
1
−
r
n
a r
n
−
1
Should be as printed but allow e.g.
S
=
but not
S
=
unless followed by
(
)
(
)
n
1
−
r
n
r
−
1
correct version
Special case:
If terms are listed rather than added and the working is otherwise correct score 110
See next page for proof by induction.
Alternative for (a):
S
=
a
+
ar
+
...
+
ar −
n
1
n
)(
)
(
1
−
r
a
+
ar
+
...
+
ar
n
−
1
)(
)
(
)
(
1
−
r S
=
1
−
r
a
+
ar
+
...
+
ar −
n
1
or
S
=
(
)
n
n
1
−
r
Writes down at least 3 correct terms of a geometric series and multiplies both sides by 1 – r or
multiplies the right hand side by 1
r
−
1
−
r
There may be extra incorrect terms but allow this mark if there are 3 correct terms
a
−
ar
n
(
)
1
−
r S
=
a
−
ar
n
or
S
=
n
n
1
r
−
Obtains the above equation where Sn had the correct first and last terms and at least one other
correct term and no incorrect terms. Right hand side must be seen unfactorised unless the “a”
was factored out earlier
(
)
(
)
a
1
−
r
n
(
)
1
−
r S
=
a
−
ar
n
=
a
1
−
r
n
⇒
S
=
*
n
n
1
−
r
or
(
)
a
−
ar
n
a
1
−
r
n
S
=
⇒
S
=
*
n
1
−
r
n
1
−
r
(
)
(
)
a
1
−
r
n
a r
n
−
1
but not
Should be as printed but allow e.g.
S
=
S
=
unless followed by
(
)
(
)
n
1
−
r
n
r
−
1
correct version
8(b)[2]
Mark (b) and (c) together
Correct strategy for r. Allow for dividing
20.48
20.48
r
3
= −
⇒
r
=
−
the 2 given terms either way round and
3
320
320
attempting to cube root.
Correct value (and no others) but allow
= −0.4
equivalents e.g. 2/5. Correct answer only
scores both marks.
512
Note that some candidates take
ar
2
= −
320 and
ar
5
=
and use these correctly to give
25
20.48
20.48
r
3
= −
⇒
r
=
−
= −
0.4
3
320
320
In such cases you can allow full marks for (b) but see note * in (c)
8(c)[3]
Correct attempt at the first term using
± their r and the −320 or the 512
−
320
. May be
r
= −
0.4
⇒
a
=
(
=
800
)
25
0.4
−
implied by their a but must be using e.g.
or
512
4
ar
= −
320 or
ar
4
=
not
512
2
(
)
r
= −
0.4
⇒
a
=
÷ −
=
800
25
25
5
512
ar
2
= −
320 or
ar
5
=
*
25
(
)
"800" 1 " 0.4"
−−
13
S
=
13
1 " 0.4"
−−
Correct attempt at the sum using their a and their r and n = 13 to find a value for S13.
Must be a fully correct attempt at the sum here using n = 13, their a and their r.
(
)
(
)
(
)
800 1
−−
0.4
13
800 1
+
0.4
13
Note that
is equivalent to
and is acceptable for this mark.
(
)
1
+
0.4
1
−−
0.4
Correct value. Note that S∞ is also 571.43
= 571.43
so working must be seen i.e. correct answer
only scores no marks.
(
)
1
a
1
−
r
so true for n = 1
n
= ⇒
1
S
=
=
a
1
1
−
r
(
)
1
k
a
−
r
Assume true for n = k so
S
=
k
1
r
−
(
)
a
1
r
k
−
1
−
ar
k
+
ar
k
−
ar
k
+
1
Add (k + 1)th term
S
=
+
ar
k
=
k
+
1
1
r
1
r
−
−
(
)
a
1
−
r
k
+
1
a
−
ar
k
+
1
=
=
1
−
r
1
−
r
So if true for n = k it has been shown true for n = k + 1 and as it is true for n = 1 it is true for (for all n)
Mark as follows:
M1: Shows true for n = 1 and assumes true for n = k and adds the (k + 1)th term
a
−
ar
k
+
1
A1(M1 on EPEN): Finds common denominator obtains
using correct algebra
1
−
r
A1: Fully correct proof reaching (
)
a
1
r
k
+
1
−
with all steps shown and conclusion
1
−
r
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