Mathematics · Oct / Nov 2022 · Paper 21 · Question 8
YMA01
As printed in the paper
The question as text
8. A geometric sequence has first term a and common ratio r Given that S∞ = 3a
Read automatically from the paper; the printed page above is exact.
Show the answer and mark scheme
8(a)[2]
a
3
a
=
r
=
...
1
−
r
2 *
r =
3
8(b)[5]
ar
−
ar
3
=
16
10
27 a
16
a
...
=
=
a =
43.2
( )
43.2(1
10
)
−
2
S
=
3
=
127.4
10
1
−
2
3
(a)
a
M1: Sets 3
a
= −
,
cancels all the a’s and attempts to rearrange to find a numerical value for r
1
r
2
a
A1*:
r =
with no errors and at least one intermediate step after 3
a
= −
3
1
r
Alternative by verification:
2 ,
a
a
2
r
3
a
=
=
=
Hence
r =
3
1
−
r
1
−
2
3
3
2
Score M1 for substituting
r =
into a correct sum to infinity formula and obtaining 3a
3
Score A1 for correct work followed by a (minimal) conclusion e.g. QED, proven, etc.
(b)
3
16
B1:
ar
−
ar
=
seen or implied.
2
M1: Proceeds to a value for a from a linear equation in a using
ar
ar
3
16
r =
and
−
=
3
2
But condone use of
r =
and
ar
2
−
ar
4
=
16
3
or any equivalent correct numerical expression e.g. 16
×
27 216
A1:
a =
43.2
,
10
5
2
dM1: Substitutes their a,
r =
and n = 10 into the correct sum formula.
3
(Also allow a correct and full attempt to calculate the sum of all 10 terms separately)
It is dependent on the previous method mark.
A1: awrt 127.4 For reference the exact answer is 92840
729
If candidates use
ar
+
ar
3
=
16
rather than
ar
−
ar
3
=
16
we will treat this as a copying slip and allow
B0M1A0dM1A0 if the subsequent work merits it.
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