Question 8:
--- 8(a) ---
8(a) | Forms an equation for sum to
infinity or 2nd term
PI by correct answer | 1.1a | M1 | ar =18
a
=96
1−r
18
r−r2 =
96
1 3
r = or
4 4
a=72 or 24
Since a < 30
3
r =
4
a= 24
Obtains both correct equations | 1.1b | A1
Solves their two equations to
find two values of a or r
PI by correct values of a or r | 3.1a | M1
Obtains two correct values of
a= 72 or 24
or
obtains two correct values of
r =
1 3 | 1.1b | A1
4 or 4
Deduces correct pair of a and r
Follow through their values of a
and r
Must have one value of a > 30 | 2.2a | A1F
--- 8(b)(i) ---
8(b)(i) | Substitutes their a and r into the
expression u =
n
𝑙𝑙−1 | 1.1b | B1F | n−1
3
u =24×  
n 4
3n−1
=3×23×
22(n−1)
3×3n−1
=
2−3×22(n−1)
3n
=
22n−5
𝑎𝑎𝑎𝑎
Writes their a or r in terms of
prime numbers fully | 3.1a | M1
Deduces that a = 24 can be
written as23×3and
r = as
3 3
2
4 2
PI by expressing all terms in
powers of 2 and 3 | 2.2a | A1
Completes reasoned argument
by expressing all terms in
powers of 2 and 3 and simplifies
to show required result
AG | 2.1 | R1
--- 8(b)(ii) ---
8(b)(ii) | Applies logarithmic subtraction
or addition law correctly | 1.1a | M1
Applies logarithmic power law to
obtain either n log 3 or
3
(2n-5) log 2
3
Condone omission of brackets | 1.1a | M1
3n
log u =log
3 n 3 22n−5
=log 3n −log 22n−5
3 3
=n−( 2n−5 ) log 2
3
=n+( 5−2n ) log 2
3
=n−2nlog 2+5log 2
3 3
=n ( 1−2log 2 )+5log 2
3 3
Total | 12
Q | Marking Instructions | AO | Marks | Typical Solution