ONLY GEOMETRIC SEQUENCES
1.
The first four terms of a sequence are 18, 54, 162, 486.
(a)
Use all four terms to show that this is a geometric sequence.
(2)
(b)
(i)
Find an expression for the nth term of this geometric sequence.
(ii)
If the nth term of the sequence is 1062 882, find the value of n.
(4)
(Total 6 marks)
2.
(a)
Consider the geometric sequence −3, 6, −12, 24, ….
(i)
Write down the common ratio.
(ii)
Find the 15th term.
Consider the sequence x − 3, x +1, 2x + 8, ….
(3)
(b)
When x = 5, the sequence is geometric.
(i)
Write down the first three terms.
(ii)
Find the common ratio.
(2)
(c)
Find the other value of x for which the sequence is geometric.
(4)
(d)
For this value of x, find
(i)
the common ratio;
(1)
(Total 10 marks)
IB Questionbank Maths SL
1
ANSWERS:
1.
(a)
(b)
For taking three ratios of consecutive terms
(M1)
54 162 486
 3


18 54 162
A1
hence geometric
AG
(i)
r=3
(A1)
un = 18  3n  1
(ii)
N0
For a valid attempt to solve 18  3 n  1 = 1062882
A1
N2
(M1)
eg trial and error, logs
n = 11
A1
N2
[6]
2.
(a)
(i)
r = 2
(ii)
u15 = 3 (2)14
A1
= 49152 (accept 49200)
(b)
(c)
N1
(A1)
A1
N2
(i)
2, 6, 18
A1
N1
(ii)
r=3
A1
N1
Setting up equation (or a sketch)
x 1 2 x  8

(or correct sketch with relevant information)
x  3 x 1
x2 + 2x + 1 = 2x2 + 2x  24
M1
A1
(A1)
x2 = 25
x = 5 or x = 5
x = 5
Notes: If “trial and error” is used, work must be
documented with several trials shown.
Award full marks for a correct answer with this
approach.
If the work is not documented, award N2 for a
correct answer.
IB Questionbank Maths SL
A1
N2
2
(d)
(i)
r=
1
2
A1
N1
[10]
IB Questionbank Maths SL
3