WorkSHEET 5.2 Arithmetic and geometric sequences Name _________________
1
Which of the following are geometric
sequences?
(a) 1, –2, 4, –8, 16, –32…
(a) The ratio of 2nd to 1st term =
ratio of 3rd to 2nd term
2
= –2
1
6
4
= –2
2
…
 32
= –2
16
So, YES this is a geometric sequence.
ratio of 6th to 5th term
(b) 1, –2, 3, –4, 5, 6…
(b) The ratio of 2nd to 1st term =
2
= –2
1
3
= –1.5
2
So, NO this is not a geometric sequence.
ratio of 3rd to 2nd term
(c) 16, 8, 4, 2, 1…
(d) 26, 18, 14, 12, 11…
1 1 1 1 1
(e) 2 , 3 , 4 , 5 , 6 …
3 3 3
3
3
(f) 2 , 6 , 18 , 54 , 96 …
2
For each of the geometric sequences in
question 1, find the first term, a, and the
common ratio, r.
3
A financial company advertises that the value
of money invested in the company doubles
every 6 years.
If Simon invests $100, how much will he have
in 24 years time?
Use this technique on the remaining parts.
(c) YES
(d) NO
(e) NO
(f) YES
2
= –2
1
8
1
(c) a = 16, r = 16 = 2
3
6
3
1
(f) a = 2 , r = 3 = 3
2
3
Determine a, r and n.
24
a = 100, r = 2, n = 6 + 1 = 5 (0 years, 6 years,
12 years, 18 years, 24 years)
Use the formula tn = ar n1
t5 = 100(2)4 = 100(16) = $1600
2
(a) a = 1, r =
4
Find the 7th term of the following sequences
where:
(a) a = 1, r = 2
(b) a = 2, r = 3
(c) a = 3, r = 2
(d) a = 3, r = 3
Use the formula tn = ar n1 to determine the 7th
term.
(a) t7 = 1(2)6 = 64
(b) t7 = 2(3)6 = 1458
(c) t7 = 3(2)6 = 192
(d) t7 = 3(3)6 = 2187
4
5
The 3rd term of a geometric sequence is 45,
while the 6th term is –1215. Find a and r.
Use the formula tn = ar n1
t3 = 45 = ar2
t6 = –1215 = ar5
Divide
t6
 1215 ar5
=
= ar2
45
t3
–27 = r3
Therefore r = –3.
Substitute into the equation for t3
45 = a(–3)2
45 = 9a
a=5
4
6
The 1st term of a geometric sequence is 1.2,
while the common ratio is 1.1. What would be
the first term to exceed 100?
a = 1.2, r = 1.1
Using tn = ar n1
tn = 1.2(1.1) n1
4
Substitute the value of 100 for tn.
100 = 1.2(1.1) n1
83.33 = (1.1) n1
Take logarithms of both sides:
1.9208 = (n – 1) 0.04139
46.4 = n – 1
n = 47.4
The first term to exceed 100 is the 48th.
Confirmation:
t47 = 1.2(1.1)46 = 96.21
t48 = 105.83
7
An ancient legend has it that a wise man once
saved a king’s life. The king said that the wise
man could have anything that the king could
give him. So, the wise man said “…give me
one grain of rice in the first square of a
chessboard, 2 grains of rice in the 2nd square, 4
grains of rice in the 3rd, and so on for all 64
squares”.
(a) Find a and r in this geometric sequence.
(a) a = 1, r = 2
2
(b) How many grains of rice in the last square? (b) t64 = 263 = 9 223 372 036 854 780 000
give or take a few grains. This was more rice
than was in the kingdom, so the wise man was
put to death!
8
Find the sum of the first 6 terms of the
following series.
Use the formula Sn =
(a) a = 5, r = 2
(a) S6 =
1
(b) a = 5, r = 2
(b)
(c) a = 10, r = 2
(c)
1
(d) a = 10, r = 2
(d)
(e) a = 6.78, r = 0.98
(e)
(f) a = 6.78, r = 0.88
(f)
a (r n  1)
r 1
5(2 6  1)
1
= 5(63)
= 315
5(0.5 6  1)
S6 =
 0.5
= –10(–0.98438)
= 9.8438
10(2 6  1)
S6 =
1
= 10(63)
= 630
10(0.5 6  1)
S6 =
 0.5
= –20(–0.98438)
= 19.6875
6.78(0.986  1)
S6 =
0.98  1
6.78(0.88584  1)
=
 0.02
= 6.78(5.708) = 38.7002
6.78(0.886  1)
S6 =
0.88  1
6.78(0.46440  1)
=
 0.12
= 6.78(4.4633)
= 30.2612
6
9
Consider a geometric series with r = 2.
(a) Show that the sum of n terms plus the first
term equals the next term in the sequence:
Sn + a = tn+1 or Sn = tn+1 – a
(a) Let tn = ar n1, so that tn+1 = ar n
a (r n  1)
Similarly, let Sn =
.
r 1
Substitute r = 2 into both equations.
tn+1 = a(2n )
a(2 n  1)
Sn =
= a(2n – 1) = a(2n) – a
2 1
Substitute tn+1 into [2]
Sn = tn+1 – a , or Sn + a = tn+1
(b) Use this formula to predict the sum of the
series
3 + 6 + 12 + 24 + 48 as 96 – 3 = 93
(b) a = 3, r = 2, so the formula can be used.
The next term (t6) in the series is 96
S5 = 96 – 3
= 93
1
[1]
[2] 1
1
1
10
Consider the geometric sequence where a = 10
and r = 0.65. By taking various sums of terms,
determine (to 2 decimal places) the largest
possible sum, regardless of the number of
terms. In other words, the sum converges to a
specific value.
a = 10, r = 0.65
Construct a table of terms and sums.
n
tn
Sn
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
10
6 .5
4.225
2.7463
1.7851
1.1603
0.7542
0.4902
0.3186
0.2071
0.1346
0.0875
0.0569
0.0370
0.0240
0.0156
0.0101
0.0066
0.0043
0.0028
0.0018
0.0012
0.0008
0.0005
10
16.5
20.725
23.4713
25.2563
26.4166
27.1709
27.6610
27.9800
28.1868
28.3214
28.4089
28.4658
28.5028
28.5268
28.5424
28.5526
28.5592
28.5635
28.5663
28.5681
28.5692
28.5700
28.5705
Observe that by term 23, the sum has
converged (to 2 decimal places) to 28.57
3