MCR 3U Functions Grade 11
SEQUENCES AND SERIES TEST 2
K/U
APP
𝟏𝟔
TIPS
𝟗
COM
𝟖
𝟏𝟓
Answers must be clear and complete for full marks. Two marks will be
given for overall form.
1. What is the 10th term of the sequence 1, 4, 7, 10, …? Explain how
you arrived at the answer. [K/U 2]
Solution
This is an arithmetic sequence with a = 1 and d = 3.
tn = a + (n – 1)d ⟹ t10 = 1 + (10 – 1) ∙ 3 = 28.
2. Which of the following does not represent an arithmetic sequence?
Explain how you arrived at the answer.
[K/U 2, COM 2]
1
a) 2tn + 3 = tn – 1, t1 = 3.
3
Solution
1
1
1
3
2tn + 3 = tn – 1 ⟹ 2tn = tn – 1 – 3 ⟹ tn = tn – 1 – .
t2 =
t3 =
1
6
1
6
3
3
1
3
2
3
6
1
2
2
6
3
1 3
6
2
t1 – = (3) – = – = –1,
3
2
2
1
9
6
6
t2 – = (–1) – = – – = –
2
2
10
6
5
2
= – = –1 ,
3
2
3
2
t2 – t1 = –1 – 3 = – 4, t3 – t2 = –1 – (–1) = –1 + 1= – ,
3
3
3
t2 – t1 ≠ t3 – t2 ∴ not an arithmetic sequence
b) f (n) = 13n – 4
Solution
d = f (n + 1) – f (n) = [13(n + 1) – 4] – [13n – 4]
= 13n + 13 – 4 – 13n + 4 = 13.
d = 13 = constant ∴ arithmetic sequence
c) 5tn – 3n + 2 = 0.
Solution
5tn – 3n + 2 = 0 ⟹ 5tn = 3n – 2 ⟹ tn = 0.6n – 0.4
d = tn + 1 – tn = [0.6(n + 1) – 0.4] – [0.6n – 0.4]
= 0.6n + 0.6 – 0.4 – 0.6n + 0.4 = 0.6.
d = 0.6 = constant ∴ arithmetic sequence
d) ntn + 7 = 17n.
Solution
7
ntn + 7 = 17n ⟹ ntn = 17n – 7 ⟹ tn = 17 – .
d = tn + 1 – tn = (17 –
7
𝑛+1
7
7
𝑛
𝑛
𝑛
) – (17 – ) = –
7
𝑛+1
=
7
𝑛(𝑛+1)
d ≠ constant ∴ not an arithmetic sequence
3. Determine the number of terms in the sequence: 5240, 4365, 3490, ...,
– 2635. Show your work. [K/U 2]
Solution
This is an arithmetic sequence with a = 5240 and d = – 875.
tn = – 2635 ⟹ a + (n – 1)d = – 2635
5240 + (n – 1)(– 875) = – 2635
(n – 1)(– 875) = – 2635 – 5240
(n – 1)(– 875) = – 7875
n – 1 = 9, n = 10.
4. The amount of money in an account earning compound interest is
modeled by the equation P (n) = Arn – 1, where P (n) is the amount in
the account at year n, A is the amount invested and r = 1 + i, where i is
the interest. If you invest $1500 in a year and earn 4% interest, how
much money will you have in year 30? [APP 4]
Solution
A = 1500, i = 0.04, r = 1.04, n = 30
P(30) = (1500)(1.04)30 – 1 = (1500)(1.04)29 = $ 4677.98
5. If the 8th term of a geometric sequence is 256 and the 5th term of the
sequence is 32, what is the common ratio? [K/U 2]
Solution
t8 = 256 ⟹ ar7 = 256 (1)
t5 = 32 ⟹ ar4 = 32
(2)
Divide (1) by (2):
𝑎𝑟 7
𝑎𝑟 4
=
256
32
⟹ 𝑟3 = 8 ⟹ r = 2
6. A brick layer building a trapezoidal wall starts with a base of 250
bricks and decreases the number of bricks by 5 each layer up. How
many bricks does he need to make the wall 26 layers high? [APP 4]
Solution
We have to find the sum of n = 26 terms of an arithmetic series with t1 =
a = 250 and d = –5.
𝑛
Sn = ∙ [2a + (n – 1) d] ∴
2
26
S26 = ∙ [2(250) + (26 – 1) (–5)]
2
= 13 ∙ 375 = $4 875.
7. Calculate the sum of the first 20 terms of an arithmetic sequence with
the third term 8 and the eighth term 143. [TIPS 3]
Solution
t8 = a + 7d ⟹ a + 7d = 143 (1)
t3 = a + 2d ⟹ a + 2d = 8
(2)
Subtract (2) from (1): 5d = 135 ⟹ d = 27 (3)
Substitute (3) into (2): a + 2(27) = 8 ⟹ a = – 46.
S20 =
20
2
∙ [2(– 46) + (20 – 1) (27)] = 10 ∙ (– 92 + 513) = 4 210.
8. Calculate the sum of the first 6 terms of the geometric series with the
2
third term 6 and common ratio of . [K/U 2]
3
Solution
2
6
r = , t3 = 6 ⟹ ar2 = 6 ⟹ a = 2 2
3
(3)
⟹a=
6
4
9
9
27
4
2
=6∙ =
= 13.5.
9. A population of rabbits, if left unchecked, triples every three month. If
there are initially 2 rabbits, how many will there be in a year and a half?
[APP 2]
Solution
18
1.5 years = 18 months ⟹ Number of tripling periods = = 6.
3
6
6
N(6) = N(0) ∙ 3 = 2 ∙ 3 = 2 ∙ 729 = 1 458.
10. Expand and simplify the first 4 terms in the binomial expansion
(3 +
𝑥
√5
6
)
Solution
Row
0
1
2
3
4
5
6
(3 +
𝑥
√5
[K/U 3]
Pascal’s Triangle
1
1
1
1
2
1
1
3
3
1
1
4
6
4
1
1
5
10
10
5
1
1
6 15
20
15 6
1
6
𝑥
𝑥
√5
√5
2
3
𝑥
) = 1∙ (3)6 + 6 ∙ (3)5( ) + 15 ∙ (3)4( ) + 20 ∙ (3)3( ) +
𝑥
4
𝑥
5
√5
𝑥
6
15 ∙ (3)2( ) + 6 ∙ (3)( ) + 1 ∙ ( )
= 729 +
√5
1458
√5
x+
√5
108
243x2 + x3
√5
√5
+
27 4
x
5
+
18 5
x
5√5
+
1
x6.
125
11. Write the following expansion in the form (a + b)n:
x8 + 8x7√𝑦 + 28x6y + ... + y4.
Explain how you arrived at your answer. [TIPS 3, COM 2]
Solution The exponent of x decreases from 8 through 0. Hence, n = 8.
Since the first term is x8, we have: a = x. The second term is 8a7b =
8x7√𝑦, hence, b = √𝑦. Thus,
x8 + 8x7√𝑦 + 28x6y + ... + y4 = (x + √𝑦)8.
Check: The third term is 28a6b2 = 28x6y
12. Explain the difference between a sequence and series. [COM 2]
Answer: A "sequence" (or "progression", in British English) is an
ordered list of numbers; the numbers in this ordered list are called
"elements" or "terms". A "series" is the value you get when you add up
all the terms of a sequence; this value is called the "sum". For instance,
"1, 2, 3, 4" is a sequence, with terms "1", "2", "3", and "4"; the
corresponding series is the sum "1 + 2 + 3 + 4", and the value of the
series is 10.
13. Explain what differentiates an arithmetic sequence from a geometric
one. [COM 2]
Answer: An arithmetic sequence goes from one term to the next by
always adding (or subtracting) the same value. For instance, 2, 5, 8, 11,
14,... and 7, 3, –1, –5,... are arithmetic, since you add 3 and subtract 4,
respectively, at each step. A geometric sequence goes from one term to
the next by always multiplying (or dividing) by the same value. So 1, 2,
4, 8, 16,... and 81, 27, 9, 3, 1, 1/3,... are geometric, since you multiply by
2 and divide by 3, respectively, at each step.
14. Your parents offer you a job for the month of July (31 days) to work
around the house for $400.00. You tell them that, instead of getting paid
$400.00, you would rather get paid 1 cent the first day, 2 cents the
second day, 4 cents the third day, 8 cents the fourth day, and so on. How
much will your parents pay you for the month of July if they accept your
offer? [APP 4]
Solution
1 + 2 + 22 + 23 + … + 231 =?
Sn =
𝑎(𝑟 𝑛 − 1)
S31 =
, a = 1, r = 2, n = 31.
𝑟−1
1 ∙ (231 − 1)
2−1
= 231 – 1 = $ 21 474 836.47
15. How many terms are there in the sequence 10, – 6,
18
5
,..., –
4 374
15 625
?
[K/U 2]
Solution
This is a geometric sequence with the first term t1 = a = 10 and common
ratio r = – 0.6.
tn = arn – 1 ⟹ –
⟹–
4 374
15 625
2 187
78125
= 10(– 0.6)n – 1
= (– 0.6)n – 1
⟹ (– 0.6)7 = (– 0.6)n – 1
⟹ 7 = n – 1, n = 8.
16. A man invested $1 000 for 10 years into an account that earns 4%
compound interest per year. How much money will he have in 10 years?
Determine the interest he will get over this period. [TIPS 3]
Solution
A = P (1 + r)t, P = $1 000 – principal amount (initial investment), r =
0.04 – interest rate, t = 10 – number of years, A – accumulated amount.
A = 1 000 (1 + 0.04)10 = $1 480.24