Predicting What Comes Next: Sequences and Progressions | Exercise 8.1

Question 1

Find the first five terms of the sequence in which the nthn^{\text{th}} term is given by, for n≥1n \ge 1:

(i) tn=3n−4t_n = 3n - 4

(ii) tn=2−5nt_n = 2 - 5n

(iii) tn=n2−2n+3t_n = n^2 - 2n + 3

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Solution
Understand the Question
  • The nthn^{\text{th}} term of a sequence, denoted as tnt_n, is a general algebraic formula that defines every term in the sequence based on its position nn.
  • To find the first five terms of each sequence, substitute the natural numbers n=1,2,3,4,5n = 1, 2, 3, 4, 5 sequentially into the given formula for tnt_n.

(i) tn=3n−4t_n = 3n - 4

Step 1 · Calculate the First Five Terms

Substitute n=1,2,3,4,5n = 1, 2, 3, 4, 5 into tn=3n−4t_n = 3n - 4:

For n=1n = 1:

t1=3(1)−4=3−4=−1\begin{aligned} t_1 &= 3(1) - 4 \\ &= 3 - 4 \\ &= -1 \end{aligned}

For n=2n = 2:

t2=3(2)−4=6−4=2\begin{aligned} t_2 &= 3(2) - 4 \\ &= 6 - 4 \\ &= 2 \end{aligned}

For n=3n = 3:

t3=3(3)−4=9−4=5\begin{aligned} t_3 &= 3(3) - 4 \\ &= 9 - 4 \\ &= 5 \end{aligned}

For n=4n = 4:

t4=3(4)−4=12−4=8\begin{aligned} t_4 &= 3(4) - 4 \\ &= 12 - 4 \\ &= 8 \end{aligned}

For n=5n = 5:

t5=3(5)−4=15−4=11\begin{aligned} t_5 &= 3(5) - 4 \\ &= 15 - 4 \\ &= 11 \end{aligned}
Answer

(i) −1,2,5,8,11-1, 2, 5, 8, 11

(ii) tn=2−5nt_n = 2 - 5n

Step 1 · Calculate the First Five Terms

Substitute n=1,2,3,4,5n = 1, 2, 3, 4, 5 into tn=2−5nt_n = 2 - 5n:

For n=1n = 1:

t1=2−5(1)=2−5=−3\begin{aligned} t_1 &= 2 - 5(1) \\ &= 2 - 5 \\ &= -3 \end{aligned}

For n=2n = 2:

t2=2−5(2)=2−10=−8\begin{aligned} t_2 &= 2 - 5(2) \\ &= 2 - 10 \\ &= -8 \end{aligned}

For n=3n = 3:

t3=2−5(3)=2−15=−13\begin{aligned} t_3 &= 2 - 5(3) \\ &= 2 - 15 \\ &= -13 \end{aligned}

For n=4n = 4:

t4=2−5(4)=2−20=−18\begin{aligned} t_4 &= 2 - 5(4) \\ &= 2 - 20 \\ &= -18 \end{aligned}

For n=5n = 5:

t5=2−5(5)=2−25=−23\begin{aligned} t_5 &= 2 - 5(5) \\ &= 2 - 25 \\ &= -23 \end{aligned}
Answer

(ii) −3,−8,−13,−18,−23-3, -8, -13, -18, -23

(iii) tn=n2−2n+3t_n = n^2 - 2n + 3

Step 1 · Calculate the First Five Terms

Substitute n=1,2,3,4,5n = 1, 2, 3, 4, 5 into tn=n2−2n+3t_n = n^2 - 2n + 3:

For n=1n = 1:

t1=12−2(1)+3=1−2+3=2\begin{aligned} t_1 &= 1^2 - 2(1) + 3 \\ &= 1 - 2 + 3 \\ &= 2 \end{aligned}

For n=2n = 2:

t2=22−2(2)+3=4−4+3=3\begin{aligned} t_2 &= 2^2 - 2(2) + 3 \\ &= 4 - 4 + 3 \\ &= 3 \end{aligned}

For n=3n = 3:

t3=32−2(3)+3=9−6+3=6\begin{aligned} t_3 &= 3^2 - 2(3) + 3 \\ &= 9 - 6 + 3 \\ &= 6 \end{aligned}

For n=4n = 4:

t4=42−2(4)+3=16−8+3=11\begin{aligned} t_4 &= 4^2 - 2(4) + 3 \\ &= 16 - 8 + 3 \\ &= 11 \end{aligned}

For n=5n = 5:

t5=52−2(5)+3=25−10+3=18\begin{aligned} t_5 &= 5^2 - 2(5) + 3 \\ &= 25 - 10 + 3 \\ &= 18 \end{aligned}
Answer

(iii) 2,3,6,11,182, 3, 6, 11, 18

Common Mistakes
  • Starting Index: Sequences start with n=1n = 1 (since n≥1n \ge 1). Starting at n=0n = 0 will yield incorrect initial terms.
  • Sign Errors in Subtraction: In part (ii), be careful with the negative signs when computing 2−5n2 - 5n (e.g. 2−10=−82 - 10 = -8, not +8+8).
  • Order of Operations: In part (iii), ensure powers and multiplications are evaluated before addition and subtraction (n2−2n+3n^2 - 2n + 3).

More questions in Exercise 8.1

Q1

Find the first five terms of the sequence in which the nthn^{\text{th}} term is given by, for n≥1n \ge 1:

(i) tn=3n−4t_n = 3n - 4

(ii) tn=2−5nt_n = 2 - 5n

(iii) tn=n2−2n+3t_n = n^2 - 2n + 3

Q2

Find the 10th10^{\text{th}} and 15th15^{\text{th}} terms of the sequence tn=5n−3t_n = 5n - 3 for n≥1n \ge 1.

Q3

Determine whether 97 and 172 are terms of the sequence tn=5n−3t_n = 5n - 3 for n≥1n \ge 1.

Q4

Which term of the sequence tn=5n−3t_n = 5n - 3 for n≥1n \ge 1 is 607?

Q5

A sequence is given by the recursive rule t1=−5t_1 = -5, tn+1=tn+3t_{n+1} = t_n + 3 for n≥1n \ge 1. Find the first five terms of the sequence. Is 5252 a term of this sequence? If so, which term is it?

Q6

Let T1=1T_1 = 1, T2=2T_2 = 2, T3=4T_3 = 4, and Tn=Tn−1+Tn−2+Tn−3T_n = T_{n-1} + T_{n-2} + T_{n-3} for n≥4n \ge 4. Find T4T_4, T5T_5, T6T_6, T7T_7, and T8T_8.

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