Predicting What Comes Next: Sequences and Progressions | Exercise 8.1

Question 5

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?

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Solution
Understand the Question
  • A recursive rule gives each subsequent term based on the preceding term. Here, each term is obtained by adding 33 to the previous term: tn+1=tn+3t_{n+1} = t_n + 3.
  • This generates an Arithmetic Progression (AP) with first term a=−5a = -5 and common difference d=3d = 3.
  • To check if a number is a term of the sequence, set the general term tn=a+(n−1)dt_n = a + (n - 1)d equal to that number and solve for nn. If nn is a positive integer (natural number), the number is a term in the sequence.

Step 1 · Find the First Five Terms

Given t1=−5t_1 = -5 and tn+1=tn+3t_{n+1} = t_n + 3 for n≥1n \ge 1:

t1=−5t2=t1+3=−5+3=−2t3=t2+3=−2+3=1t4=t3+3=1+3=4t5=t4+3=4+3=7\begin{aligned} t_1 &= -5 \\[0.6em] t_2 &= t_1 + 3 = -5 + 3 = -2 \\[0.6em] t_3 &= t_2 + 3 = -2 + 3 = 1 \\[0.6em] t_4 &= t_3 + 3 = 1 + 3 = 4 \\[0.6em] t_5 &= t_4 + 3 = 4 + 3 = 7 \end{aligned}

Step 2 · Check if 52 is a Term of the Sequence

The sequence forms an arithmetic progression with first term a=−5a = -5 and common difference d=3d = 3.

The nthn^{\text{th}} term formula is: tn=a+(n−1)dt_n = a + (n-1)d

tn=−5+(n−1)3=−5+3n−3=3n−8\begin{aligned} t_n &= -5 + (n-1)3 \\ &= -5 + 3n - 3 \\ &= 3n - 8 \end{aligned}

Set tn=52t_n = 52 to find nn:

52=3n−852+8=3n60=3nn=603n=20\begin{aligned} 52 &= 3n - 8 \\ 52 + 8 &= 3n \\ 60 &= 3n \\[0.6em] n &= \dfrac{60}{3} \\[0.6em] n &= 20 \end{aligned}

Since n=20n = 20 is a natural number, 5252 is a term of the sequence.

Answer

The first five terms are −5,−2,1,4,7-5, -2, 1, 4, 7. Yes, 5252 is the 20th20^{\text{th}} term of the sequence.

Common Mistakes
  • Arithmetic Sign Errors: Making a sign error when adding 33 to negative numbers, e.g., calculating −5+3-5 + 3 as −8-8 instead of −2-2.
  • Non-integer nn Verification: Forgetting that nn must be a positive integer. If solving for nn results in a fraction or negative number, the given value is not part of the sequence.

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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