Predicting What Comes Next: Sequences and Progressions | Exercise 8.3

Question 4

Which term of the GP: 2,6,18,2, 6, 18, \dots is 43744374? Write the explicit formula as well as the recursive formula for the nthn^{\text{th}} term.

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Solution
Understand the Question
  • A Geometric Progression (GP) is a sequence where each term after the first is found by multiplying the previous term by a fixed non-zero number called the common ratio (rr).
  • Given GP: 2,6,18,2, 6, 18, \dots
    • First term: a=2a = 2
    • Common ratio: r=62=3r = \dfrac{6}{2} = 3
  • To solve the problem:
    1. Find the explicit formula for the general term: tn=arn1t_n = a r^{n-1}.
    2. Equate tn=4374t_n = 4374 and solve for nn using powers of 33.
    3. Write the recursive formula, which defines the first term t1t_1 and gives the rule tn=rtn1t_n = r \cdot t_{n-1} for n2n \ge 2.

Step 1 · Find the Explicit Formula

Diagram 1

First term a=2a = 2

Common ratio rr:

r=second termfirst term=62=3\begin{aligned} r &= \dfrac{\text{second term}}{\text{first term}} \\[0.6em] &= \dfrac{6}{2} \\[0.6em] &= 3 \end{aligned}

The explicit formula for the nthn^{\text{th}} term is tn=arn1t_n = ar^{n-1}: tn=2×3n1t_n = 2 \times 3^{n-1}

Step 2 · Find the Term Number for 4374

Set tn=4374t_n = 4374:

2×3n1=43743n1=437423n1=2187\begin{aligned} 2 \times 3^{n-1} &= 4374 \\[0.6em] 3^{n-1} &= \dfrac{4374}{2} \\[0.6em] 3^{n-1} &= 2187 \end{aligned}

Express 21872187 as a power of 33:

31=332=933=2734=8135=24336=72937=2187\begin{aligned} 3^1 &= 3 \\ 3^2 &= 9 \\ 3^3 &= 27 \\ 3^4 &= 81 \\ 3^5 &= 243 \\ 3^6 &= 729 \\ 3^7 &= 2187 \end{aligned}

Comparing exponents:

3n1=37n1=7n=7+1n=8\begin{aligned} 3^{n-1} &= 3^7 \\ n - 1 &= 7 \\ n &= 7 + 1 \\ n &= 8 \end{aligned}

Therefore, 43744374 is the 8th8^{\text{th}} term.

Step 3 · Write the Recursive Formula

A recursive formula requires the first term and the relation between consecutive terms:

t1=2tn=3×tn1for n2\begin{aligned} t_1 &= 2 \\ t_n &= 3 \times t_{n-1} \quad \text{for } n \ge 2 \end{aligned}
Answer

(i) 43744374 is the 8th8^{\text{th}} term.

(ii) Explicit formula: tn=2×3n1t_n = 2 \times 3^{n-1}

(iii) Recursive formula: t1=2t_1 = 2, tn=3tn1 for n2\, t_n = 3t_{n-1} \text{ for } n \ge 2

Common Mistakes
  • Exponent Off-by-One Error: Writing the explicit formula as tn=arnt_n = ar^n instead of tn=arn1t_n = ar^{n-1}, which incorrectly gives n=7n = 7.
  • Incomplete Recursive Formula: Omitting the base term t1=2t_1 = 2 or the domain n2n \ge 2. A recursive definition is incomplete without specifying the starting value.
  • Power Factorisation Mistake: Making an arithmetic error while expressing 21872187 as 373^7.

More questions in Exercise 8.3

Q1

Find the 12th12^{\text{th}} term of a GP with common ratio 2, whose 8th8^{\text{th}} term is 192.

Q2

Find the 10th10^{\text{th}} and nthn^{\text{th}} terms of the GP: 5,25,125,5, 25, 125, \dots.

Q3

A sequence is given by the recursive rule t1=2t_1 = 2, tn+1=3tn2t_{n+1} = 3t_n - 2 for n1n \ge 1. Which term of the sequence is 730?

Q4

Which term of the GP: 2,6,18,2, 6, 18, \dots is 43744374? Write the explicit formula as well as the recursive formula for the nthn^{\text{th}} term.

Q5

A ball is dropped from a height of 80 metres. After hitting the ground, it bounces back to 60% of the height from which it fell. It continues bouncing in this way—each time rising to 60% of the previous height.

(i) What height does the ball reach after the 5th5^{\text{th}} bounce? (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th6^{\text{th}} time?

Q6

Which term of the sequence 2,22,4,2, 2\sqrt{2}, 4, \dots is 128?

Q7

Fig. 8.12 shows Stages 0 to 3 of the Sierpiński square carpet. Stage 0 of this fractal is a square sheet of paper. To construct Stage 1, each side of the square is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares. The centre square is then removed and the 8 smaller squares are retained, leaving a square hole in the centre. The same process is repeated on the eight smaller shaded squares to obtain Stage 2 and so on.

Look at Fig. 8.12 and try to answer the following questions.

(i) How many red squares are there in Stages 0 to 3? (ii) Can you predict the number of red squares in Stages 4 and 5? (iii) Can you find a rule for the number of red squares at the nthn^{\text{th}} stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage. (iv) Suppose the area of the square in Stage 0 is 1 square unit. What is the area of the red region in Stages 1, 2 and 3? What will be the area of the red region in Stages 4 and 5? Find the explicit as well as the recursive formula for the area of the red region at the nthn^{\text{th}} stage. What happens to this area as nn, the number of stages, goes on increasing?

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