Which term of the sequence is 128?

Understand the Question The given sequence is a Geometric Progression (GP) where each term is multiplied by a constant factor called the common ratio (). To determine which term is , we first identify the first term and the common ratio . Then, we use the formula for the -th term of a GP, , and solve for the term index . Step 1 · Find the First Term and Common Ratio Given sequence: First term, Common ratio, : Verifying with the next terms: Thus, the sequence is a GP with and . Step 2 · Find the Term Number Let the -th term be . Using the formula : Equating exponents of on both sides: Answer > term Common Mistakes Index Exponent Confusion: Confusing with by forgetting that , which leads to . Skipping Division by : Forgetting to divide by the first term before converting to powers of .

Predicting What Comes Next: Sequences and Progressions | Exercise 8.3

Question 6

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

Solution

Understand the Question

The given sequence is a Geometric Progression (GP) where each term is multiplied by a constant factor called the common ratio ( $r$ ).
To determine which term is $128$ , we first identify the first term $a$ and the common ratio $r$ .
Then, we use the formula for the $n$ -th term of a GP, $t_n = a r^{n-1}$ , and solve for the term index $n$ .

Step 1 · Find the First Term and Common Ratio

Given sequence: $2, 2\sqrt{2}, 4, \dots$

First term, $a = 2$

Common ratio, $r$ :

r = \dfrac{2\sqrt{2}}{2} = \sqrt{2}

Verifying with the next terms:

\begin{aligned} r &= \dfrac{4}{2\sqrt{2}} \\[0.6em] &= \dfrac{2}{\sqrt{2}} \\[0.6em] &= \dfrac{2 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}} \\[0.6em] &= \dfrac{2\sqrt{2}}{2} = \sqrt{2} \end{aligned}

Thus, the sequence is a GP with $a = 2$ and $r = \sqrt{2}$ .

Step 2 · Find the Term Number $n$

Let the $n$ -th term be $t_n = 128$ .

Using the formula $t_n = ar^{n-1}$ : $128 = 2(\sqrt{2})^{n-1}$

\begin{aligned} \dfrac{128}{2} &= (\sqrt{2})^{n-1} \\[0.6em] 64 &= (\sqrt{2})^{n-1} \\[0.6em] 2^6 &= (2^{1/2})^{n-1} \\[0.6em] 2^6 &= 2^{\frac{n-1}{2}} \end{aligned}

Equating exponents of $2$ on both sides:

\begin{aligned} 6 &= \dfrac{n-1}{2} \\[0.6em] 12 &= n - 1 \\[0.6em] n &= 12 + 1 \\[0.6em] n &= 13 \end{aligned}

Answer

$13^{\text{th}}$ term

Common Mistakes

Index Exponent Confusion: Confusing $(\sqrt{2})^{n-1}$ with $2^{n-1}$ by forgetting that $\sqrt{2} = 2^{1/2}$ , which leads to $2^{\frac{n-1}{2}}$ .
Skipping Division by $a$ : Forgetting to divide $128$ by the first term $a = 2$ before converting to powers of $2$ .

More questions in Exercise 8.3

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

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

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

Which term of the GP: $2, 6, 18, \dots$ is $4374$ ? Write the explicit formula as well as the recursive formula for the $n^{\text{th}}$ term.

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 $5^{\text{th}}$ bounce? (ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the $6^{\text{th}}$ time?

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

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 $n^{\text{th}}$ 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 $n^{\text{th}}$ stage. What happens to this area as $n$ , the number of stages, goes on increasing?