Which term of the sequence is 128?

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

We need to find the position of 128 in the given sequence.

Step 1 — Find the first term and common ratio

Let's look at the given sequence. The sequence is $2, 2\sqrt{2}, 4, \dots$ . The first term, $a$ , is 2. We can find the common ratio, $r$ . Divide the second term by the first term.

$r = \frac{2\sqrt{2}}{2}$

$r = \sqrt{2}$

Let's check with the next terms. Divide the third term by the second term.

$r = \frac{4}{2\sqrt{2}}$

$r = \frac{2}{\sqrt{2}}$

$r = \frac{2 \times \sqrt{2}}{\sqrt{2} \times \sqrt{2}}$

$r = \frac{2\sqrt{2}}{2}$

$\boxed{r = \sqrt{2}}$

Since the ratio is constant, this is a geometric progression.

Step 2 — Find the term number

Let the $n$ -th term, $t_n$ , be 128. The formula for the $n$ -th term of a GP is $t_n = ar^{n-1}$ . We know $a = \mathbf{2}$ and $r = \sqrt{\mathbf{2}}$ . Let's substitute these values into the formula.

$128 = 2(\sqrt{2})^{n-1}$

Divide both sides by 2.

$\frac{128}{2} = (\sqrt{2})^{n-1}$

$64 = (\sqrt{2})^{n-1}$

We can write $\sqrt{2}$ as $2^{1/2}$ . We can write 64 as $2^6$ .

$2^6 = (2^{1/2})^{n-1}$

$2^6 = 2^{\frac{n-1}{2}}$

Now, compare the powers of 2 on both sides.

$6 = \frac{n-1}{2}$

Multiply both sides by 2.

$12 = n-1$

Add 1 to both sides.

$n = 12 + 1$

$\boxed{n = 13}$

So, 128 is the 13th term of the sequence.

Answer

128 is the 13th term of the sequence.

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, ... .

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, ... 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?