Find the term of a GP with common ratio 2, whose term is 192.

We need to find the first term and then the 12th term of the GP. Step 1 — Find the first term Let's write the formula for the term of a GP. The formula is . We are given the common ratio . We know the term is . So, we can write . > Step 2 — Calculate the 12th term Now we know the first term . We also know the common ratio . Let's find the term using the same formula . For the term, will be . > Answer > (i) The 12th term of the GP is 3072.

Predicting What Comes Next: Sequences and Progressions | Exercise 8.3

Question 1

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

Solution

We need to find the first term and then the 12th term of the GP.

Step 1 — Find the first term

Let's write the formula for the $n^{\text{th}}$ term of a GP. The formula is $t_n = ar^{n-1}$ . We are given the common ratio $\mathbf{r = 2}$ . We know the $8^{\text{th}}$ term is $\mathbf{192}$ . So, we can write $t_8 = ar^{8-1}$ .

$t_8 = ar^7$

$192 = a(2)^7$

$192 = a \times 128$

$a = \frac{192}{128}$

$a = \frac{3}{2}$

$\boxed{a = \frac{3}{2}}$

Step 2 — Calculate the 12th term

Now we know the first term $\mathbf{a = \frac{3}{2}}$ . We also know the common ratio $\mathbf{r = 2}$ . Let's find the $12^{\text{th}}$ term using the same formula $t_n = ar^{n-1}$ . For the $12^{\text{th}}$ term, $n$ will be $\mathbf{12}$ .

$t_{12} = ar^{12-1}$

$t_{12} = ar^{11}$

$t_{12} = \frac{3}{2} \times (2)^{11}$

$t_{12} = 3 \times \frac{2^{11}}{2^1}$

$t_{12} = 3 \times 2^{10}$

$t_{12} = 3 \times 1024$

$t_{12} = 3072$

$\boxed{t_{12} = 3072}$

Answer

(i) The 12th term of the GP is 3072.

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?