NCERT Solutions for Class 9 Maths Chapter 8 — Exercise Set 8.3#

For a geometric progression (GP) with first term $a$ and common ratio $r$:

$$t_n = a\,r^{\,n-1}$$

$$\text{Recursive rule: } t_1 = a, \quad t_{n+1} = r \times t_n$$

1. Find the 12th term of a GP with common ratio 2, whose 8th term is 192. #

Solution

From the 8th term to the 12th term we multiply by $r$ four times:

$$t_{12} = t_8 \times r^4$$

$$t_{12} = 192 \times 2^4$$

$$t_{12} = 192 \times 16$$

$$t_{12} = 3072$$

∴ The 12th term is 3072.

2. Find the 10th and nth terms of the GP: 5, 25, 125, … . #

Solution

Here $a = 5$ and $r = \dfrac{25}{5} = 5$. We find the $n^{th}$ term:

$$t_n = 5 \times 5^{\,n-1}$$

$$t_n = 5^n$$

So:

$$t_{10} = 5^{10} = 9765625$$

∴ $t_{10} = $ 9765625 and $t_n = $ $5^n$.

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

Solution

We generate terms until we reach 730:

$$t_1 = 2$$

$$t_2 = 3(2) - 2 = 4$$

$$t_3 = 3(4) - 2 = 10$$

$$t_4 = 3(10) - 2 = 28$$

$$t_5 = 3(28) - 2 = 82$$

$$t_6 = 3(82) - 2 = 244$$

$$t_7 = 3(244) - 2 = 730$$

∴ 730 is the 7th term.

(Each term is 1 more than a power of 3: $t_n = 3^{\,n-1} + 1$. Then $3^{\,n-1} + 1 = 730$ gives $3^{\,n-1} = 729 = 3^6$, so $n = 7$.)

4. Which term of the GP: 2, 6, 18, … is 4374? Write the explicit formula as well as the recursive formula for the nth term. #

Solution

Here $a = 2$ and $r = 3$:

$$t_n = 2 \times 3^{\,n-1}$$

We set it equal to 4374:

$$2 \times 3^{\,n-1} = 4374$$

$$\Rightarrow 3^{\,n-1} = 2187$$

$$\Rightarrow 3^{\,n-1} = 3^7$$

$$\Rightarrow n = 8$$

∴ 4374 is the 8th term.

  • Explicit formula: $t_n = 2 \times 3^{\,n-1}$
  • Recursive formula: $t_1 = 2$, $t_{n+1} = 3t_n$ for $n \geq 1$

5. 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 5th bounce?

The heights after each bounce form a GP with first term $80 \times 0.6 = 48$ and ratio $0.6$:

$$h_5 = 80 \times 0.6^5$$

$$h_5 = 80 \times 0.07776$$

$$h_5 = 6.2208$$

∴ After the 5th bounce, the ball rises to 6.2208 m (about 6.22 m).

(ii) What is the total vertical distance the ball has travelled by the time it hits the ground for the 6th time?

The ball first falls 80 m (1st hit). Then it goes up and comes down again for each of the bounces 1 to 5, hitting the ground for the 2nd to 6th times. The bounce heights are:

$$48, \quad 28.8, \quad 17.28, \quad 10.368, \quad 6.2208$$

Their sum is:

$$48 + 28.8 + 17.28 + 10.368 + 6.2208 = 110.6688$$

Each bounce height is travelled twice (up and down):

$$\text{Total} = 80 + 2 \times 110.6688$$

$$\text{Total} = 80 + 221.3376$$

$$\text{Total} = 301.3376$$

∴ The ball travels 301.3376 m (about 301.34 m).

6. Which term of the sequence $2, 2\sqrt{2}, 4, \ldots$ is 128? #

Solution

Here $a = 2$ and $r = \dfrac{2\sqrt{2}}{2} = \sqrt{2}$:

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

We set it equal to 128:

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

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

$64 = 2^6 = (\sqrt{2})^{12}$, so:

$$(\sqrt{2})^{\,n-1} = (\sqrt{2})^{12}$$

$$\Rightarrow n - 1 = 12$$

$$\Rightarrow n = 13$$

∴ 128 is the 13th term.

7. 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. #

(i) How many red squares are there in Stages 0 to 3?

At each stage, every red square is replaced by 8 smaller red squares:

Stage 0 1 2 3
Red squares 1 8 64 512
(ii) Can you predict the number of red squares in Stages 4 and 5?
$$R_4 = 512 \times 8 = 4096$$

$$R_5 = 4096 \times 8 = 32768$$
(iii) Can you find a rule for the number of red squares at the nth stage? Write the explicit formula as well as the recursive formula for the number of red squares at any stage.

The numbers 1, 8, 64, 512, … form a GP with first term 1 and ratio 8.

  • Explicit formula: $R_n = 8^n$
  • Recursive formula: $R_0 = 1$, $R_{n+1} = 8R_n$
(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 nth stage. What happens to this area as n, the number of stages, goes on increasing?

Each step keeps 8 of the 9 equal pieces, so the red area is multiplied by $\dfrac{8}{9}$:

Stage 1 2 3 4 5
Red area $\dfrac{8}{9}$ $\dfrac{64}{81}$ $\dfrac{512}{729}$ $\dfrac{4096}{6561}$ $\dfrac{32768}{59049}$
  • Explicit formula: $A_n = \left(\dfrac{8}{9}\right)^n$
  • Recursive formula: $A_0 = 1$, $A_{n+1} = \dfrac{8}{9}A_n$

As $n$ increases, $\left(\dfrac{8}{9}\right)^n$ keeps getting smaller, and gets closer and closer to 0 (for example $A_{20} \approx 0.095$ and $A_{50} \approx 0.003$). The number of red squares grows without limit, but their total area shrinks towards 0.

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