NCERT Solutions Class 9 Maths Exercise 8.3 — Predicting What Comes Next: Exploring Sequences and Progressions

NCERT Class 9 Mathematics — Ganita Manjari Part I, Predicting What Comes Next: Exploring Sequences and Progressions, Exercise Set 8.3. All 7 questions solved.

A geometric progression multiplies by a fixed common ratio $r$ at each step, where an AP adds a fixed amount. One formula does most of the work:

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

and its recursive twin, $t_1 = a$ with $t_{n+1} = r\,t_n$.

Key insight. Where an AP question ends in a linear equation, a GP question ends in an exponential one — $r^{\,n-1} = k$ — and you solve it by writing $k$ as a power of $r$ and matching the indices. $4^{\,n-1} = 65536$ is impossible to solve by inspection until you notice $65536 = 4^8$, at which point it is immediate.

Question 1

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

Solution. There is no need to find $a$ at all. Going from the $8$th term to the $12$th is four steps, each multiplying by $2$:

$$t_{12} = t_8 \times r^{4} = 192 \times 2^4 = 192 \times 16 = 3072$$

The longer route, for comparison. From $t_8 = a r^7 = 192$:

$$a \times 128 = 192 \ \Longrightarrow \ a = \frac{192}{128} = \frac{3}{2}$$

$$t_{12} = a r^{11} = \frac{3}{2} \times 2048 = 3072 \ \checkmark$$

Both give $3072$, but the first avoids the fraction entirely. In a GP, $t_q = t_p \times r^{\,q-p}$ is worth remembering in its own right.

Question 2

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

Solution.

$$a = 5, \qquad r = \frac{25}{5} = 5$$

(Check with the next pair: $\frac{125}{25} = 5$ ✓ — in a GP it is the ratios that must match, not the differences.)

The $n$th term.

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

That tidy form is worth noticing: this GP is simply the powers of $5$.

The $10$th term.

$$t_{10} = 5^{10} = 9\,765\,625$$

Question 3

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 $730$?

Solution. This rule is not a GP — the “$-2$” spoils it — so start by generating terms and looking for the structure:

$$t_1 = 2, \quad t_2 = 3(2) – 2 = 4, \quad t_3 = 3(4) – 2 = 10, \quad t_4 = 3(10) – 2 = 28$$ $$t_5 = 3(28) – 2 = 82, \quad t_6 = 3(82) – 2 = 244, \quad t_7 = 3(244) – 2 = 730$$

So $730$ is the $7$th term.

Why it works out so neatly. Subtract $1$ from every term: $1, 3, 9, 27, 81, 243, 729$ — the powers of $3$. That is not a coincidence. If $u_n = t_n – 1$ then

$$u_{n+1} = t_{n+1} – 1 = 3t_n – 3 = 3(t_n – 1) = 3u_n$$

so $u_n$ is a GP, with $u_1 = 1$ and ratio $3$, giving $u_n = 3^{\,n-1}$ and hence

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

Now the question is one equation: $3^{\,n-1} + 1 = 730$, so $3^{\,n-1} = 729 = 3^6$, and $n = 7$ ✓

Spotting that shift is the difference between listing seven terms and listing seventy, had the question asked about a larger value.

Question 4

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

Solution.

$$a = 2, \qquad r = \frac{6}{2} = 3$$

Explicit formula.

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

Recursive formula.

$$t_1 = 2, \qquad t_{n+1} = 3\,t_n \quad \text{for } n \ge 1$$

Which term is $4374$?

$$2 \times 3^{\,n-1} = 4374 \ \Longrightarrow \ 3^{\,n-1} = 2187$$

Now write $2187$ as a power of $3$: $3^5 = 243$, $3^6 = 729$, $3^7 = 2187$. So

$$n – 1 = 7 \ \Longrightarrow \ n = 8$$

Check: $2 \times 3^7 = 2 \times 2187 = 4374$ ✓

Question 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.

Solution. The rebound heights form a GP with ratio $r = 0.6 = \frac{3}{5}$, starting from the first rebound:

$$h_1 = 80 \times 0.6 = 48, \quad h_2 = 28.8, \quad h_3 = 17.28, \quad h_4 = 10.368, \quad h_5 = 6.2208$$

all in metres.

(i) What height does the ball reach after the $5$th bounce?

$$h_5 = 80 \times (0.6)^5 = 80 \times 0.07776 = 6.2208 \text{ m}$$

So about $6.22$ m — exactly $\frac{3888}{625}$ m.

Note the index. The height after the $n$th bounce is $80 \times (0.6)^n$, with the exponent equal to the bounce number, because the drop from $80$ m is not itself a bounce.

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

This part rewards careful counting rather than a formula. Trace the journey:

Journey	Distance
Drop from $80$ m → 1st landing	$80$
Up $48$, down $48$ → 2nd landing	$2 \times 48$
Up $28.8$, down $28.8$ → 3rd landing	$2 \times 28.8$
Up $17.28$, down $17.28$ → 4th landing	$2 \times 17.28$
Up $10.368$, down $10.368$ → 5th landing	$2 \times 10.368$
Up $6.2208$, down $6.2208$ → 6th landing	$2 \times 6.2208$

So the $6$th landing follows the initial drop and five complete up-and-down bounces:

$$D = 80 + 2\,(48 + 28.8 + 17.28 + 10.368 + 6.2208)$$

$$= 80 + 2 \times 110.6688 = 80 + 221.3376 = 301.3376 \text{ m}$$

About $301.34$ m — exactly $\frac{188336}{625}$ m.

The two traps here. The first is forgetting the doubling: every rebound is travelled twice, up and then down. The second is counting six bounces instead of five — the first landing costs no bounce at all, so reaching the $6$th landing takes only five.

Question 6

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

Solution. First confirm it is a GP by testing the ratios:

$$\frac{2\sqrt2}{2} = \sqrt2, \qquad \frac{4}{2\sqrt2} = \frac{4}{2\sqrt2} \times \frac{\sqrt2}{\sqrt2} = \frac{4\sqrt2}{4} = \sqrt2 \ \checkmark$$

So $a = 2$ and $r = \sqrt2$, giving

$$t_n = 2\,(\sqrt2)^{\,n-1}$$

Set this equal to $128$:

$$2(\sqrt2)^{\,n-1} = 128 \ \Longrightarrow \ (\sqrt2)^{\,n-1} = 64$$

The trick is to put both sides in the same base. Since $\sqrt2 = 2^{1/2}$ and $64 = 2^6$:

$$\left(2^{1/2}\right)^{n-1} = 2^{6} \ \Longrightarrow \ 2^{\frac{n-1}{2}} = 2^{6} \ \Longrightarrow \ \frac{n-1}{2} = 6 \ \Longrightarrow \ n = 13$$

Check: $t_{13} = 2(\sqrt2)^{12} = 2 \times 2^6 = 2 \times 64 = 128$ ✓

It is easy to under-count here. Because the ratio is only $\sqrt2$, each pair of steps doubles the term — so the sequence reaches $128$ in thirteen terms where a ratio of $2$ would have taken seven.

Question 7

Fig. 8.12 shows Stages $0$ to $3$ of the Sierpiński square carpet. Stage $0$ is a square sheet of paper. To construct Stage $1$, each side is trisected and the points of trisection of opposite sides are joined to obtain nine smaller squares; the centre square is 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.

Solution.

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

At each stage every red square is replaced by $8$ red squares one-third the size, because the middle one of its nine parts is thrown away.

• Stage $0$: the whole sheet — $1$ red square.
• Stage $1$: $1 \times 8 = 8$.
• Stage $2$: $8 \times 8 = 64$.
• Stage $3$: $64 \times 8 = 512$.

$$1, \ 8, \ 64, \ 512$$

(ii) Can you predict the number of red squares in Stages $4$ and $5$?

Keep multiplying by $8$:

$$\text{Stage } 4: \ 512 \times 8 = 4096, \qquad \text{Stage } 5: \ 4096 \times 8 = 32\,768$$

(iii) Can you find a rule for the number of red squares at the $n^{th}$ stage? Write the explicit formula as well as the recursive formula

The counts $1, 8, 64, 512, \dots$ form a GP with common ratio $8$, beginning at Stage $0$:

$$\textbf{Explicit: } \ R_n = 8^{\,n}$$

$$\textbf{Recursive: } \ R_0 = 1, \qquad R_{n+1} = 8\,R_n$$

(If you prefer to number the stages from $1$, the explicit rule reads $R_n = 8^{\,n-1}$ — the formula depends on where you start counting, which is why the starting index must always be stated.)

(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 it be in Stages $4$ and $5$? Find the explicit and recursive formulae, and say what happens as $n$ increases

Two things change together, and they pull in opposite directions. At each stage the number of red squares is multiplied by $8$, but each square’s side is divided by $3$, so its area is divided by $9$. The total area is therefore multiplied by

$$\frac{8}{9}$$

each time. Starting from $A_0 = 1$:

Stage	Red squares	Area of each	Total red area
$0$	$1$	$1$	$1$
$1$	$8$	$\frac19$	$\frac{8}{9} \approx 0.889$
$2$	$64$	$\frac{1}{81}$	$\frac{64}{81} \approx 0.790$
$3$	$512$	$\frac{1}{729}$	$\frac{512}{729} \approx 0.702$
$4$	$4096$	$\frac{1}{6561}$	$\frac{4096}{6561} \approx 0.624$
$5$	$32\,768$	$\frac{1}{59049}$	$\frac{32768}{59049} \approx 0.555$

$$\textbf{Explicit: } \ A_n = \left(\frac{8}{9}\right)^{\!n} \qquad \textbf{Recursive: } \ A_0 = 1, \ \ A_{n+1} = \frac{8}{9}\,A_n$$

What happens as $n$ increases. Since $\frac89 < 1$, each stage keeps only about $89\%$ of the previous area, and the powers $\left(\frac89\right)^n$ shrink towards zero. By Stage $20$ the red area is down to about $0.095$, and by Stage $50$ to about $0.003$.

So in the limit the carpet has zero area — the paper has been eaten away entirely — even though at every finite stage there are red squares everywhere you look, and their number is growing without bound. That combination of infinitely many pieces with vanishing total area is what makes this shape a fractal, and it is the most surprising thing in the chapter.

Common mistakes

• Testing differences instead of ratios. $2, 6, 18$ has differences $4$ and $12$, which do not match; its ratios are both $3$. A GP is defined by division, not subtraction.
• Question 1, using $r^{5}$ instead of $r^{4}$. From the $8$th term to the $12$th is four steps, not five and not twelve.
• Question 3, assuming the rule is a GP. $t_{n+1} = 3t_n – 2$ multiplies and subtracts. Treating it as $t_n = 2 \times 3^{n-1}$ gives $1458$ at $n = 7$, not $730$.
• Question 5(i), using $(0.6)^6$. The exponent matches the bounce number: after the $5$th bounce the height is $80(0.6)^5$.
• Question 5(ii), counting six bounces or forgetting to double. The $6$th landing needs the first drop plus five complete bounces, and every rebound is travelled twice. Missing either gives $80 + 2(110.6688 + 3.73) \approx 309$ m or $190.67$ m — both plausible-looking and both wrong.
• Question 6, answering $7$. With ratio $\sqrt2$ it takes two steps to double, so reaching $128$ takes thirteen terms. Always convert to a common base before matching exponents.
• Question 7(iv), multiplying the area by $\frac89$ the wrong way round. Each little square has $\frac19$ of the area and there are $8$ of them, so the factor is $\frac{8}{9}$ — less than $1$, so the area must fall. A factor of $\frac98$ would make the paper grow, which is impossible.

Practise next

• The End-of-Chapter Exercises, Questions 5, 7, 10, 11 and 13 — five more GP questions, including one where the common ratio turns out to be negative.
• Exercise Set 8.2 — revise the AP formulae alongside, since the End-of-Chapter set mixes the two without warning.