Predicting What Comes Next Class 9 Mathematics

20 Questions | 56 Marks

The first term of an arithmetic progression is

a = 5

, the common difference is

d = 3

, and the last term is

l = 104

. Find the number of terms and the sum of all the terms. 3marks

Consider the sequence where the

n^{t h}

term is given by

t_{n} = 4 n + 1

. Find the first five terms of this sequence. Is this sequence an arithmetic progression? Justify your answer. 3marks

A company manufactures motorcycles. The cost of production for the first 20 motorcycles is

50, 000 a n d t h e c o s t o f p r o d u c t i o n f o r t h e n e x t 20 m o t o r c y c l e s i s

60,000. The cost of production for the next 20 motorcycles is

70,000 and so on. (a) If the cost of production of motorcycles forms an arithmetic progression, what is the cost of production for the 100 motorcycles? [2 marks] (b) What is the total cost of production of 200 motorcycles? [3 marks] 4\. A geometric progression has its first term

a=2

a n d c o m m o n r a t i o

r=3

. Find the 5th term and the sum of the first 5 terms. [2 marks] 5\. The sum of the first three terms of a geometric progression is 39, and their product is 729. Find the terms of the GP and the common ratio. [3 marks] 6\. A ball is dropped from a height of 100 feet. After each bounce, it rises to

\frac{2}{3}

of its previous height. (a) Write down the heights reached by the ball after the 1st, 2nd, and 3rd bounces. [2 marks] (b) Is this sequence of heights an arithmetic progression or a geometric progression? Justify. [1 mark] (c) What is the height of the ball after the 6th bounce? [2 marks] 7\. The sum of the first

n a t u r a l n u m b e r s i s g i v e n b y t h e f o r m u l a

S_n = \frac{n(n+1)}{2}

. Using this formula, find the sum of the first 50 natural numbers. [1 mark] 8\. Find the sum of the first 10 terms of the arithmetic progression:

1, 4, 7, 10, \dots

[2 marks] 9\. In an AP, the sum of the first 10 terms is 205 and the sum of the first 20 terms is 810. Find the first term and the common difference. [3 marks] 10\. A student saves money from her pocket money. She saves ₹10 on the first day, ₹15 on the second day, ₹20 on the third day, and so on. (a) If she continues this pattern, what is the amount she will save on the 25th day? [2 marks] (b) What is the total amount she will have saved by the end of the 25th day? [2 marks] 11\. Consider the sequence defined by the recursive rule

u_1 = 3

a n d

u_n = 2u_{n-1} + 1

f o r

n \geq 2

. Find the first four terms of this sequence. [2 marks] 12\. The

4^{th}

t e r m o f a n A P i s 0 a n d t h e

11^{th}

term is -24. Find the first term and the common difference. [3 marks] 13\. The sum of the first three terms of a GP is 13, and the sum of the first six terms is 364. Find the first term and the common ratio. [3 marks] 14\. A fractal known as the Sierpiński carpet is constructed by repeatedly removing the central ninth of a square. If the initial square has an area of 1 square unit: (a) What is the area of the black region after the first step (Stage 1)? [1 mark] (b) What is the area of the black region after the second step (Stage 2)? [1 mark] (c) Write a formula for the area of the black region at the

n^{th}

stage. [1 mark] 15\. A sequence is given by

t_n = n^2 - 3n + 2

. F i n d t h e

1^{st}

,

5^{th}

, a n d

10^{th}

terms of this sequence. [2 marks] 16\. For the AP: 10, 7, 4, 1, ..., which term is -41? [2 marks] 17\. Find the sum of the first 30 terms of the AP: 8, 6, 4, 2, ... [2 marks] 18\. The first term of a GP is 27 and the common ratio is

\frac{1}{3}

. F i n d t h e

7^{th}$ term. 2marks

A sum of money is invested at compound interest. The amount after 1 year is ₹10,000 and the amount after 2 years is ₹12,000.(a) Show that the amounts at the end of each year form a geometric progression. [1 mark](b) Find the amount after 3 years. 2marks

Consider the sequence of prime numbers: 2, 3, 5, 7, 11, 13, ...(a) Is this sequence an arithmetic progression? Justify. [1 mark](b) Is this sequence a geometric progression? Justify. 1mark

Answers

The sequence is an AP with

a = 5

d = 3

l = 104

.We use the formula

l = a + (n - 1) d

104 = 5 + (n - 1) 3

99 = (n - 1) 3

33 = n - 1

$n = 34$ .
The number of terms is 34.
The sum $S_{n} = \frac{n}{2} (a + l)$ .
$S_{34} = \frac{34}{2} (5 + 104) = 17 (109) = 1853$ .
The sum of all terms is 1853.

The terms of the sequence are:For

n = 1

t_{1} = 4 (1) + 1 = 5

.For

n = 2

t_{2} = 4 (2) + 1 = 9

.For

n = 3

t_{3} = 4 (3) + 1 = 13

.For

n = 4

t_{4} = 4 (4) + 1 = 17

.For

n = 5

t_{5} = 4 (5) + 1 = 21

.The first five terms are 5, 9, 13, 17, 21.To check if it is an AP, we find the difference between consecutive terms:

t_{2} - t_{1} = 9 - 5 = 4

t_{3} - t_{2} = 13 - 9 = 4

t_{4} - t_{3} = 17 - 13 = 4

t_{5} - t_{4} = 21 - 17 = 4

.Since the difference between consecutive terms is constant (4), the sequence is an arithmetic progression.

The costs of production for successive batches of 20 motorcycles form an AP: 50000, 60000, 70000, ...The first term

a = 50000

.The common difference

d = 60000 - 50000 = 10000

.(a) We need to find the cost of production for the 100th motorcycle. This is equivalent to finding the 5th term of the sequence (since each term represents a batch of 20 motorcycles, the 5th batch corresponds to motorcycles from

4 \times 20 + 1 = 81

5 \times 20 = 100

).The

n^{t h}

term of an AP is given by

t_{n} = a + (n - 1) d

.For the 5th term (cost of the 5th batch of 20 motorcycles):

t_{5} = 50000 + (5 - 1) 10000 = 50000 + 4 \times 10000 = 50000 + 40000 = 90000

.The cost of production for the 100th motorcycle (i.e., the 5th batch) is ₹90,000.

(b) We need to find the total cost of production of 200 motorcycles. This means we need to find the sum of the first 10 terms of the AP (since $200 / 20 = 10$ ).
The sum of the first $n$ terms of an AP is $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .

S_{10} = \frac{10}{2} [2 (50000) + (10 - 1) 10000]

S_{10} = 5 [100000 + 9 \times 10000]

S_{10} = 5 [100000 + 90000]

$S_{10} = 5 [190000] = 950000$ .
The total cost of production of 200 motorcycles is ₹9,50,000.

The first term of the GP is

a = 2

.The common ratio is

r = 3

.The

n^{t h}

term of a GP is given by

t_{n} = a r^{n - 1}

.The 5th term is

t_{5} = 2 \times 3^{5 - 1} = 2 \times 3^{4} = 2 \times 81 = 162

.The sum of the first

n

terms of a GP is given by

S_{n} = \frac{a (r^{n} - 1)}{r - 1}

.The sum of the first 5 terms is

S_{5} = \frac{2 (3^{5} - 1)}{3 - 1} = \frac{2 (243 - 1)}{2} = 242

.The 5th term is 162 and the sum of the first 5 terms is 242.

Let the terms of the GP be

\frac{a}{r}, a, a r

.Sum of the first three terms:

\frac{a}{r} + a + a r = 39

a (\frac{1}{r} + 1 + r) = 39

(Equation 1)Product of the first three terms:

(\frac{a}{r}) \times a \times (a r) = 729

a^{3} = 729

a = \sqrt[3]{729} = 9

.Substitute

a = 9

into Equation 1:

9 (\frac{1}{r} + 1 + r) = 39

\frac{1}{r} + 1 + r = \frac{39}{9} = \frac{13}{3}

.Multiply by

3 r

to clear denominators:

3 + 3 r + 3 r^{2} = 13 r

3 r^{2} + 3 r - 13 r + 3 = 0

3 r^{2} - 10 r + 3 = 0

.Factor the quadratic equation:

(3 r - 1) (r - 3) = 0

.So,

r = \frac{1}{3}

r = 3

Case 1: $a = 9$ , $r = 3$ .
The terms are $\frac{9}{3}, 9, 9 \times 3$ , which are 3, 9, 27.

Case 2: $a = 9$ , $r = \frac{1}{3}$ .
The terms are $\frac{9}{1 / 3}, 9, 9 \times \frac{1}{3}$ , which are 27, 9, 3.

The terms of the GP are 3, 9, 27 or 27, 9, 3. The common ratio is 3 or $\frac{1}{3}$ .

The sequence of heights forms a GP.The initial height is 100 feet.The common ratio

r = \frac{2}{3}

.(a) Height after 1st bounce:

t_{1} = 100 \times \frac{2}{3} = \frac{200}{3}

feet.Height after 2nd bounce:

t_{2} = 100 \times (\frac{2}{3})^{2} = 100 \times \frac{4}{9} = \frac{400}{9}

feet.Height after 3rd bounce:

t_{3} = 100 \times (\frac{2}{3})^{3} = 100 \times \frac{8}{27} = \frac{800}{27}

feet.

(b) The sequence of heights is $\frac{200}{3}, \frac{400}{9}, \frac{800}{27}, \dots$ .
The ratio of consecutive terms is $\frac{400 / 9}{200 / 3} = \frac{400}{9} \times \frac{3}{200} = \frac{2}{3}$ .
The ratio of consecutive terms is $\frac{800 / 27}{400 / 9} = \frac{800}{27} \times \frac{9}{400} = \frac{2}{3}$ .
Since the ratio of consecutive terms is constant ( $\frac{2}{3}$ ), this sequence is a geometric progression. It is not an arithmetic progression because the difference between consecutive terms is not constant.

(c) The height after the $n^{t h}$ bounce is given by $t_{n} = a \times r^{n - 1}$ , where $a$ is the height after the first bounce.
However, if we consider the initial height as the 0th term ( $t_{0} = 100$ ), then the height after the $n^{t h}$ bounce is $t_{n} = 100 \times (\frac{2}{3})^{n}$ .
Height after the 6th bounce: $t_{6} = 100 \times (\frac{2}{3})^{6} = 100 \times \frac{64}{729} = \frac{6400}{729}$ feet.

Using the formula

S_{n} = \frac{n (n + 1)}{2}

for the sum of the first

n

natural numbers:For

n = 50

S_{50} = \frac{50 (50 + 1)}{2} = \frac{50 \times 51}{2} = 25 \times 51 = 1275

.The sum of the first 50 natural numbers is 1275.

The AP is

1, 4, 7, 10, \dots

.The first term

a = 1

.The common difference

d = 4 - 1 = 3

.We need to find the sum of the first 10 terms,

S_{10}

.Using the formula

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

S_{10} = \frac{10}{2} [2 (1) + (10 - 1) 3]

S_{10} = 5 [2 + 9 \times 3]

S_{10} = 5 [2 + 27]

$S_{10} = 5 [29] = 145$ .
The sum of the first 10 terms is 145.

Let the first term be

a

and the common difference be

d

.The sum of the first

n

terms of an AP is

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

.Given

S_{10} = 205

\frac{10}{2} [2 a + (10 - 1) d] = 205

5 [2 a + 9 d] = 205

$2 a + 9 d = 41$ (Equation 1)

Given $S_{20} = 810$ :

\frac{20}{2} [2 a + (20 - 1) d] = 810

10 [2 a + 19 d] = 810

$2 a + 19 d = 81$ (Equation 2)

Subtract Equation 1 from Equation 2:

(2 a + 19 d) - (2 a + 9 d) = 81 - 41

10 d = 40

$d = 4$ .

Substitute $d = 4$ into Equation 1:

2 a + 9 (4) = 41

2 a + 36 = 41

2 a = 5

$a = \frac{5}{2} = 2.5$ .

The first term is 2.5 and the common difference is 4.

This is an arithmetic progression.The amount saved on the first day is

a = 10

.The common difference

d = 15 - 10 = 5

.(a) We need to find the amount saved on the 25th day, which is the 25th term of the AP.

t_{n} = a + (n - 1) d

t_{25} = 10 + (25 - 1) 5 = 10 + 24 \times 5 = 10 + 120 = 130

.She will save ₹130 on the 25th day.

(b) We need to find the total amount saved by the end of the 25th day, which is the sum of the first 25 terms of the AP.
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .

S_{25} = \frac{25}{2} [2 (10) + (25 - 1) 5]

S_{25} = \frac{25}{2} [20 + 24 \times 5]

S_{25} = \frac{25}{2} [20 + 120]

$S_{25} = \frac{25}{2} [140] = 25 \times 70 = 1750$ .
The total amount saved by the end of the 25th day is ₹1750.

The recursive rule is

u_{1} = 3

and

u_{n} = 2 u_{n - 1} + 1

for

n \geq 2

.The first term is given:

u_{1} = 3

.For

n = 2

u_{2} = 2 u_{2 - 1} + 1 = 2 u_{1} + 1 = 2 (3) + 1 = 6 + 1 = 7

.For

n = 3

u_{3} = 2 u_{3 - 1} + 1 = 2 u_{2} + 1 = 2 (7) + 1 = 14 + 1 = 15

.For

n = 4

u_{4} = 2 u_{4 - 1} + 1 = 2 u_{3} + 1 = 2 (15) + 1 = 30 + 1 = 31

.The first four terms of the sequence are 3, 7, 15, 31.

Let the AP be

a, a + d, a + 2 d, \dots

.The

4^{t h}

term is

t_{4} = a + (4 - 1) d = a + 3 d

. We are given

t_{4} = 0

.So,

a + 3 d = 0

(Equation 1).The

11^{t h}

term is

t_{11} = a + (11 - 1) d = a + 10 d

. We are given

t_{11} = - 24

.So,

a + 10 d = - 24

(Equation 2).

Subtract Equation 1 from Equation 2:

(a + 10 d) - (a + 3 d) = - 24 - 0

7 d = - 24

$d = - \frac{24}{7}$ .

Substitute $d = - \frac{24}{7}$ into Equation 1:

a + 3 (- \frac{24}{7}) = 0

a - \frac{72}{7} = 0

$a = \frac{72}{7}$ .

The first term is $\frac{72}{7}$ and the common difference is $- \frac{24}{7}$ .

Let the terms of the GP be

a

a r

a r^{2}

.Sum of the first three terms:

a + a r + a r^{2} = 13

a (1 + r + r^{2}) = 13

(Equation 1).Sum of the first six terms:

a + a r + a r^{2} + a r^{3} + a r^{4} + a r^{5} = 364

a (1 + r + r^{2} + r^{3} + r^{4} + r^{5}) = 364

.We can write

1 + r + r^{2} + r^{3} + r^{4} + r^{5} = (1 + r + r^{2}) + r^{3} (1 + r + r^{2}) = (1 + r + r^{2}) (1 + r^{3})

.So,

a (1 + r + r^{2}) (1 + r^{3}) = 364

.Substitute

a (1 + r + r^{2}) = 13

into this equation:

13 (1 + r^{3}) = 364

1 + r^{3} = \frac{364}{13} = 28

r^{3} = 27

r = 3

Substitute $r = 3$ into Equation 1:
$a (1 + 3 + 3^{2}) = 13$ .
$a (1 + 3 + 9) = 13$ .
$a (13) = 13$ .
$a = 1$ .

The first term is 1 and the common ratio is 3.
The terms of the GP are: $a = 1$ , $a r = 1 \times 3 = 3$ , $a r^{2} = 1 \times 3^{2} = 9$ .
The terms are 1, 3, 9.

The construction process involves replacing a square with smaller squares, and this pattern repeats.(a) Stage 0: The initial square has area 1.Stage 1: The square is divided into 9 smaller equal squares, and the central one is removed. So, 8 smaller squares remain. The area of the black region is

8 \times \frac{1}{9} = \frac{8}{9}

square units.

(b) Stage 2: Each of the 8 remaining squares is divided into 9 smaller squares, and the central one is removed. So, for each of the 8 squares, $\frac{8}{9}$ of its area remains black.
The area of the black region at Stage 2 is $\frac{8}{9} \times (Area at Stage 1) = \frac{8}{9} \times \frac{8}{9} = (\frac{8}{9})^{2} = \frac{64}{81}$ square units.

(c) The area of the black region at Stage $n$ is given by $(\frac{8}{9})^{n}$ .
This is a geometric progression with first term $a = \frac{8}{9}$ (for $n = 1$ ) and common ratio $r = \frac{8}{9}$ .
The area at the $n^{t h}$ stage is $A_{n} = 1 \times (\frac{8}{9})^{n} = (\frac{8}{9})^{n}$ .

The formula for the

n^{t h}

term is

t_{n} = n^{2} - 3 n + 2

.For

n = 1

t_{1} = 1^{2} - 3 (1) + 2 = 1 - 3 + 2 = 0

.For

n = 5

t_{5} = 5^{2} - 3 (5) + 2 = 25 - 15 + 2 = 12

.For

n = 10

t_{10} = 10^{2} - 3 (10) + 2 = 100 - 30 + 2 = 72

.The 1st term is 0, the 5th term is 12, and the 10th term is 72.

The AP is

10, 7, 4, 1, \dots

.The first term

a = 10

.The common difference

d = 7 - 10 = - 3

.We need to find which term is -41. Let

t_{n} = - 41

t_{n} = a + (n - 1) d

- 41 = 10 + (n - 1) (- 3)

- 41 - 10 = (n - 1) (- 3)

- 51 = - 3 (n - 1)

\frac{- 51}{- 3} = n - 1

17 = n - 1

n = 18

.The 18th term is -41.

The AP is

8, 6, 4, 2, \dots

.The first term

a = 8

.The common difference

d = 6 - 8 = - 2

.We need to find the sum of the first 30 terms,

S_{30}

S_{n} = \frac{n}{2} [2 a + (n - 1) d]

S_{30} = \frac{30}{2} [2 (8) + (30 - 1) (- 2)]

S_{30} = 15 [16 + 29 (- 2)]

S_{30} = 15 [16 - 58]

$S_{30} = 15 [- 42] = - 630$ .
The sum of the first 30 terms is -630.

The GP has first term

a = 27

.The common ratio

r = \frac{1}{3}

.We need to find the

7^{t h}

term,

t_{7}

t_{n} = a r^{n - 1}

t_{7} = 27 \times (\frac{1}{3})^{7 - 1} = 27 \times (\frac{1}{3})^{6}

t_{7} = 3^{3} \times \frac{1}{3^{6}} = \frac{1}{3^{3}} = \frac{1}{27}

.The

7^{t h}

term is

\frac{1}{27}

Let

A_{n}

be the amount after

n

years.

A_{1} = 10000

A_{2} = 12000

.(a) To check if it's a GP, we find the ratio of consecutive terms.

\frac{A_{2}}{A_{1}} = \frac{12000}{10000} = \frac{12}{10} = \frac{6}{5} = 1.2

.If the amount after 3 years is

A_{3}

, and the sequence is a GP, then

\frac{A_{3}}{A_{2}}

should also be 1.2.This means the investment grows by a constant factor each year, which is the characteristic of compound interest. Therefore, the amounts at the end of each year form a geometric progression with common ratio

r = 1.2

(b) To find the amount after 3 years, we use the common ratio.
$A_{3} = A_{2} \times r = 12000 \times 1.2 = 12000 \times \frac{6}{5} = 2400 \times 6 = 14400$ .
The amount after 3 years is ₹14,400.

The sequence of prime numbers is 2, 3, 5, 7, 11, 13, ...

(a) To check if it's an arithmetic progression, we find the difference between consecutive terms:
$3 - 2 = 1$ .
$5 - 3 = 2$ .
$7 - 5 = 2$ .
$11 - 7 = 4$ .
The difference between consecutive terms is not constant. Therefore, the sequence of prime numbers is not an arithmetic progression.

(b) To check if it's a geometric progression, we find the ratio of consecutive terms:
$\frac{3}{2} = 1.5$ .
$\frac{5}{3} \approx 1.67$ .
$\frac{7}{5} = 1.4$ .
The ratio of consecutive terms is not constant. Therefore, the sequence of prime numbers is not a geometric progression.

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