Predicting What Comes Next Class 9 — Competency Based Questions

20 Questions | 50 Marks

A farmer is planting a new orchard. He plants 5 apple trees in the first row. He decides to plant 3 more apple trees in each subsequent row.(a) Write down the number of apple trees in the first 4 rows as a sequence. [1 mark](b) Write down the explicit formula for the nth row. [1 mark](c) If the farmer plants 15 rows, how many apple trees will be in the 15th row? [1 mark](d) If the farmer wants to plant a total of 180 apple trees, how many rows will he need to plant? 2marks

Consider a sequence where the first term is 7 and each subsequent term is found by subtracting 4 from the previous term.Assertion (A): This sequence represents an Arithmetic Progression.Reason (R): In this sequence, the difference between consecutive terms is constant. [1 mark](a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is true.

Anjali is saving money for a new bicycle. She saves ₹100 in the first week and decides to increase her savings by ₹20 each week.(a) Write down the amount of money Anjali saves in the first 5 weeks. [1 mark](b) Find the explicit formula for the amount of money saved in the nth week. [1 mark](c) How much money will Anjali save in the 12th week? [1 mark](d) If the bicycle costs ₹2500, how many weeks will it take Anjali to save enough money? 2marks

A scientist is observing the growth of a bacterial colony. The colony starts with 50 bacteria. The number of bacteria triples every hour.(a) Write down the number of bacteria after 0, 1, 2, and 3 hours. [1 mark](b) Write down the explicit formula for the number of bacteria after n hours. [1 mark](c) How many bacteria will be present after 6 hours? [1 mark](d) If the culture dish can hold a maximum of 40,500 bacteria, how many hours will it take for the colony to reach its maximum capacity? 2marks

A runner is training for a marathon. On the first day, she runs 5 km. She plans to increase her daily running distance by 1.5 km each day.Assertion (A): The daily distances run by the athlete form a geometric progression.Reason (R): The distance increases by a constant factor each day. [1 mark](a) Both A and R are true and R is correct explanation of A.(b) Both A and R are true but R is not correct explanation of A.(c) A is true but R is false.(d) A is false but R is true.

A father is teaching his son about sequences. He gives him a sequence defined by the recursive formula:

t_{1} = 4

$t_{n} = 2 t_{n - 1} + 1$ for $n \geq 2$
(a) Find the first 4 terms of this sequence. [2 marks]
(b) Determine if this sequence is an arithmetic progression or a geometric progression. Justify your answer. [2 marks]

Consider the sequence: 1, 1/2, 1/4, 1/8, ...(a) Is this sequence an arithmetic progression or a geometric progression? [1 mark](b) If it is a GP, what is its first term and common ratio? [1 mark](c) Write the explicit formula for the nth term of this sequence. [1 mark](d) Find the 10th term of this sequence. [1 mark](e) Which term of this sequence is

1 / 1024

? 1mark

The sum of the first n natural numbers is given by the formula

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

.(a) Calculate the sum of the first 20 natural numbers. [1 mark](b) If the sum of the first n natural numbers is 210, find the value of n. [2 marks](c) Explain with a small example how this formula is derived using the method of pairing terms. 2marks

A company manufactures bulbs. The production of bulbs increases by a constant amount each month. In January, they produced 5000 bulbs. In February, they produced 5500 bulbs, and in March, 6000 bulbs.(a) Write the sequence representing the production of bulbs for the first three months. [1 mark](b) Identify the type of sequence and state its common difference. [1 mark](c) Write the explicit formula for the number of bulbs produced in the nth month. [1 mark](d) How many bulbs will be produced in the 10th month? [1 mark](e) In which month will the production reach 10,000 bulbs? 2marks

A savings scheme offers an initial deposit of ₹5000. For every subsequent year, the deposit increases by ₹1000.Assertion (A): The annual deposits form an arithmetic progression.Reason (R): The difference between consecutive deposits is constant. [1 mark](a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is true.

A pattern of dots is arranged in a specific way. The first row has 2 dots, the second row has 4 dots, the third row has 8 dots, and so on.(a) Write the number of dots in the first 5 rows. [1 mark](b) Identify the type of sequence and find its common ratio. [1 mark](c) Write the explicit formula for the number of dots in the nth row. [1 mark](d) How many dots will be in the 8th row? 1mark

A quadratic sequence is given by the formula

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

.(a) Find the first 4 terms of this sequence. [2 marks](b) Calculate the differences between consecutive terms. Are they constant? [1 mark](c) Calculate the differences between the consecutive differences (second differences). What do you observe? 2marks

A school is organizing a fundraising event. The number of participants is expected to increase by 15% each week. If 100 participants are registered in the first week.(a) Write down the number of participants in the first 3 weeks. [1 mark](b) Is this an arithmetic progression or a geometric progression? Justify your answer. [1 mark](c) Write the explicit formula for the number of participants in the nth week. [1 mark](d) How many participants are expected in the 5th week? 1mark

A staircase has steps such that the number of steps in each successive flight forms an arithmetic progression. The first flight has 10 steps, and the second flight has 15 steps.(a) Write the sequence for the first 3 flights. [1 mark](b) Find the common difference of this AP. [1 mark](c) Write the explicit formula for the number of steps in the nth flight. [1 mark](d) How many steps will be in the 8th flight? 1mark

A pattern of squares is created such that each subsequent square has half the side length of the previous square. If the first square has a side length of 16 cm.(a) Write the sequence of side lengths for the first 4 squares. [1 mark](b) Identify the type of sequence and state its common ratio. [1 mark](c) Write the explicit formula for the side length of the nth square. [1 mark](d) What is the side length of the 7th square? 1mark

The Sierpiński triangle is formed by repeatedly removing the central triangle from a larger triangle. The number of black triangles at each stage forms a GP.Assertion (A): The number of black triangles in the Sierpiński triangle follows a GP.Reason (R): Each black triangle in a stage is replaced by 3 smaller black triangles in the next stage. [1 mark](a) Both A and R are true and R is the correct explanation of A.(b) Both A and R are true but R is not the correct explanation of A.(c) A is true but R is false.(d) A is false but R is true.

A person saves ₹50 in the first month, ₹100 in the second month, ₹150 in the third month, and so on.(a) Write the sequence of savings for the first 5 months. [1 mark](b) Find the explicit formula for the amount saved in the nth month. [1 mark](c) Calculate the total savings after 12 months. 2marks

In a GP, the first term is 2 and the common ratio is 3.(a) Write down the first 4 terms of the GP. [1 mark](b) Write the explicit formula for the nth term. [1 mark](c) Find the 6th term of the GP. [1 mark](d) If a term in this GP is 4374, find its position in the sequence. 2marks

A student observes a pattern of dots. The first pattern has 3 dots. The second pattern has 7 dots. The third pattern has 11 dots.(a) Write down the sequence of dots for the first 4 patterns. [1 mark](b) What is the rule for finding the number of dots in the nth pattern? [1 mark](c) How many dots will be in the 15th pattern? 1mark

A radioactive substance decays such that half of the amount remains after every year. If the initial amount is 640 grams.(a) Write down the amount of substance remaining after 1, 2, and 3 years. [1 mark](b) Identify the type of sequence and state its common ratio. [1 mark](c) Write the explicit formula for the amount of substance remaining after n years. [1 mark](d) How much substance will remain after 5 years? 1mark

Answers

(a) 5, 8, 11, 14(b)

t_{n} = 5 + (n - 1) 3

t_{n} = 3 n + 2

(c)

t_{15} = 3 (15) + 2 = 45 + 2 = 47

apple trees(d)

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

where

a = 5

and

d = 3

180 = \frac{n}{2} (2 (5) + (n - 1) 3)

360 = n (10 + 3 n - 3)

360 = n (3 n + 7)

3 n^{2} + 7 n - 360 = 0

Using the quadratic formula: $n = \frac{- 7 \pm \sqrt{7^{2} - 4 (3) (- 360)}}{2 (3)} = \frac{- 7 \pm \sqrt{49 + 4320}}{6} = \frac{- 7 \pm \sqrt{4369}}{6} = \frac{- 7 \pm 66.1}{6}$
Taking the positive value: $n = \frac{59.1}{6} \approx 9.85$ . Since the number of rows must be a whole number, and the farmer wants to plant at least 180 trees, he needs to plant 10 rows.
$S_{10} = \frac{10}{2} (2 (5) + (10 - 1) 3) = 5 (10 + 9 \times 3) = 5 (10 + 27) = 5 (37) = 185$ . So, 10 rows.

(a) Both A and R are true and R is the correct explanation of A.

(a) 100, 120, 140, 160, 180(b)

t_{n} = 100 + (n - 1) 20

t_{n} = 20 n + 80

(c)

t_{12} = 20 (12) + 80 = 240 + 80 = 320

(d)

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

where

a = 100

and

d = 20

2500 = \frac{n}{2} (2 (100) + (n - 1) 20)

5000 = n (200 + 20 n - 20)

5000 = n (180 + 20 n)

20 n^{2} + 180 n - 5000 = 0

n^{2} + 9 n - 250 = 0

Using the quadratic formula: $n = \frac{- 9 \pm \sqrt{9^{2} - 4 (1) (- 250)}}{2 (1)} = \frac{- 9 \pm \sqrt{81 + 1000}}{2} = \frac{- 9 \pm \sqrt{1081}}{2} = \frac{- 9 \pm 32.88}{2}$
Taking the positive value: $n = \frac{23.88}{2} \approx 11.94$ . So, it will take 12 weeks.

(a) 50, 150, 450, 1350(b)

t_{n} = 50 \times 3^{n}

(where n is the number of hours elapsed, starting with n=0 for the initial amount)(c)

t_{6} = 50 \times 3^{6} = 50 \times 729 = 36450

bacteria(d)

40500 = 50 \times 3^{n}

810 = 3^{n}

$3^{6} = 729$ , $3^{7} = 2187$ .
Wait, let's recheck the calculation. $40500 / 50 = 810$ .
$3^{n} = 810$ . $3^{1} = 3, 3^{2} = 9, 3^{3} = 27, 3^{4} = 81, 3^{5} = 243, 3^{6} = 729$ .
The number of bacteria will slightly exceed 40,500 after 6 hours ( $50 \times 3^{6} = 36450$ ). It will reach the capacity at some point during the 7th hour. To be precise, we need to solve $3^{n} = 810$ . $n = \log_{3} (810) \approx 6.14$ hours. If the question asks "how many hours will it take", it implies a continuous growth or the point it reaches capacity. If it implies at the end of which hour it reaches capacity, then it would be 7 hours. Given the context of discrete hourly checks, it would be after 7 hours that the capacity is exceeded. However, if we consider the continuous growth model, it's 6.14 hours. Let's assume it's about the end of the hour.
After 6 hours: 36450 bacteria.
After 7 hours: $36450 \times 3 = 109350$ bacteria.
So, it reaches capacity during the 7th hour. If the question means "at the end of which hour will it exceed capacity", then it is 7 hours. Let's assume it asks for the smallest integer number of hours after which the capacity is reached or exceeded.
It reaches capacity at approximately 6.14 hours. So, after 7 hours, the capacity will have been reached and exceeded. The question is a bit ambiguous here. Let's interpret it as "when does it first exceed or reach capacity". So, 7 hours.

(d) A is false but R is true. (The increase is by a constant addition, not multiplication, so it's AP, not GP).

(a)

t_{1} = 4

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

t_{3} = 2 (9) + 1 = 19

t_{4} = 2 (19) + 1 = 39

The first 4 terms are: 4, 9, 19, 39.
(b) This is neither an AP nor a GP.
For AP, the difference between consecutive terms should be constant: $9 - 4 = 5$ , $19 - 9 = 10$ , $39 - 19 = 20$ . The differences are not constant.
For GP, the ratio of consecutive terms should be constant: $9 / 4 = 2.25$ , $19 / 9 \approx 2.11$ , $39 / 19 \approx 2.05$ . The ratios are not constant.

(a) Geometric Progression(b) First term

a = 1

, common ratio

r = \frac{1 / 2}{1} = \frac{1}{2}

(c)

t_{n} = a \times r^{n - 1} = 1 \times (\frac{1}{2})^{n - 1} = (\frac{1}{2})^{n - 1}

(d)

t_{10} = (\frac{1}{2})^{10 - 1} = (\frac{1}{2})^{9} = \frac{1}{512}

(e)

t_{n} = (\frac{1}{2})^{n - 1} = \frac{1}{1024}

2^{n - 1} = 1024

2^{n - 1} = 2^{10}

$n - 1 = 10 \Rightarrow n = 11$ . The 11th term.

(a)

S_{20} = \frac{20 (20 + 1)}{2} = \frac{20 \times 21}{2} = 10 \times 21 = 210

.(b)

S_{n} = 210

\frac{n (n + 1)}{2} = 210

n (n + 1) = 420

n^{2} + n - 420 = 0

(n + 21) (n - 20) = 0

Since n must be positive, $n = 20$ .
(c) Let $S = 1 + 2 + 3 + . . . + n$ .
Write the sum in reverse: $S = n + (n - 1) + (n - 2) + . . . + 1$ .
Add the two equations:

2 S = (1 + n) + (2 + (n - 1)) + (3 + (n - 2)) + . . . + (n + 1)

$2 S = (n + 1) + (n + 1) + (n + 1) + . . . + (n + 1)$ (n times)

2 S = n (n + 1)

$S = \frac{n (n + 1)}{2}$ .
For example, $S = 1 + 2 + 3 + 4 + 5$ . $S = 5 + 4 + 3 + 2 + 1$ . $2 S = 6 + 6 + 6 + 6 + 6 = 5 \times 6 = 30$ . $S = 15$ . Formula: $S_{5} = \frac{5 (5 + 1)}{2} = \frac{5 \times 6}{2} = 15$ .

(a) 5000, 5500, 6000(b) Arithmetic Progression (AP), common difference

d = 500

.(c)

t_{n} = a + (n - 1) d = 5000 + (n - 1) 500

t_{n} = 500 n + 4500

.(d)

t_{10} = 500 (10) + 4500 = 5000 + 4500 = 9500

bulbs.(e)

10000 = 500 n + 4500

5500 = 500 n

$n = \frac{5500}{500} = 11$ . The 11th month (which is November).

(a) Both A and R are true and R is the correct explanation of A.

(a) 2, 4, 8, 16(b) Geometric Progression (GP), common ratio

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

.(c)

t_{n} = a \times r^{n - 1} = 2 \times 2^{n - 1} = 2^{n}

.(d)

t_{8} = 2^{8} = 256

dots.

(a)

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

t_{2} = 2^{2} + 2 (2) = 4 + 4 = 8

t_{3} = 3^{2} + 2 (3) = 9 + 6 = 15

t_{4} = 4^{2} + 2 (4) = 16 + 8 = 24

The first 4 terms are: 3, 8, 15, 24.
(b) First differences: $8 - 3 = 5$ , $15 - 8 = 7$ , $24 - 15 = 9$ . No, the differences are not constant.
(c) Second differences: $7 - 5 = 2$ , $9 - 7 = 2$ . The second differences are constant and equal to 2.

(a) Week 1: 100Week 2:

100 \times 1.15 = 115

Week 3:

115 \times 1.15 = 132.25

(or 132 participants, assuming whole numbers)(b) Geometric Progression. The number of participants is multiplied by a constant factor (1.15) each week.(c)

t_{n} = 100 \times (1.15)^{n - 1}

(d)

t_{5} = 100 \times (1.15)^{5 - 1} = 100 \times (1.15)^{4} = 100 \times 1.74900625 \approx 174.9

or 175 participants.

(a) 10, 15, 20(b) Common difference

d = 15 - 10 = 5

.(c)

t_{n} = a + (n - 1) d = 10 + (n - 1) 5

t_{n} = 5 n + 5

.(d)

t_{8} = 5 (8) + 5 = 40 + 5 = 45

steps.

(a) 16, 8, 4, 2(b) Geometric Progression (GP), common ratio

r = \frac{8}{16} = \frac{1}{2}

.(c)

t_{n} = a \times r^{n - 1} = 16 \times (\frac{1}{2})^{n - 1}

.(d)

t_{7} = 16 \times (\frac{1}{2})^{7 - 1} = 16 \times (\frac{1}{2})^{6} = 16 \times \frac{1}{64} = \frac{16}{64} = \frac{1}{4}

cm.

(a) Both A and R are true and R is the correct explanation of A.

(a) 50, 100, 150, 200, 250(b)

t_{n} = 50 + (n - 1) 50

t_{n} = 50 n

.(c)

S_{12} = \frac{12}{2} (2 a + (12 - 1) d)

where

a = 50, d = 50

S_{12} = 6 (2 (50) + 11 (50))

S_{12} = 6 (100 + 550)

$S_{12} = 6 (650) = 3900$ . Total savings = ₹3900.

(a)

a = 2, r = 3

. Terms: 2,

2 \times 3 = 6

6 \times 3 = 18

18 \times 3 = 54

. So, 2, 6, 18, 54.(b)

t_{n} = a \times r^{n - 1} = 2 \times 3^{n - 1}

.(c)

t_{6} = 2 \times 3^{6 - 1} = 2 \times 3^{5} = 2 \times 243 = 486

.(d)

t_{n} = 4374

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

3^{n - 1} = \frac{4374}{2} = 2187

We know $3^{1} = 3, 3^{2} = 9, 3^{3} = 27, 3^{4} = 81, 3^{5} = 243, 3^{6} = 729, 3^{7} = 2187$ .
So, $n - 1 = 7 \Rightarrow n = 8$ . The 8th term.

(a) 3, 7, 11, 15(b) This is an AP with first term

a = 3

and common difference

d = 4

.The rule is

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

t_{n} = 4 n - 1

.(c)

t_{15} = 4 (15) - 1 = 60 - 1 = 59

dots.

(a) Initial amount: 640gAfter 1 year:

640 \times \frac{1}{2} = 320

gAfter 2 years:

320 \times \frac{1}{2} = 160

gAfter 3 years:

160 \times \frac{1}{2} = 80

g(b) Geometric Progression (GP), common ratio

r = \frac{1}{2}

.(c)

t_{n} = a \times r^{n - 1} = 640 \times (\frac{1}{2})^{n - 1}

(where n is the year number, starting with n=1 for after 1 year). Alternatively, if

n

represents the number of years elapsed:

t_{n} = 640 \times (\frac{1}{2})^{n}

. Let's use the latter for simplicity.(d)

t_{5} = 640 \times (\frac{1}{2})^{5} = 640 \times \frac{1}{32} = \frac{640}{32} = 20

grams.

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Predicting What Comes Next: Exploring Sequences and Progressions — Competency Based Questions

Competency based questions

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