Predicting What Comes Next Exploring Sequences and Progressions Class 9 Extra Questions Maths Chapter 8

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Class 9 Predicting What Comes Next Exploring Sequences and Progressions Extra Questionsi

Extra Questions of Predicting What Comes Next Exploring Sequences and Progressions

Class 9 Maths Chapter 8 Extra Questions – Predicting What Comes Next Exploring Sequences and Progressions Extra Questions Class 9

Predicting What Comes Next Exploring Sequences and Progressions Class 9 Short Question Answer

Question 1.
Which term of the A.P. 4, 9, 14, is 109?
Solution:
Let 109 is the nth term,
∴ Using T_n = a + (n – 1) d, we have:
109 = 4 + (n – 1)5 [∵ a = 4 and d = 9 – 4 = 5]
⇒ n = 21 + 1 = 22
Thus, the 22nd term is 109.

Question 2.
If a, (a – 2) and 3 a are in A.P. then what is the value of a?
Solution:
∵ a, (a – 2) and 3a are in A.P.
∴ (a – 2) – a = 3a – (a – 2)
⇒ a – 2 – a = 3a – a + 2
⇒ -2 = 2a + 2
⇒ 2a = – 2 – 2 = – 4
⇒ a = \(\frac{-4}{2}\)
⇒ a = -2
Thus, the required value of a is -2.

Question 3.
Find the 5th term of the geometric progression: 3, 6, 12, 24,…
Solution:
Here, a = 3, r = 2, n = 5.
Using formula: a_n = a. r^n-1
a₅ = (3)(2)^5-1 = (3)(2)⁴ = 3 × 16 = 48
So. the 5th term is 48.

Question 4.
Write down the 20th term of the GP 1, – 1. 1, – 1….
Solution:
Here, a = 1 st term = 1
r = common ratio = – 1
Since a_n= ar^n-1
⇒ a₂₀ = 1.(- 1)^20-1 = 1.(-1)¹⁹ = – 1

Question 5.
What is the sum of all the natural numbers from 1 to 50?
Solution:
We have:
1, 2, 3, 4, ………. , 50 are in an A.P. such that
S_n = \(\frac{n(n+1)}{2}=\frac{50 \times(50+1)}{2}\)
= 25 × 51
= 1275

Question 2.
If the nth term of an A.P. is (7n – 5). Find its 100th term.
Solution:
Here, T_n = 7n – 5
T₁ = 7 (1) — 5 = 2
T₂ = 7 (2) – 5 = 9
T₃ = 7(3) – 5 = 16
∴ a = 2 and d = T₂ – T₁ = 9 – 2 = 7
Now T₁₀₀ = 2 + (100 – 1)7 [using T_n = a + (n – 1 )d]
= 2 + 99 × 7
= 2 + 693
= 695.

Question 3.
The fifth term of GP is 81 and the second term is 24, find the GP.
Solution:
Let a₁ be the first term and r be the common ratio of GP.
∵ a_n = ar^n-1
And a₅ = 81 [Given]
⇒ ar⁴ = 81
and a² = 24 [Given]
⇒ ar = 24
Dividing (1) by (2), we have
\(\frac{81}{24}\)
i.e r = \(\frac{27}{8}=\left(\frac{3}{2}\right)^3\)
∴ r = \(\frac{3}{2}\)
Putting this value of r in (2), we get (3
a(\(\frac{3}{2}\)) = 24
a = \(\frac{24 \times 2}{3}\) = 16
Hence, the G.P. is 16, 24, 36, 54, 81,…

Question 4.
If 4th and 8th terms of GP are 24 and 384 respectively, then find out first term and common ratio.
Solution:
Let a be the first term and r be the common ratio of GP.
∵ 4th term = 24
a₄ = ar³ = 24
And 8th term = 384
⇒ a₈ = ar⁷ = 384
Dividing (2) by (1), we get
\(\frac{1}{2}\)
⇒ r⁴ = 16 = 2⁴
⇒ r = 2
Substituting r = 2 in (1), we get
a(2)³ = 24
⇒ a = 24 – 5 – 8 = 3
Hence, first term = 3, common ratio = 2.

Predicting What Comes Next Exploring Sequences and Progressions Class 9 Long Question Answer

Question 1.
If 9th term of an A.P. is zero, prove that its 29th term is double of its 19th term.
Solution:
Let ‘a’ be the first term and ‘d’ be the common difference.
Now, Using T_n = a + (n – 1) d,
we have
⇒ T₉ = a + 8d
⇒ a + 8d = 0 …(1) [∵ T₉ = 0 Given]
T₁₉ = a + 18 d = z(a + 8 d) + 10 d
= (0)+ 10d = 10d …(2) [∵ a + 8d = 0]
T₂₉ = a + 28d
= (a + 8d) + 20d
= 0 + 20d = 20d [∵ a + 8d = 0]
= 2 x (10d) = 2(T₁₉) [∵ T₁₉ = 10d]
⇒ T₂₉ = 2 (T₁₉).
Hence proved.

Question 2.
The sum of 4th and 8th terms of an A.P. is 24, and the sum of 6th and 10th terms is 44. Find the A.P.
Solution:
Let, the first term = a
Common difference be = d
Using Tn = a + (n – 1) d, we have
T₄ = a + 3d
T₆ = a + 5d
T₈ = a + 7d
T₁₀ = a + 9d
∵ T₄ + T₈ = 24 [Given]
(a + 3d) + (a + 7d) = 24
⇒ 2a + 10d = 24
⇒ a + 5d = 12
Also T₆ + T₁₀ = 44
(a + 5 d) + (a + 9 d) = 44
⇒ 2a + 14d = 44
⇒ a + 7d = 22 [Dividing by 2] …(2)
Subtracting (1) from (2), we have:
(a + 7d) – (a + 5d) = 22 – 12
⇒ 2d = 10
⇒ d = 5
From (1), a + 5 (5) = 12
⇒ a = 12 – 25 = – 13
Since, the A.P. is given by:
a, a + d, a + 2d, …………..
∴ We have the required A.P. as:
-13, (-13 + 5), [-13 + 2(5)], …………. or -13, -8, -3,

Question 3.
Find four numbers forming a GP in which the third term is greater than the first by 9, and the second term is greater than the fourth by 18.
Solution:
Let 1 st term, be a and r be the common ratio of GP.
According to the question, we have
a₃ = a + 9
⇒ ar² = a + 9
⇒ a(r – 1) = 9 …(1)
Also, a₂ = a4 + 18
⇒ ar = ar³ + 18
⇒ ar(r² – 1) = -18 …(2)
Dividing (2) by (1), we have r = – 2
Substituting r = – 2 in (1),
we get o(4 – 1) = 9
⇒ 3a = 9
⇒ a = 3
Hence, the four numbers are a, ar, ar², ar³, i.e., 3, – 6, 12,-24.

Question 4.
A fractal tree has an initial branch length of 50 cm. Each successive branch is half the length of the previous branch. What will be the length of the 6th branch?
Solution:
Here, the length of the branches in the fractal tree follows a Geometric Progression (GP), where each successive branch is a constant fraction of the previous branch.
Here, First term (initial branch length) a = 50 cm
Common ratio (r) = \(\frac{1}{2}\) (Since each branch is half the length of the previous one)
Number of term (n) = 6
nth term of GP is: t_n = a × r^n-1
Substitute the known value in the above formula:
t₆ = 50 × \(\left(\frac{1}{2}\right)^{6-1}\)
t₆ = 50 × \(\left(\frac{1}{2}\right)^5\)
t₆ = 50 × \(\frac{1}{32}\)
⇒ t₈ = 1.5625
Thus, the length of the 6th branch is 1.5625 cm.

Predicting What Comes Next Exploring Sequences and Progressions Class 9 Case Based Questions

Question 1.
Amit was playing a number card game. In the game, some number cards (having +ve or -ve numbers) are arranged in a row such that they are following an arithmetic progression. On his first turn, Amit picks up 6th and 14th card and finds their sum to be -76. On the second turn he picks up 8th and 16th card and finds their sum to be -96. Based on the above information, answer the following questions.
(i) What is the difference between the numbers on any two consecutive cards?
(ii) What is the number of the 19th card?
Solution:
(i) Let the numbers on cards be a, a + d, a + 2d, …
According to question, we have (a + 5d) + (a + 13 d) = -76
⇒ 2a + 18d = -76
⇒ a + 9d = -38 …(1)

And (a + 7d) + (a + 15d) = -96
⇒ 2a + 22d = -96
⇒ a + 11d = -48 …(2)
From (1) and (2), we get
2d = -10
⇒ d = -5
Thus, the difference between the numbers on any two consecutive cards = common difference of the A.P. = -5.

(ii) From (1), a + 9(-5) = -38
⇒ a = 1
Number of 19th card T₁₉ = a + (19 – 1)d
= 7+ 18 × (-5)
= -83

Question 2.
The diagram shows the first three stages in a sequence that goes on forever and, in the limit, forms the Sierpinski triangle.

(i) How many white triangles are there at stage n, that is, how many triangles have been removed altogether?
(ii) What is the area of the white triangles at stage n?
Solution:
(i) For each shaded triangle at stage n, one additional
white triangle appears at stage (n + 1). Equivalently the number of white triangles added at each stage is three times the number of white triangles added at the previous stage. Therefore, the number of white triangles removed at stage n is 1 + 3 + 3² + 3³ + … + 3^n-1, where n ≥ 1.

(ii) The total area of all the white triangles is [1 – (total area of shaded triangles)]. Therefore, the total area of the white triangles at stage n is 1 – (\(\frac{3}{4}\))ⁿ sq. units.

Predicting What Comes Next Exploring Sequences and Progressions Class 9 Competency Based Questions

Question 1.
4 groups in a class were asked to come up with an arithmetic progression (AP). Shown below are their responses:

Which of these groups correctly came up with an AP?
(a) only groups M and O
(b) only groups N and O
(c) only groups M, N and O
(d) all groups – M, N, O and P
Solution:
(d) all groups – M, N, O and P
We check if the common difference is constant:
Group M:
4, 2, 0, -2,… → Common difference = -2
Group N:
41,38.5, 36, 33.5,… → Common difference = -2.5
Group O:
-19, —21, -23, -25,… → Common difference = -2
Group P:
—3, -3, —3, -3,… → Constant sequence (difference = 0)
All groups – M, N, O, and P follow the AP pattern.
Hence, (d) is the correct answer.

Question 2.
If an is the nth term of an arithmetic progression whose common difference is d, then which of the following statements is valid?
(a) a₂₄ = a₁ + 24d
(b) a₂₅ – a₂ + 24d
(c) a₂₆ = a₂ + 24d
(d) None of these
Solution:
(c) a₂₆ = a₂ + 24d
Finding the nth term equation
General formula:
a_n = a + (n – 1)d
Checking the given choices,
a₂₆ = a₁ + (26 – 1)d
= a₁ + 25d …(1)
By putting a₁ = a₂ – d in (1)
a₂₆ = a₂ – d + 25 d
= a₂ + 24d
a₂₆ = a₂ + 24d
Hence, (c) is the correct answer.

Question 3.
Determine whether the following sequence is an arithmetic progression or not.
(-12 + 12a), (-11 + 1 la), (-10 + 10a), … where a is any rational number.
Show your work.
Solution:
Identifies the two sets of consecutive terms and finds the difference between the terms in each set by subtracting a term from its next term.
For example,
Second term – First term
= (-11 + 11a) – (-12+ 12a)
= -11 + 11a + 12 – 12a = (1 – a)
Third term – Second term
= (-10 + 10a) – (-11 + 11a)
= -10 + 10a + 11 – 11a = (1 – a)
Compares the difference and concludes that the given sequence is an arithmetic progression.

Question 4.
If the nth term of a GP is 128 and both the first term a and the common ratio r are 2. Find the number of terms in the GP.
Solution:
nth term of a GP, a_n = 128 [Given]
First term of GP, a = 2, Common ratio, r = 2
nth term of G.P., a_n = a × r^n-1
⇒ 128 = 2 × 2^n-1
⇒ 64= 2^n-1
⇒ 26 = 2^n-1
⇒ n – 1 = 6
⇒ n = 7
Therefore, there are 7 terms in GP.

Question 5.
The cannon fires every 2 minutes, with the first shot occurring 10 minutes after the start of the fight. Additionally, the weight of each cannonball increases by 0.5 kg with each successive shot, starting with the first ball weighing 0.5 kg.
(i) How many balls are fired after the first 30 minutes of fight?
(ii) What is the ball’s weight when the 12th ball is fired?
(iii) After how much time will the ball of 10 kg be fired? Show your work.
Solution:
(i) First term (a) = 10 minutes and common difference (d) – 2 minutes
t_n = 30 minutes
Assumes n as the number of balls fired in 30 minutes. 30 = 10 + (n – 1) × 2
⇒ n = 11
Hence 11 balls have been fired after the first 30 minutes of fight.

(ii) First term (a) = 0.5 kg and common difference (d) = 0.5 kg.
∴ Required weight of 12th ball,
t₁₂ = 0.5 + (12 – 1) × 0.5 = 6
Thus weight of the 12th ball fired is 6 kg.

(iii) Assumes that after «th ball, the 10 kg ball is fired and using formula we can write vGo= aw + (n-l)dw 10 = 0.5 + (n – 1) × 0.5
On solving, we get n = 20 and hence the 20th ball would weigh 10 kg.
Uses the n = 20 to evaluate the time as:
10 + (20 – 1) × 2 = 48
So, after 48 mins of the fight starting, 10 kg ball will be fired.

Predicting What Comes Next Exploring Sequences and Progressions Class 9 Extra Questions for Practice

Multiple Choice Questions

Question 1.
The 11th term from the end of the A.P. 10, 7, 4,…, – 62 is
(a) 25
(b) 30
(c) -32
(d) 35

Question 2.
The number of all two digit positive numbers divisible by 3 is
(a) 15
(b) 30
(c) 45
(d) 60

Question 3.
Which of the following is the nth tenn of the geometric progression 3, 6, 12, 24,…?
(a) 3.2ⁿ
(b) 3.2ⁿ⁺¹
(c)6.2^n-1
(d) 3.2^n-1

Assertion Reason Questions

Two statements are given, one labelled as Assertion (A) and the other labelled as Reason (R). Select the correct answer from the options (a), (b), (c), and (d) given below.
(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
(c) Assertion (A) is true, but Reason (R) is false.
(d) Assertion (A) is false, but Reason (R) is true.

Question 1.
Assertion (A): The common ratio of a geometric progression can never be zero.
Reason (R): In a geometric progression, each term is obtained by multiplying the previous term by the common ratio.

Question 2.
Assertion (A): The difference between any two consecutive terms in the sequence of numbers √6, √24, √54, √96,… is 3√6 .
Reason (R): The sequence of numbers √6, √24, √54, √96,… form an arithmetic progression.

Short Answer Type Questions

Question 1.
Find three numbers in A.R whose sum is 21 and their product is 231. (Hint: Assume three terms a – d, a, a + d)

Question 2.
In a geometric progression, the common ratio is -2. If the first term is 3, what is the 5th term?

Question 3.
The sum of the first 3 terms of a geometric progression is 14, and the common ratio is 2. What is the first term?

Question 4.
The first and the last term of an A.P. are 17 and 350 respectively. If the common difference is 9 how many terms are there in the A.P.?

Question 5.
The 5th term of a GP is 2, find the product of its first 9 terms.

Question 6.
In a geometric progression, the 2nd term is 9 and the 5th term is 72. Find the first term and the common ratio.

Long Answer Type Questions

Question 1.
In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?
OR
The sum of the second and third terms of a GP is 280 and the sum of the 5th and 6th terms is 4375. Find the 4th term of GP.

Question 2.
If pth and qth terms of a GP are q and p respectively, show that (p + q)^th term is \(\left(\frac{q^p}{p^q}\right)^{\frac{1}{p-q}}\)
OR
The area of the “Sierpinski carpet” decreases in each iteration by a factor of \(\frac{8}{9}\). If the be the area after 4 iterations?

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