Practicing Class 9 Maths MCQ and Ganita Manjari Class 9 Maths Chapter 8 Predicting What Comes Next Exploring Sequences and Progressions MCQ Questions Online Test with Answers daily helps in time management.
MCQ on Predicting What Comes Next Exploring Sequences and Progressions Class 9
Predicting What Comes Next Exploring Sequences and Progressions MCQ Class 9
Class 9 Maths Predicting What Comes Next Exploring Sequences and Progressions MCQ
Question 1.
The value of x for which 2x, (x + 10) and (3x + 2) are the three consecutive terms of an AP is
(a) 6
(b) -6
(c) -2
(d) 2
Answer:
(a) 6
Explanation:
Since, 2x, (x + 10), (3x + 2) are in AP.
∴ x + 10 – 2x – (3x + 2) – (x + 10)
⇒ —x + 10 = 2x – 8
⇒ 3x = 18
⇒ x = 6
Question 2.
Which of the following sequence is not an AP?
(a) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\)……….
(b) -1.2, -3.2, -5.2, -7.2,………
(c) √2, √8, √18, ……….
(d) 12, 32, 52, 72, …….
Answer:
(a) 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\)……….
Explanation:
Option (a),
We have 2, \(\frac{5}{2}\), 3, \(\frac{7}{2}\)……….
Here, a2 – a1 = \(\frac{5}{2}\) – 2
= \(\frac{1}{2}\)
a3 – a2 = 3 – \(\frac{5}{2}\)
= \(\frac{1}{2}\)
a4 – a3 = \(\frac{7}{2}\) – 3
= \(\frac{1}{2}\)
Since, the difference of any two consecutive terms is same.
So, the given sequence is in AP.
Option (b),
We have -1.2, -3.2, -5.2, -7.2,….
Here, a2 – a1 = -3.2 – (-1.2) = -2
a3 – a2 = -5.2 – (-3.2) = -2
a4 – a3 = -7.2 – (-5.2) = -2
Since, the difference of any two consecutive terms is same.
So, the given sequence is in AP.
Option (c),
We have \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \ldots \ldots\) i.e. \(\sqrt{2}, 2 \sqrt{2}, 3 \sqrt{2}, \ldots\)
Here, a2 – a1 = 2√2 – √2 = √2
a3 – a2 = 3√2 – 2√2 = √2
Since, the difference of any two consecutive terms is same.
So, the given sequence is in AP.
Option (D),
We have 12, 32, 52, 72,…i.e. 1, 9, 25, 49, …
Here, a2 – a1 = 9 – 1 = 8,
a3 – a2 = 25 – 9 = 16
a4 – a3 = 49 – 25 = 24
Since, the difference of any two consecutive terms is not same.
So, the given sequence is not in AP.
Question 3.
The next term of the AP \(\sqrt{5}, \sqrt{20}, \sqrt{45},\) …………. is
(a) \(\sqrt{60}\)
(b) 4√5
(c) \(\sqrt{75}\)
(d) \(\sqrt{125}\)
Answer:
(b) 4√5
Explanation:
Given, AP is \(\sqrt{5}, \sqrt{20}, \sqrt{45},\), … or √5, 2√5, 3√5
Here, common difference = 2√5 – √5 = √5
So, next term is 3√5 + √5 = 4√5.
Question 4.
In an AP, if a = 8 and a10 = -19 then value ofd is
(a) 3
(b) \(\frac{-11}{9}\)
(c) \(\frac{-27}{10}\)
(d) -3
Answer:
(d) -3
Explanation:
Given, a = 8 and a10 = -19
⇒ a + 9d = -19 [∵ an = a + (n – 1)d]
⇒ 8 + 9d = -19
⇒ 9d = -27
⇒ d = -3
Question 5.
nth term of an AP is 5n – 15. The common difference of the AP is
(a) 5n
(b) 5
(c) -5
(d) 10
Answer:
(b) 5
Explanation:
Given, nth term of an AP.
an = 5n – 15
For the first term, n = 1
a1 = 5(1) – 15
⇒ a2 = 5 – 15
⇒ a1 = —10
For the second term, n = 2
n2 = 5(2) – 15
⇒ a2 = 10 – 15
⇒ a2 =-5
Then, the common difference,
d = a2 – a1
⇒ d = -5 – (-10)
⇒ d = -5 + 10
⇒ d = 5
Question 6.
nth term of the AP \(-\frac{3}{2}, \frac{3}{2}, \frac{9}{2}, \ldots\) is
(a) \(\frac{3n}{2}\) – 3
(b) 3n – \(\frac{9}{2}\)
(c) \(\frac{3 n-9}{2}\)
(d) 3n + \(\frac{3}{2}\)
Answer:
(b) 3n – \(\frac{9}{2}\)
Explanation:
Given, AP is \(-\frac{3}{2}, \frac{3}{2}, \frac{9}{2}, \ldots .\)
The first term, a = –\(\frac{3}{2}\) and common difference,
d = \(\frac{3}{2}-\left(-\frac{3}{2}\right)=\frac{3}{2}+\frac{3}{2}=3\) = 3
Then, the nth term of AP,
an = a + (n – 1)d
⇒ an = –\(\frac{3}{2}\) + (n – 1)3
⇒ an = –\(\frac{3}{2}\) + 3n – 3
⇒ an = 3n – \(\frac{9}{2}\)
Question 7.
The 10th term of the AP \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \ldots\) is
(a) \(\sqrt{162}\)
(b) \(\sqrt{200}\)
(c) \(\sqrt{54}\)
(d) \(\sqrt{94}\)
Answer:
(b) \(\sqrt{200}\)
Explanation:
Given, AP is √2, √8 , √18, …………
Now, common difference
= √8 – √2 = √2(√4 – 1)
= √2 (2 – 1) = √2
Now, the general form for the nth term of an AP is an = a + (n – 1)d
For the 10th term (n = 10), we get
a10 = √2 + (10 – 1) x √2 = √2 + 9 x √2
= 10√2 = √200
Therefore, the 10th term is √200.
Question 8.
Which term of the AP 5, 15, 25,… will be 130 more than its 31st term?
(a) 42
(b) 44
(c) 46
(d) 48
Answer:
(b) 44
Question 9.
Two APs have the same common difference. The first term of one of these is -1 and that of the other is -8 Then, the difference between their 4th terms is
(a) 1
(b) 8
(c) 7
(d) 9
Answer:
(c) 7
Explanation:
Let the common difference of two APs be d.
Given, first term of one AP (a1) = – 1
and first term of second AP (b1) = – 8.
∴ Fourth term of 1st AP (a4) = a1 + (4 – 1)d
= – 1 + 3d
and fourth term of 2nd AP (b4) = b1 + (4 – 1)d
= – 8 + 3d
∴ a4 – b4 = (- 1 + 3d) – (- 8 + 3d)
= -1 + 3d + 8 – 3d
= 7
Question 10.
Find the sum of the first 25 natural numbers.
(a) 300
(b) 325
(c) 350
(d) 375
Answer:
(b) 325
Explanation:
Given, n = 25
The sum of the first 25 natural numbers,
S = \(\frac{25(25+1)}{2}\) [S = \(\frac{n(n+1)}{2}\)]
= \(\frac{25 \times 26}{2}\) = 325
Question 11.
If 1 + 2 + 3 + 4 +…..+ 50 = 25k then k is equa to
(a) 50
(b) 51
(c) 49
(d) 26
Answer:
(b) 51
Explanation:
Given, AP is 1 + 2 + 3 + 4 +…..+ 50
∴ S50 = \(\frac{1}{2}\)(50 + 1)
[∵ sum of the first n natural numbers, Sn = \(\frac{1}{2}\)]
= 25 × 51 = 1275
Also, given S50 = 25k
From Eqs. (i) and (ii), we have
25k = 1275
⇒ k = \(\frac{1275}{25}\) = 51
Question 12.
If the sequence an = 3(2n),∀ n ∈ N is GP then its common ratio is
(a) -2
(b) 2
(c) 3
(d) \(\frac{1}{2}\)
Answer:
(b) 2
Explanation:
We have, an = 3(2n)
and an+1 = 3(2n+1)
Now, common ratio = \(\frac{1}{2}\)
Question 13.
The third term of GP is 4, the product of the first five terms is
(a) 1024
(b) 512
(c) 256
(d) None of these
Answer:
(a) 1024
Explanation:
Let the first five term of GP be
\(\frac{a}{r^2}, \frac{a}{r}\), a, ar, ar2, where r is common ratio. r2 r
Given, third term = 4
i.e. a = 4
∴ Product of first five term
= \(\frac{a}{r^2}, \frac{a}{r}\), a, ar, ar2
= a5 = (4)5 [∵ using Eq. (i)]
= 1024
Question 14.
The 9th term of the GP 2, 2√2, 4, 8√2. … is
(a) 16
(b) 64
(c) 32
(d) 128
Answer:
(c) 32
Explanation:
Given GP is 2, 2√2, 4, 8√2. …
∴ 9th term, T9 = a.rn-1
= 2(√2)8 [∵ r = \(\frac{2 \sqrt{2}}{2}\) = √2]
Question 15.
Which term of the sequence 3, 6, 12, …………. is 768?
(a) 10
(b) 8
(c) 9
(d) None of these
Answer:
(c) 9
Explanation:
Hint We have, 3, 6, 12,…
Let nth term be 768.
∴ Tn = a.rn-1 = 768
Here, a = 3 and r = 2
∴ 3.(2)n-1 = 768
⇒ n = 9
Question 16.
Which of the following is a GP?
(a) 2, 4, 8, 16, …
(b) \(\frac{1}{9}, \frac{-1}{27}, \frac{1}{81}, \frac{-1}{243}, \ldots\)
(c) 0.01,0.0001,0.000001,…
(d) All of the above
Answer:
(d) All of the above
Explanation:
(A) We have,
a1 = 2, \(\frac{a_2}{a_1}\) = 2. \(\frac{a_3}{a_2}\) = 2, \(\frac{a_4}{a_3}\) = 2 and soon.
(B) We observe, a1 = \(\frac{1}{9}, \frac{a_2}{a_1}=\frac{-1}{3}\)
\(\frac{a_3}{a_2}=\frac{-1}{3}, \frac{a_4}{a_3}=\frac{-1}{3}\) and so on.
(C) We have, a = 0.01, \(\frac{a_2}{a_1}\) = 0.01,
\(\frac{a_3}{a_2}\) = 0.01, \(\frac{a_4}{a_3}\) = 0.01 and so on.
It is observed that in each case, every term except the first term bears a constant ratio to the term immediately preceding it.
Thus, options (A), (B) and (C) all are in GP.
Question 17.
The nth and 10th terms ola GP 5, 25, 125, …, are respectively
(a) 5n-1, 59
(b) 5n-2, 58
(c) 5n, 510
(d) None of these
Answer:
(c) 5n, 510
Explanation:
Here, a = 5 and r = 5.
Thus, the nth term of the given GP is given by
an = arn-1
⇒ an = 5(5)n-1 = 5n
Now, for n = 10, we have a1 = 510
Question 18.
The third term of a geometric progression is 4. The product of the first five terms is
(a) 43
(b) 45
(c) 44
(d) None of these
Answer:
(b) 45
Explanation:
Here, T3 = 4 ⇒ ar2 = 4
∴ Product of first five terms = a. ar. ar2. ar3. ar4
= a5.r10 = (ar2)5 = 45
Question 19.
In a GP of positive terms. if any term is equal to the sum of next two terms then the common ratio of GP is
(a) \(\frac{-1 \pm \sqrt{5}}{2}\)
(b) \(\frac{-1-\sqrt{5}}{2}\)
(c) -1 + √5
(d) \(\frac{-1+\sqrt{5}}{2}\)
Answer:
(d) \(\frac{-1+\sqrt{5}}{2}\)
Explanation:
Let a be the first term and r be the common ratio of GP.
According to the question,
a = ar + ar2 ⇒ 1 = r + r2
⇒ r2 + r – 1 = 0
Now, r = \(\frac{-1 \pm \sqrt{1^2-4(1)(-1)}}{2 \times 1}=\frac{-1 \pm \sqrt{1+4}}{2}=\frac{-1 \pm \sqrt{5}}{2}\)
Since, the terms of the GP are positive, the common ratio r must be positive.
So, r = \(\frac{-1+\sqrt{5}}{2}\)
Question 20.
If the sum of three numbers in GP is 38 and their product is 1728 then numbers are
(a) 8, 12, 18
(b) 8, 16, 32
(c) 18, 12, 6
(d) None of these
Answer:
(a) 8, 12, 18
Explanation:
Let the three numbers be \(\frac{a}{r}\), a and ar.
Then, product = 1728
\(\frac{a}{r}\).a.ar = 1728
⇒ a3 = 1728
⇒ a = 12
Sum = 38
\(\frac{a}{r}\) + a + ar = 38
⇒ a(\(\frac{1}{r}\) + 1 + r) = 38
⇒ 12\(\left(\frac{1+r+r^2}{r}\right)\) = 38
⇒ 6 + 6r + 6r2 = 19r
⇒ 6r2 – 13r + 6 = 0
⇒ (3r – 2) (2r – 3) = 0
⇒ r = \(\frac{3}{2}\) and \(\frac{2}{3}\)
Hence, on putting the values of a and r, the required numbers are 8, 12, 18.
Question 21.
In a square fractal pattern, the number of smaller squares formed at each step follows a GP with first term 1 and common ratio 8.
The number of squares in the 5th step is
(a) 512
(b) 4096
(c) 32768
(d) 64
Answer:
(b) 4096
Explanation:
a = 1, r = 8
∴ S5 = 1 × 84 = 4096
Question 22.
In a fractal, the number of elements triples at each step. lithe tìrst step has 4 elements then the number of elements in the 4th step
(a) 87
(b) 108
(c) 99
(d) 101
Answer:
(b) 108
Explanation:
a4 = 4 × 33
= 4 × 27
= 108
Question 23.
The minimum number of moves required to solve the Tower of Hanoi puzzle with 8 discs is
(a) 127
(b) 255
(c) 256
(d) 511
Answer:
(b) 255
Explanation:
We have, Tn = 2n – 1
T8 = 28 – 1
= 255
Predicting What Comes Next Exploring Sequences and Progressions Class 9 Assertion and Reason Questions
Direction (Q. Nos. 1-6) In the questions given below, there are two statements marked as Assertion (A) and Reason (R). Read the statements and choose the correct option.
Question 1.
Assertion (A): – 5, –\(\frac{5}{2}\) , 0, \(\frac{5}{2}\), ………… is an arithmetic progression.
Reason (R): The terms of an arithmetic progression cannot have both positive and negative rational numbers.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true hut R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(c) A is true but R is false.
Explanation:
A is true because common difference is same but R is false.
Question 2.
Assertion (A): The sequence -1, -1, -1, …, -1 is an AP.
Reason (R): In an AP, an – an-1, is constant, where n ≥ 2 and n ∈ N.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true hut R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
Given, sequence is – 1, – 1, – 1,…., – 1
For above sequence, an – an-1 = 0 = constant.
Here, an – an-1l is constant, where n > 2 and n ∈ N.
So, both A and R are true and R is the correct explanation of A.
Question 3.
Assertion (A): It the numbers –\(\frac{3}{7}\), k, –\(\frac{7}{3}\) are in GP then k = ± 1.
Reason (R): If a1, a2, a3 are in GP then \(\frac{a_2}{a_1}=\frac{a_3}{a_2} \)
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true hut R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
–\(\frac{3}{7}\), k, –\(\frac{7}{3}\) are in G.P
If a1, a2, a3 are in GP then
\(\frac{a_2}{a_1}=\frac{a_3}{a_2} \)
∴ \(\frac{k}{-\frac{3}{7}}=\frac{-\frac{7}{3}}{k}\)
⇒ k = ± 1
Question 4.
Assertion (A: 1, 2, 4, 8, … is a GP, 4, 8, 16, 32, … is a GP and 1 + 4, 2 + 8, 4 + 16, 8 + 32… is also a GP.
Reason (R): Let general term of a GP with common ratio r be Tk + 1 and general term of another GP with common ratio r be Tk + 1 then the series whose general term Tk + 1 = Tk + 1, + Tk + 1 is also a GP with common ratio r.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true hut R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
Sequences a1, a2,…,an are in GP.
The common ratio r is taken as \(\frac{a_{k+1}}{a_k}\).
Thus, the general term of a GP with common ratio r be take Tk + 1 and general term of another GP with common ratio r be Tk + 1 then the series, whose general term Tk + 1 = Tk + 1 + Tk + 1 is also a GP with common ratio r.
Question 5.
Assertion (A): The number of squares in a Sierpinski carpet at each step form a GP.
Reason (R) At each step, every square produces a fixed number of smaller squares.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true hut R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation:
A is true, since, the number of squares increases multiplicatively and R is also true because “each square generates a constant number of new squares.
Since, this constant multiplication leads to a GP, R correctly explains A.
Question 6.
Assertion (A): The minimum numher of moves required in the Tower of Hanoi puzzle for n discs is 2n – 1
Reason (R): The number of moves increases by a constant number at each stage.
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true hut R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.
Answer:
(c) A is true but R is false.
Explanation:
A is true, since the formula for minimum moves is Tn = 2n – 1.
R is false, because the number of moves does not increase by a constant values instead, it follows. Tn = 2Tn-1 + 1
Which shows exponential growth not linear growth.