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

By using Extra Questions for Class 9 Maths and Ganita Manjari Class 9 Maths Chapter 8 Predicting What Comes Next Exploring Sequences and Progressions Extra Questions, students can improve their problem-solving skills.

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

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

Class 9 Ganita Manjari Chapter 8 Extra Questions

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

Question 1.
Find the next term of the A.P.
\(\sqrt{7}, \sqrt{28}, \sqrt{63}, \ldots\)
Solution:
The given A.P. is
\(\sqrt{7}, \sqrt{28}, \sqrt{63}, \ldots\)
Here,
a = √7
d= \(\sqrt{28}\) – √7
= \(\sqrt{4 \times 7}\) – √7
= √4√7 – √7
= 2√7 – √7
= (2 – 1)√7 = √7
Next term, i.e., 4^th term = a4
= a + (4 – 1)d [∵ a_n = a + (n – 1)d]
= a + 3d
= √7 + 3√7
= (1 + 3)√7 = 4√7
= √16√7
= \(\sqrt{16 \times 7}=\sqrt{112}\)

Question 2.
Find the common difference of the A.P.
\(\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p}, \ldots\)
Solution:
The given A.P is \(\frac{1}{p}, \frac{1-p}{p}, \frac{1-2 p}{p}\), ………….
Here, a₁ = \(\frac{1}{p}\)
a₂ = \(\frac{1-p}{p}\)
∴ Common difference = a₂ – a₁
= \(\frac{1-p}{p}-\frac{1}{p}\)
= \(\frac{1-p-1}{p}=\frac{-p}{p}\) = -1

Question 3.
Find the common difference of the A.P.
\(\frac{1}{2 q}, \frac{1-2 q}{2 q}, \frac{1-4 q}{2 q}, \ldots\)
Solution:
The given A.P is \(\frac{1}{2 q}, \frac{1-2 q}{2 q}, \frac{1-4 q}{2 q}, \ldots\)
Here, a₁ = \(\frac{1}{2q}\)
a₂ = \(\frac{1-2q}{2q}\)
∴ Common difference = a₂ – a₁

Question 4.
Find the common difference of the A.P.
\(\frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-9 q}{3 q}, \ldots\)
Solution:
The given A.P is \(\frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-9 q}{3 q}, \ldots\)
Here, a₁ = \(\frac{1}{3q}\)
a₂ = \(\frac{1-6q}{3q}\)
∴ Common difference = a₂ – a₁

Question 5.
If k, 2k – 1 and 2k + 1 are three consecutive terms of an A.P., find the value of k.
Solution:
∵ k, 2k – 1 and 2k + 1 are three consecutive terms of an A.P.
∴ a₂ – a₁ = a₃ – a₂
⇒ (2k – 1) – k = (2k + 1) – (2k – 1)
⇒ k – 1 = 2k + 1 – 2k + 1
⇒ k – 1 =2
⇒ k = 1 + 2
⇒ k = 3.

Question 6.
Find the 9th term of A.P. 3, 2\(\frac{1}{2}\), 2, …
Solution:
The given A.P is 3, 2\(\frac{1}{2}\), 2, …
Here, a = 3
d = 2\(\frac{1}{2}\) – 3
= 2 + \(\frac{1}{2}\) – 3
= \(\frac{4+1-6}{2}=\frac{-1}{2}\)
∴ 9^th term = a₉ = a + (9 – 1)d [∵ a_n = a + (n – 1)d]
= a + 8d
= 3 + 8(-\(\frac{1}{2}\))
= 3 – 4 = -1

Question 7.
Find the 30th term of A.P. 10, 7, 4, …
Solution:
The given A.P. is 10, 7, 4, …
Here, a = 10
d = 7 – 10 = -3
30^th term = a₃₀
= a + (30 + 1)d [∵a_n = a + (n – 1 )d]
= a + 29d
= 10 + 29 × (-3)
= 10 – 87
= -77

Question 8.
If the first term of an A.P. is p and the common difference is q, find its 10^th term.
Solution:
Here, a = p, d = q
10^th term = a₁₀
= a + (10 – 1)d [∵ a_n = a + (n – 1)d]
= a + 9d
= p + 9q.

Question 9.
Find the sum of first 15 multiples of 8.
Solution:
First 15 multiples of 8 are
8 × 1, 8 × 2, 8 × 3, 8 × 4, …, 8 × 15
i.e., 8, 16, 24, 32, …,120
This is an A.P. with common difference 8.
Here, a = 8, l = 120, n = 15
Sum of first 15 multiples of 8
= S₁₅
= \(\frac{15}{2}\)(8 + 120)
= \(\frac{15}{2}\) × 128
= 15 × 64
= 960.

Question 10.
Which term of the GP., 5, 10, 20, 40,… is 5120?
Solution:
Given G.P. has a = 5, r = 2.
Let a_n = 5120 ⇒ a . r^n-1 = 5120
⇒ 5 × 2^n-1 = 5120
⇒ n – 1 = 10 ⇒ n = 11
The 11th term of the given G.P. is 5120.

Question 11.
The third term of a G.P. is 4. Find the product of first 5 terms.
Solution:
Let a be the first term and r be the common ratio.
Given a₃ = 4 ⇒ a.r² = 4
The product of first 5 terms = a .ar.ar².ar³.ar⁴ = a⁵r¹⁰ = (ar²)⁵ = 4⁵ = 1024.

Question 12.
The 5^th, 8^th and 11^th terms of a G.P. are p, q and s, respectively. Show that q² = ps.
Solution:
Let a be the first term and r be the common ratio of the G.P.
Given t₅ = p ⇒ ar⁴ = p ………..(1)
Similarly, t₈ = q ⇒ ar⁷ ……….(2)
and t₁₁ = s ⇒ ar¹⁰ = s ………(3)
Now, (1) x (3) ⇒ ps = ar⁴ . ar¹⁰
= a²r¹⁴ = (ar⁷)² = q²
Hence, q² = ps

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

Question 1.
Which term of the A.P. 84, 80, 76, 12, … will be the first negative term?
Solution:
The given A.P. is 84, 80, 76, 72, …
Here, a = 84
d= 80 – 84 = -4
Let nth term of the given A.P. be the first negative term.
Then, a_n < 0
⇒ a + (n – 1 )d < 0
⇒ 84 + (n – 1)(-4) < 0
⇒ 84 – 4n + 4 < 0
⇒ 88 < 4n ⇒ 4n > 88
⇒ n > \(\frac{88}{4}\)
⇒ n > 22
Least integral value of n = 23.
Hence, 23^rd term of the given A.P. is the first negative term.

Question 2.
How many multiples of 4 lie between 10 and 260?
Solution:
Multiples of 4 that lie between 10 and 260 are 12, 16, 20, …, 256 This is an A.P.
Here, a = 12
d = 16 – 12 = 4
l = 256
Let the number of terms of this A.P. be n.
Then, a_n = l = a + (n – 1)d
⇒ 256 = 12 + (n – 1)4
⇒ 256 = 12 + 4n – 4 = 8 + 4n
⇒ 256 – 8 = 4n
⇒ 4n = 248
⇒ n = \(\frac{248}{4}\) = 62
Hence, 62 multiples of 4 lie between 10 and 260.
Remark. Please note: Here 260 is also a multiple of 4 but we have not included it as a term of A.P. because between 10 and 260.

Question 3.
The 9th term of an A.P. is equal to 6 times the second term. If its 5th term is 22, find the A.P.
Solution:
Let a and d be the first term and the common difference of the A.P. respectively.
According to the question,
a₉ = 6 a₂
⇒ a + (9 – 1)d = 6(a + (2 – 1)d) [∵ a_n = a + (n – 1 )d]
⇒ a + 8d = 6a + 6d
⇒ 5a – 2d = 0
⇒ 5a = 2d
⇒ a = \(\frac{2d}{5}\)
Again, according to the question,
a₅ = 22
⇒ a + (5 – 1)d = 22 [∵ a_n = a + (n – 1 )d]
⇒ a + 4d = 22

Multiplying every term by 5,
2d + 20d = 22 × 5
⇒ 22d= 22 × 5
⇒ d = \(\frac{22 \times 5}{22}\) = 5

Putting d = 5 in (1),
a = \(\frac{2 d}{5}=\frac{2 \times 5}{5}\) = 2
Hence, the required A.P. is
2, 2 + 5, 2 + 2(5), 2 + 3(5), … i.e., 2, 7, 12, 17, …

Question 4.
If the seventh term of an A.P. is \(\frac{1}{9}\) and its ninth term is \(\frac{1}{7}\), find its (63)^rd term.
Solution:
Let a and d be the first term and the common difference of the A.P. respectively.
According to the question,
a₇ = \(\frac{1}{9}\)
⇒ a + (7 – 1)d = \(\frac{1}{9}\) [∵ a_n = a + (n – 1 )d]
⇒ a + 6d = \(\frac{1}{9}\) ….(1)

Again, according to the question,
a₉ = \(\frac{1}{7}\)
a + (9 – 1)d = \(\frac{1}{7}\) [∵ a_n = a + (n – 1 )d]
⇒ a + 8d = \(\frac{1}{7}\) …..(2)

Let us solve (1) and (2) for a and d.
Subtracting (1) from (2), we get
2d = \(\frac{1}{7}-\frac{1}{9}=\frac{9-7}{63}\)
⇒ 2d = \(\frac{2}{63}\)
⇒ d = \(\frac{2}{63 \times 2}=\frac{1}{63}\)

Putting d = \(\frac{1}{63}\) in (1), we get

Question 5.
The sum of the first three terms of a G.P. is \(\frac{39}{10}\) and their product is 1. Find the first term, the common ratio and the terms of the G.P.
Solution:
Let the three numbers be \(\frac{a}{r}\) a, ar.
Given product = 1

When, r = \(\frac{2}{5}\), the numbers are \(\frac{5}{2}\), 1, \(\frac{2}{5}\) .
When, r = \(\frac{2}{5}\), the numbers are \(\frac{2}{5}\), 1, \(\frac{5}{2}\)
Hence, the terms of the G.P. are
\(\frac{5}{2}, 1, \frac{2}{5}\) … or \(\frac{5}{2}, 1, \frac{2}{5}\) ………….

Question 6.
If a, b, c are in G.P., prove that a² + b², ab + bc, b² + c² are also in G.P.
Solution:
Given, a, b, c are in G.P.

Hence, a² + b², ab + bc, b² + c² are also in G.P.

Question 7.
If p^th, q^th and r^th terms of a G.P are a, b, c, respectively, prove that a^q-r.b^r-p.c^p-q = 1.
Solution:
Let A be the first term and R be the common ratio of the G.P.
Given, a_p = a ⇒ AR^p-1 = a.
a_q = b ⇒ AR^q-1 = b.
a_r = c ⇒ AR^r-1 = c.
Hence, a^q-rb^r-pc^p-q = AR^p-1.AR^q-1.AR^r-1
= A^{q-r + r-p + p-q.R(q-1)(q-r)+(q-1)(r-p)+(r-)(p-q)}
= A° R^{pq-pr-q++r+ qr-pq-r+p+pr-qr-p+q}
= A°R° = 1

Question 8.
If a, b, c, d are in G.P., prove that aⁿ + bⁿ, bⁿ + cⁿ, cⁿ + dⁿ are in G.P.
Solution:
Let r be the common ratio of the G.P. a, b, c, d.
Then, b = ar, c = ar² and d = ar³.

Question 9.
If a, b, c, d are in G.P., prove that a² – b², b² – c², c² – d² are in G.P.
Solution:
Let r be the common ratio of the given G.P. Then
\(\frac{b}{a}=\frac{c}{b}=\frac{d}{c}\) = r
⇒ b = ar, c = br = ar², d = cr = ar³
Now, a² – b² = a² — a²r² = a²(1 – r²)
b² – c² = a²r² – a²r² = a²r²(1 – r²)
and c² – d² = a²r⁴ – a²r⁶ = a²r⁴ (1 – r²)
Therefore, \(\frac{b^2-c^2}{a^2-b^2}=\frac{c^2-d^2}{b^2-c^2}\) = r²
Hence, a² – b², b² – c², c² – d² are in G.P.

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

Question 1.
Ramkali would need ₹ 1800 for admission fee and books etc. for her daughter to start going to school from next year. She saved ₹ 50 in the first month of this year and increased her monthly saving by ₹ 20. After a year how much money will she have? Will she be able to fulfill her dream of sending her daughter to school?
Solution:
Since Ramkali increased her monthly saving by equal amount of ₹ 20 every month, so her savings form an AP with
d= ₹ 20,
Here, a = ₹ 50, d = ₹ 20,
n = 1 year = 12 months.
Total savings of Ramkali in 12 months (i.e., in one year)
= S₁₂ = \(\frac{n}{2}\)[2a + (n – 1)d]
= \(\frac{12}{2}\) [2(50) + (12 – 1)20]
= 6 [100 + 220]
= 6 × 320
= ₹ 1920
This amount is more than the required amount of ₹ 1800.
Since, she needs ₹ 1800 for admission fee and books etc. for her daughter to start going to school from next year, therefore, she will be able to fulfill her dream of sending her daughter to school.

Question 2.
The sum of first q terms of an A.P. is 63q – 3q². If its p^th term is – 60, find the value of p. Also, find the 11th term of this A.P.
Solution:
S_q = 63q – 3q² (given) …(1)
Putting q = p – 1 ,p successively in (1), we get
S_p-1 = 63(p – 1) – 3(p – 1)²
= 63(p – 1) – 3(/p² – 2p + 1)
= 63p – 63 – 3p² + 6p – 3
= – 3p² + 69p – 66
S_p = 63p – 3p²

a_p = S_p – S_p-1
= (63p – 3p²) – (- 3p² + 69p – 66)
= 63p – 3p² + 3p2 – 69p + 66
= – 6p + 66

According to the question,
a_p = – 60
⇒ -6p + 66 = – 60
⇒ -6p = -60 – 66
⇒ -6p = -126
⇒ p = \(\frac{-126}{-6}\) = 21
Hence, the required value of p is 21.

Again, putting q = 10, 11 successively in (1), we get
S₁₀ = 63 × 10 – 3(10)²
= 630 – 300
= 330
S₁₁ =63 × 11 – 3(11)²
= 693 – 363
= 330

11th term of the A.P.
= S₁₁ – S₁₀
= 330 – 330
= 0

Question 3.
Prove that the nth term of an A.P. cannot be n² + 1. Justify your answer.
Solution:
a_n = n² + 1 …(1)
Putting n = 1, 2, 3 in (1), we get
a₁ = 1² + 1 = 1 + 1 = 2
a₂ = 2² + 1 = 4 + 1 = 5
a₃ = 3² + 1 = 9 + 1 = 10
∴ a₂ – a₁ = 5 – 2 = 3
a₁ – a₂ = 10 – 5 = 5
∵ a₂ – a₁ ≠ a₃ – a₂
The n^th term of an A.P. cannot be n2 + 1.

Question 4.
The 6th term of an A.P. is zero. Prove that its 21st term is triple its 11^th term.
Solution:
Let the first term and the common difference of the A.P. be a and d respectively.
Then, according to the question,
a₆ = 0
⇒ a + (6 – 1)d = 0 [∵ a_n = a + (n – 1 )d]
⇒ a + 5d = 0
⇒ a = – 5d
We are to prove that
a₂₁ = 3a₁₁
L.H.S. = a₂₁ = a + (21 – 1)d [∵ a_n = a + (n – 1 )d]
= a + 20d
= -5d + 20d [By (1)]
= 15 d …(2)

R.H.S. = 3a₁₁ = 3[a + (11 – 1)d] [∵ an = a + (n – 1)d]
= 3 (a + 10d)
= 3(- 5d + 10d) [By (1)]
= 3(5d) = 15d …(3)
From (2) and (3) L.H.S. = R.H.S.
Hence, its 21st term is triple its 11th term.

Question 5.
In an increasing G.P., the sum of the first and last term is 66 and the product of the second and last but one term is 128. If the sum of the series is 126, find the number of terms of the series.
Solution:
Let the increasing G.P be a, ar, ar², …, ar^n-1, where a is the first term, r is the common ratio and ar^n-1 is the nth term or last term.
Given, the sum of first and last term = 66
⇒ a + ar^n-1 = 66
⇒ ar^n-1 = 66 – a …(i)

Also, product of second and last but one term = 128
⇒ a_r.ar^n-2 = 128
a₂.ar^n-2 = 128 …(ii)

Dividing (ii) by (i), we get.
\(\frac{1}{2}\)
⇒ a(66 – a) = 128
⇒ a² – 66a + 128=0
⇒ (a – 64) (a – 2) = 0
⇒ a = 64 or a = 2

On substituting a = 2 in eq. (ii), we get
4r.ar^n-1 = 128
r^n-1 = 32 …(iii)

AIso,given S_n = 126 ⇒ \(\frac{a\left(r^n-1\right)}{r-1}\) = 126

⇒ 64r – 2 = 126r – 126
⇒ 62r = 124
⇒ r = 2.
Substituting in (iii), we get
2^n-1 = 32 = 2⁵ ⇒ n – 15 ⇒ n = 6.
If we take a = 64 and substituting in (I), we get
64r^n-1 = 2 ⇒ r^n-1 = \(\frac{1}{32}\)
Also, S_n =126

⇒ 2r – 64 = 126r – 126
⇒ 124r = 62
⇒ r = \(\frac{1}{2}\)

When a = 64 and r = \(\frac{1}{2}\), the series is a decreasing series which we cannot accept.
∴ n = 6.
Hence, the number of terms of the series is 6.

Question 6.
If p, q, r are in G.P. and the equation px² + 2qx + r = 0 and dx² + 2ex + f = 0 have a common root, then show that \(\frac{d}{p}, \frac{e}{q}, \frac{f}{r}\) are in A.P.
Solution:
Given, p. q. r are in GP. ⇒ q² = pr.
Now, consider the equation px² + 2qx + r = 0.
Solving, we get

Also, given that px² + 2qx = r = 0 and
dx² + 2ex + f = 0 have a common root.
Hence, is the root of the equation
dx² + 2ex + f = 0.

⇒ \(\frac{1}{2}\) are in A.P.

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

A telecom company offers a data pack where the first month gives 2 GB, the second month 4 GB, the third month 8 GB, and so on.

Question 1.
Does this form an arithmetic progression or a geometric progression?
Answer:
GP

Question 2.
What is the common ratio?
Answer:
2

Question 3.
How much data will be given in the 6th month?
Answer:
64

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

Very Short Answer Type Questions

Question 1.
Find the common difference of the A.P.
\(\frac{1}{3}, \frac{1-3 b}{3}, \frac{1-6 b}{3}\)……………..
Solution:
-b

Question 2.
Find the common difference of the A.P.
\(\frac{1}{2 b}, \frac{1-6 b}{2 b}, \frac{1-12 b}{2 b}\) ……….
Solution:
-3

Question 3.
If the numbers 2n – 1, 3n + 2 and 6n – 1 are in A.P, find n and hence find the numbers.
Solution:
3; 5, 11, 17

Question 4.
Find whether 0(zero) is a term of the A.P.
40, 37, 34, 31,….
Solution:
No

Question 5.
Find x, so that 2x + 1, x² + x + 1 and 3x² – 3x + 3 are consecutive terms of an A.P.
Solution:
1, 2

Short Answer Type Questions

Question 1.
Ram Prasad saved ₹ 10 in the first week of a year and then increased his weekly savings by ₹ 2.75. If in the nth week, his savings become ₹ 59.50, find n.
Solution:
19

Question 2.
The sum of 5th and 7th terms of an A.P. is 52 and the 10th term is 46. Find the common difference.
Solution:
5

Question 3.
The 10th term of an A.P. is equal to three times its sixth term. If its 9th term is 10, find the A.P.
Solution:
-6, -4, -2, 0, 2, ………….

Question 4.
Find the sum of all odd numbers between 10 and 200.
Solution:
9975

Question 5.
If S_n denotes the sum of the first n terms of an A.P, prove that S₃₀ = 3(S₂₀ – S₁₀).
Solution:
Do it yourself

Question 6.
How many two digit numbers are divisible by 7?
Solution:
11^th

Question 7.
If a, b, c are in G.P. and a^1/x = b^1/y = c^1/z, prove that x, y, z are in A.P.
Solution:
Do it yourself

Question 8.
The first term of a G.P. is 1. The sum of the third and fifth terms is 90. Find the common ratio of the G.P.
Solution:
r = ±3

Long Answer Type Questions

Question 1.
If the 10th term of an A.P. is 21 and the sum of its first ten terms is 120, find its «th term.
Solution:
2n + 1

Question 2.
The sum of first n terms of an A.P. is 5n² + 3n. If its m^th term is 168, find the value of m. Also, find the 20^th term of this A.P.
Solution:
17; 198

Question 3.
The sum of first m terms of an A.P. is 4m² – m. If its n^th term is 107, find the value of n. Also find the 21^st term of this A.P.
Solution:
14; 163

Question 4.
Ramkali required ₹ 2500 after 12 weeks to send her daughter to school. She saved ₹ 100 in the first week and increased her weekly saving by ₹ 20 every week. Find whether she will be able to send her daughter to school after 12 weeks. What value is generated in the above situation?
Solution:
Yes! The importance and utility of savings is generated in the given question.

Question 5.
The 24^th term of an A.P. is twice its 10^th term. Show that its 72^nd term is 4 times its 15^th term.
Solution:
Do it Yourself

Question 6.
The sum of first three terms of a G.P. is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.
Solution:
r = 2, a = 16/7, S_n = \(\frac{16}{7}\) (2ⁿ – 1)

Case Study Based Questions

A stadium has rows of seats such that the first row has 100 seats, the second row has 105 seats, the third row has 110 seats, and so on.
Question 1.
What is the common difference?
Solution:
5

Question 2.
How many seats are there in the 25^th row?
Solution:
220

Question 3.
Find the total number of seats in the first 30 rows.
Solution:
5175