In this post, we provide the Karnataka SSLC Class 10 Maths Solutions for Chapter 1: Arithmetic Progressions, Exercise 1.1, along with detailed solutions in high-quality PDF format.
Class 10 Maths Solutions Chapter 1 : Arithmetic Progressions in PDF Format
Class 10 Maths Solutions Chapter 1 : Arithmetic Progressions Exercise 1.1
Question 1.
In which of the following situation, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is ₹15 for the first km and ₹8 for each additional km.
Answer:
The taxi fare structure can be analyzed as follows:
The fare for the first 1 km is ₹15.
The fare for the first 2 km is ₹15 + ₹8 = ₹23.
The fare for the first 3 km is ₹23 + ₹8 = ₹31.
The fare for the first 4 km is ₹31 + ₹8 = ₹39.
Thus, the sequence of fares — 15, 23, 31, 39, and so on — forms an arithmetic progression (A.P.), where each term increases by ₹8 from the previous one.
(ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air remaining in the cylinder at a time.
Answer:
Let the initial amount of air in the cylinder be 𝑉
V litres. Each time the vacuum pump removes 1/4 of the remaining air, meaning 3/4 of the air stays behind.
So, the air remaining after each stroke will be:
V, 3V/4 , (3V/4)2 , (3V/4)3…
This pattern shows that the amount of air is decreasing in a geometric progression (G.P.), not an arithmetic progression (A.P.), because the difference between each term is not constant
(iii) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre.
Answer:
The cost of digging can be understood as follows:
The cost of digging for the first meter is ₹150.
The cost of digging for the first 2 meters is ₹150 + ₹50 = ₹200.
The cost of digging for the first 3 meters is ₹200 + ₹50 = ₹250.
The cost of digging for the first 4 meters is ₹250 + ₹50 = ₹300.
Thus, the costs — 150, 200, 250, 300, and so on — form an arithmetic progression (A.P.) because each term increases by ₹50 from the previous one.
(iv) The amount of money in the account every year, when Rs 10000 is deposited at compound interest at 8% per annum.
Answer:
The formula for compound interest is: p(1 + r/100)power t
A is the amount in the account after t years,
P is the principal amount (initial deposit),
r is the rate of interest per annum,
t is the time in years.
In this case:
P=10000,
r=8,
t=1,2,3,….
So, the amount in the account after each year will be:
After 1 year: =10000×1.08=10800.
After 2 years: =10000×1.1664=11664.
After 3 years: =10000×1.2597=12597.
So, the amount of money in the account after each year will grow as follows:
After 1 year: ₹10,800
After 2 years: ₹11,664
After 3 years: ₹12,597
And so on, increasing every year due to compound interest.
Question 2.
Write first four terms of the AP, when the first term a and the common difference d are given as follows:
(i) a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = 1/2
(v) a = -1.25, d = -0.25
Answer:
(i) a = 10, d = 10
The first four terms of the A.P. are:
First term: a₁ = 10
Second term: a₂ = a₁ + d = 10 + 10 = 20
Third term: a₃ = a₂ + d = 20 + 10 = 30
Fourth term: a₄ = a₃ + d = 30 + 10 = 40
Thus, the first four terms of the A.P. are: 10, 20, 30, 40.
(ii) a = -2, d = 0
Since the common difference is 0, every term will be the same:
First term: a₁ = -2
Second term: a₂ = a₁ + d = -2 + 0 = -2
Third term: a₃ = a₂ + d = -2 + 0 = -2
Fourth term: a₄ = a₃ + d = -2 + 0 = -2
Thus, the first four terms of the A.P. are: -2, -2, -2, -2.
(iii) a = 4, d = -3
The first four terms of the A.P. are:
First term: a₁ = 4
Second term: a₂ = a₁ + d = 4 – 3 = 1
Third term: a₃ = a₂ + d = 1 – 3 = -2
Fourth term: a₄ = a₃ + d = -2 – 3 = -5
Thus, the first four terms of the A.P. are: 4, 1, -2, -5.
(iv) a = -1, d = 1/2
The first four terms of the A.P. are:
First term: a₁ = -1
Second term: a₂ = a₁ + d = -1 + 1/2 = -1/2
Third term: a₃ = a₂ + d = -1/2 + 1/2 = 0
Fourth term: a₄ = a₃ + d = 0 + 1/2 = 1/2
Thus, the first four terms of the A.P. are: -1, -1/2, 0, 1/2.
(v) a = -1.25, d = -0.25
The first four terms of the A.P. are:
First term: a₁ = -1.25
Second term: a₂ = a₁ + d = -1.25 – 0.25 = -1.50
Third term: a₃ = a₂ + d = -1.50 – 0.25 = -1.75
Fourth term: a₄ = a₃ + d = -1.75 – 0.25 = -2.00
Thus, the first four terms of the A.P. are: -1.25, -1.50, -1.75, -2.00.
Question 3.
For the following A.P.s, write the first term and the common difference.
(i) 3, 1, – 1, – 3 …
(ii) -5, – 1, 3, 7 …
(iii) 1/3, 5/3, 9/3, 13/3 ….
(iv) 0.6, 1.7, 2.8, 3.9 …
Answer:
(i) 3, 1, -1, -3, …
First term (a) = 3
Common difference (d) = 1 – 3 = -2
(ii) -5, -1, 3, 7, …
First term (a) = -5
Common difference (d) = -1 – (-5) = 4
(iii) 1/3, 5/3, 9/3, 13/3, …
First term (a) = 1/3
Common difference (d) = 5/3 – 1/3 = 4/3
(iv) 0.6, 1.7, 2.8, 3.9, …
First term (a) = 0.6
Common difference (d) = 1.7 – 0.6 = 1.1
Question 4.
Which of the following are APs? If they form an A.P. find the common difference d and write three more terms.
(i) 2, 4, 8, 16 …
(ii) 2, 5/2, 3, 7/2 ….
(iii) -1.2, -3.2, -5.2, -7.2 …
(iv) -10, – 6, – 2, 2 …
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2
(vi) 0.2, 0.22, 0.222, 0.2222 ….
(vii) 0, – 4, – 8, – 12 …
(viii) -1/2, -1/2, -1/2, -1/2 ….
(ix) 1, 3, 9, 27 …
(x) a, 2a, 3a, 4a …
(xi) a, a², a³, a⁴…..
(xii) √2, √8, √18, √32 …
(xiii) √3, √6, √9, √12 …
(xiv) 1², 3², 5², 7², …
(xv) 1², 5², 7², 7³, …
Answer:
(i) 2, 4, 8, 16, …
This is not an arithmetic progression (A.P.). It forms a geometric progression because the ratio between consecutive terms is constant (each term is multiplied by 2).
(ii) 2, 5/2, 3, 7/2, …
This is an A.P. because the difference between consecutive terms is constant.
First term: 2
Common difference (d): 5/2 – 2 = 1/2, 3 – 5/2 = 1/2, 7/2 – 3 = 1/2
The common difference is 1/2.
Next three terms:
7/2 + 1/2 = 4
4 + 1/2 = 9/2
9/2 + 1/2 = 5
So, the next three terms are: 4, 9/2, 5.
(iii) -1.2, -3.2, -5.2, -7.2, …
This is an A.P. because the difference between consecutive terms is constant.
First term: -1.2
Common difference (d): -3.2 – (-1.2) = -2, -5.2 – (-3.2) = -2, -7.2 – (-5.2) = -2
The common difference is -2.
Next three terms:
-7.2 + (-2) = -9.2
-9.2 + (-2) = -11.2
-11.2 + (-2) = -13.2
So, the next three terms are: -9.2, -11.2, -13.2.
(iv) -10, -6, -2, 2, …
This is an A.P. because the difference between consecutive terms is constant.
First term: -10
Common difference (d): -6 – (-10) = 4, -2 – (-6) = 4, 2 – (-2) = 4
The common difference is 4.
Next three terms:
2 + 4 = 6
6 + 4 = 10
10 + 4 = 14
So, the next three terms are: 6, 10, 14.
(v) 3, 3 + √2, 3 + 2√2, 3 + 3√2, …
This is an A.P. because the difference between consecutive terms is constant.
First term: 3
Common difference (d): (3 + √2) – 3 = √2, (3 + 2√2) – (3 + √2) = √2, (3 + 3√2) – (3 + 2√2) = √2
The common difference is √2.
Next three terms:
3 + 4√2
3 + 5√2
3 + 6√2
So, the next three terms are: 3 + 4√2, 3 + 5√2, 3 + 6√2.
(vi) 0.2, 0.22, 0.222, 0.2222, …
This is not an A.P. because the difference between terms is not constant. This sequence is a pattern of adding more decimal places of 2.
(vii) 0, -4, -8, -12, …
This is an A.P. because the difference between consecutive terms is constant.
First term: 0
Common difference (d): -4 – 0 = -4, -8 – (-4) = -4, -12 – (-8) = -4
The common difference is -4.
Next three terms:
-12 + (-4) = -16
-16 + (-4) = -20
-20 + (-4) = -24
So, the next three terms are: -16, -20, -24.
(viii) -1/2, -1/2, -1/2, -1/2, …
This is an A.P. because all the terms are the same, so the common difference is 0.
First term: -1/2
Common difference (d): 0 (all terms are the same)
Next three terms:
The sequence stays the same: -1/2, -1/2, -1/2.
(ix) 1, 3, 9, 27, …
This is not an A.P. because the difference between consecutive terms is not constant. This sequence is a geometric progression, where each term is multiplied by 3.
(x) a, 2a, 3a, 4a, …
This is an A.P. because the difference between consecutive terms is constant.
First term: a
Common difference (d): 2a – a = a, 3a – 2a = a, 4a – 3a = a
The common difference is a.
Next three terms:
4a + a = 5a
5a + a = 6a
6a + a = 7a
So, the next three terms are: 5a, 6a, 7a.
(xi) a, a², a³, a⁴, …
This is not an A.P. because the terms are powers of ‘a’, and the difference between them is not constant.
(xii) √2, √8, √18, √32, …
This is not an A.P. because the difference between consecutive terms is not constant. The terms are based on square roots, and the differences are not fixed.
(xiii) √3, √6, √9, √12, …
This is not an A.P. because the difference between consecutive terms is not constant. The terms are square roots of multiples of 3, and the differences do not form a constant value.
(xiv) 1², 3², 5², 7², …
This is not an A.P. because the terms are squares of odd numbers, and the difference between consecutive terms is not constant.
(xv) 1², 5², 7², 7³, …
This is not an A.P. because the terms involve powers of numbers, and the difference between them is not constant.
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