NCERT Class 10 Maths Chapter 5 ARITHMETIC PROGRESSIONS 5.1 Solutions

Question 1: In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?

(i) The taxi fare after each km when the fare is \( \text{₹}\,15 \) for the first km and \( \text{₹}\,8 \) for each additional km.

Solution:

• Taxi fare for 1 km = \( \text{₹}\,15 \)
• Taxi fare for 2 km = \( \text{₹}\,15 \) + \( \text{₹}\,8 \) = \( \text{₹}\,23 \)
• Taxi fare for 3 km = \( \text{₹}\,23 \) + \( \text{₹}\,8 \) = \( \text{₹}\,31 \)
• Taxi fare for 4 km = \( \text{₹}\,31 \) + \( \text{₹}\,8 \) = \( \text{₹}\,39 \)

Sequence: 15, 23, 31, 39, …

The common differences are:

a₂-a₁= 23 – 15 = 8

a₃-a₂= 31 – 23 = 8

a₄-a₃= 39 – 31 = 8

The common difference is constant (d = 8). Hence, this forms an arithmetic progression because each term is obtained by adding a fixed number (8) to the preceding term.

(ii) The amount of air present in a cylinder when a vacuum pump removes \( \frac{1}{4} \) of the air remaining in the cylinder at a time.

Solution:

Let the initial volume of air be \( V \) liters.

After 1st stroke: Remaining air = \( V – \frac{1}{4}V = \frac{3}{4}V \)

After 2nd stroke: Remaining air = \( \frac{3}{4}V – \frac{1}{4} \cdot \frac{3}{4}V = \frac{3}{4} \cdot \frac{3}{4}V = \frac{9}{16}V \)

After 3rd stroke: Remaining air = \( \frac{9}{16}V – \frac{1}{4} \cdot \frac{9}{16}V = \frac{9}{16} \cdot \frac{3}{4}V = \frac{27}{64}V \)

Therefore, the volumes will be \( V, \frac{3}{4}V, \frac{9}{16}V, \frac{27}{64}V, … \) …and so on.

The differences are-

• \( \frac{3}{4}V – V = -\frac{1}{4}V \)
• \( \frac{9}{16}V – \frac{3}{4}V = \frac{9}{16}V – \frac{12}{16}V = -\frac{3}{16}V \)

The differences are not constant \(( -\frac{1}{4}V \neq -\frac{3}{16}V )\). we can clearly see that the adjacent terms of this series do not have common difference. Hence, this does not form an arithmetic progression.

(iii) The cost of digging a well after every meter of digging, when it costs \( \text{₹}\,150 \) for the first meter and rises by \( \text{₹}\,50 \) for each subsequent meter.

Solution:

• Cost for 1st meter = \( \text{₹}\,150 \)
• Cost for 2nd meter = \( \text{₹}\,150 \) + \( \text{₹}\,50 \) = \( \text{₹}\,200 \)
• Cost for 3rd meter = \( \text{₹}\,200 \) + \( \text{₹}\,50 \) = \( \text{₹}\,250 \)
• Cost for 4th meter = \( \text{₹}\,250 \) + \( \text{₹}\,50 \) = \( \text{₹}\,300 \)
• Sequence: 150, 200, 250, 300, …

The common differences are:

• a₂-a₁= 200 – 150 = 50
• a₃-a₂= 250 – 200 = 50
• a₄-a₃= 300 – 250 = 50

The common difference is constant (d = 50). Hence, this forms an arithmetic progression because each term increases by a fixed amount (50).

(iv) The amount of money in the account every year, when \( \text{₹}\,10,000 \) is deposited at compound interest at 8% per annum.

Solution:

• Amount after 1 year = \( 10000 \left(1 + \frac{8}{100}\right) = 10000 \times 1.08 = 10800 \)
• Amount after 2 years = \( 10000 \left(1 + \frac{8}{100}\right)^2 = 10000 \times (1.08)^2 = 11664 \)
• Amount after 3 years = \( 10000 \left(1 + \frac{8}{100}\right)^3 = 10000 \times (1.08)^3 \approx 12597.12 \)
• Sequence: 10000, 10800, 11664, 12597.12, …

The common differences are:

• ( 10800 – 10000 = 800 )
• ( 11664 – 10800 = 864 )
• ( 12597.12 – 11664 \approx 933.12 )

The differences are not constant \(( 800 \neq 864 \neq 933.12 )\). Hence, this does not form an arithmetic progression because the amounts grow exponentially due to compound interest.

Question 2: Write the first four terms of the AP, when the first term (a) and the common difference (d) are given as follows:

(i) \( a = 10, d = 10 \)

Solution:

• First term \(( a_1 )\) = \( a = 10 \)
• Second term \(( a_2 )\) = \( a_1 + d = 10 + 10 = 20 \)
• Third term \(( a_3 )\) = \( a_2 + d = 20 + 10 = 30 \)
• Fourth term \(( a_4 )\) = \( a_3 + d = 30 + 10 = 40 \)

Answer: The first four terms are \( 10, 20, 30, 40 \).

(ii) \( a = -2, d = 0 \)

Solution:

• First term \(( a_1 )\) = \( a = -2 \)
• Second term \(( a_2 )\) = \( a_1 + d = -2 + 0 = -2 \)
• Third term \(( a_3 )\) = \( a_2 + d = -2 + 0 = -2 \)
• Fourth term \(( a_4 )\) = \( a_3 + d = -2 + 0 = -2 \)

Answer: The first four terms are \( -2, -2, -2, -2 \).

(iii) \( a = 4, d = -3 \)

Solution:

• First term \(( a_1 )\) = \( a = 4 \)
• Second term \(( a_2 )\) = \( a_1 + d = 4 + (-3) = 1 \)
• Third term \(( a_3 )\) = \( a_2 + d = 1 + (-3) = -2 \)
• Fourth term \(( a_4 )\) = \( a_3 + d = -2 + (-3) = -5 \)

Answer: The first four terms are \( 4, 1, -2, -5 \).

(iv) \( a = -1, d = \frac{1}{2} \)

Solution:

• First term \(( a_1 )\) = \( a = -1 \)
• Second term \(( a_2 )\) = \( a_1 + d = -1 + \frac{1}{2} = -\frac{1}{2} \)
• Third term \(( a_3 )\) = \( a_2 + d = -\frac{1}{2} + \frac{1}{2} = 0 \)
• Fourth term \(( a_4 )\) = \( a_3 + d = 0 + \frac{1}{2} = \frac{1}{2} \)

Answer: The first four terms are \( -1, -\frac{1}{2}, 0, \frac{1}{2} \).

(v) \( a = -1.25, d = -0.25 \)

Solution:

• First term \(( a_1 )\) = \( a = -1.25 \)
• Second term \(( a_2 )\) = \( a_1 + d = -1.25 + (-0.25) = -1.50 \)
• Third term \(( a_3 )\) = \( a_2 + d = -1.50 + (-0.25) = -1.75 \)
• Fourth term \(( a_4 )\) = \( a_3 + d = -1.75 + (-0.25) = -2.00 \)

Answer: The first four terms are \( -1.25, -1.50, -1.75, -2.00 \).

Question 3: For the following APs, write the first term and the common difference:

(i) \( 3, 1, -1, -3, … \)

Solution:

• First term \(( a )\) = \( 3 \)
• Common difference \(( d )\) = Second term(a₂) – First term(a₁) = \( 1 – 3 = -2 \)

Answer: First term = \( 3 \), Common difference = \( -2 \).

(ii) \( -5, -1, 3, 7, … \)

Solution:

• First term \(( a )\) = \( -5 \)
• Common difference \(( d )\) = Second term(a₂) – First term(a₁) = \( -1 – (-5) = -1 + 5 = 4 \)

Answer: First term = \( -5 \), Common difference = \( 4 \).

(iii) \( \frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, … \)

Solution:

• First term \(( a )\) = \( \frac{1}{3} \)
• Common difference \(( d )\) = Second term(a₂) – First term(a₁) = \( \frac{5}{3} – \frac{1}{3} = \frac{4}{3} \)

Answer: First term = \( \frac{1}{3} \), Common difference = \( \frac{4}{3} \).

(iv) \( 0.6, 1.7, 2.8, 3.9, … \)

Solution:

• First term \(( a )\) = \( 0.6 \)
• Common difference \(( d )\) = Second term(a₂) – First term(a₁) = \( 1.7 – 0.6 = 1.1 \)

Answer: First term = \( 0.6 \), Common difference = \( 1.1 \).

Question 4: Which of the following are APs? If they form an AP, find the common difference ( d ) and write three more terms.

(i) \( 2, 4, 8, 16, … \)

Solution:

The differences are:

• \( a_2 – a_1 = 4 – 2 = 2 \)
• \( a_3 – a_2 = 8 – 4 = 4 \)
• \( a_4 – a_3 = 16 – 8 = 8 \)

Since \( 2 \neq 4 \neq 8 \), the differences are not constant.

Answer: This is not an AP.

(ii) \( 2, \frac{5}{2}, 3, \frac{7}{2}, … \)

Solution:

The differences:

• \( a_2 – a_1 = \frac{5}{2} – 2 = \frac{5}{2} – \frac{4}{2} = \frac{1}{2} \)
• \( a_3 – a_2 = 3 – \frac{5}{2} = \frac{6}{2} – \frac{5}{2} = \frac{1}{2} \)
• \( a_4 – a_3 = \frac{7}{2} – 3 = \frac{7}{2} – \frac{6}{2} = \frac{1}{2} \)

The common difference is constant \(( d = \frac{1}{2} )\). This is an AP.

Next three terms are:

• \( a_5 = \frac{7}{2} + \frac{1}{2} = \frac{8}{2} = 4 \)
• \( a_6 = 4 + \frac{1}{2} = \frac{9}{2} \)
• \( a_7 = \frac{9}{2} + \frac{1}{2} = \frac{10}{2} = 5 \)

Answer: This is an AP with \( d = \frac{1}{2} \). Next three terms are: \( 4, \frac{9}{2}, 5 \).

(iii) \( -1.2, -3.2, -5.2, -7.2, … \)

Solution:

The differences are:

• \( a_2 – a_1 = -3.2 – (-1.2) = -3.2 + 1.2 = -2 \)
• \( a_3 – a_2 = -5.2 – (-3.2) = -5.2 + 3.2 = -2 \)
• \( a_4 – a_3 = -7.2 – (-5.2) = -7.2 + 5.2 = -2 \)

The common difference is constant \(( d = -2 )\). This is an AP.

Next three terms are:

• \( a_5 = -7.2 + (-2) = -9.2 \)
• \( a_6 = -9.2 + (-2) = -11.2 \)
• \( a_7 = -11.2 + (-2) = -13.2 \)

Answer: This is an AP with \( d = -2 \). Next three terms are: \( -9.2, -11.2, -13.2 \).

(iv) \( -10, -6, -2, 2, … \)

Solution:

The differences are:

• \( a_2 – a_1 = -6 – (-10) = -6 + 10 = 4 \)
• \( a_3 – a_2 = -2 – (-6) = -2 + 6 = 4 \)
• \( a_4 – a_3 = 2 – (-2) = 2 + 2 = 4 \)

The common difference is constant \(( d = 4 )\). This is an AP.

Next three terms are:

• \( a_5 = 2 + 4 = 6 \)
• \( a_6 = 6 + 4 = 10 \)
• \( a_7 = 10 + 4 = 14 \)

Answer: This is an AP with \( d = 4 \). Next three terms are: \( 6, 10, 14 \).

(v) \( 3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, … \)

Solution:

The differences of the following sequence are:

• \( a_2 – a_1 = (3 + \sqrt{2}) – 3 = \sqrt{2} \)
• \( a_3 – a_2 = (3 + 2\sqrt{2}) – (3 + \sqrt{2}) = \sqrt{2} \)
• \( a_4 – a_3 = (3 + 3\sqrt{2}) – (3 + 2\sqrt{2}) = \sqrt{2} \)

The common difference is constant \(( d = \sqrt{2} )\). This is an AP.

Next three terms are:

• \( a_5 = (3 + 3\sqrt{2}) + \sqrt{2} = 3 + 4\sqrt{2} \)
• \( a_6 = (3 + 4\sqrt{2}) + \sqrt{2} = 3 + 5\sqrt{2} \)
• \( a_7 = (3 + 5\sqrt{2}) + \sqrt{2} = 3 + 6\sqrt{2} \)

Answer: This is an AP with \( d = \sqrt{2} \). Next three terms are: \( 3 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2} \).

(vi) \( 0.2, 0.22, 0.222, 0.2222, … \)

Solution:

The differences of the sequence are:

• \( a_2 – a_1 = 0.22 – 0.2 = 0.02 \)
• \( a_3 – a_2 = 0.222 – 0.22 = 0.002 \)
• \( a_4 – a_3 = 0.2222 – 0.222 = 0.0002 \)

Since \( 0.02 \neq 0.002 \neq 0.0002 \), the differences are not constant.

Answer: This is not an AP.

(vii) \( 0, -4, -8, -12, … \)

Solution:

The differences of the given sequence are:

• \( a_2 – a_1 = -4 – 0 = -4 \)
• \( a_3 – a_2 = -8 – (-4) = -8 + 4 = -4 \)
• \( a_4 – a_3 = -12 – (-8) = -12 + 8 = -4 \)

The common difference is constant \(( d = -4 )\). This is an AP.

Next three terms are:

• \( a_5 = -12 + (-4) = -16 \)
• \( a_6 = -16 + (-4) = -20 \)
• \( a_7 = -20 + (-4) = -24 \)

Answer: This is an AP with \( d = -4 \). Next three terms are: \( -16, -20, -24 \).

(viii) \( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, … \)

Solution:

The differences of the sequence are:

• \( a_2 – a_1 = -\frac{1}{2} – (-\frac{1}{2}) = 0 \)
• \( a_3 – a_2 = -\frac{1}{2} – (-\frac{1}{2}) = 0 \)
• \( a_4 – a_3 = -\frac{1}{2} – (-\frac{1}{2}) = 0 \)

The common difference is constant \(( d = 0 )\). This is an AP.

Next three terms are:

• \( a_5 = -\frac{1}{2} + 0 = -\frac{1}{2} \)
• \( a_6 = -\frac{1}{2} + 0 = -\frac{1}{2} \)
• \( a_7 = -\frac{1}{2} + 0 = -\frac{1}{2} \)

Answer: This is an AP with \( d = 0 \). Next three terms are: \( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2} \).

(ix) \( 1, 3, 9, 27, … \)

Solution:

The differences of the given sequence are:

• \( a_2 – a_1 = 3 – 1 = 2 \)
• \( a_3 – a_2 = 9 – 3 = 6 \)
• \( a_4 – a_3 = 27 – 9 = 18 \)
• Since \( 2 \neq 6 \neq 18 \), the differences are not constant.

Answer: This is not an AP.

(x) \( a, 2a, 3a, 4a, … \)

Solution:

The differences of the given sequence are:

• \( a_2 – a_1 = 2a – a = a \)
• \( a_3 – a_2 = 3a – 2a = a \)
• \( a_4 – a_3 = 4a – 3a = a \)

The common difference is constant \(( d = a )\). This is an AP.

Next three terms are:

• \( a_5 = 4a + a = 5a \)
• \( a_6 = 5a + a = 6a \)
• \( a_7 = 6a + a = 7a \)

Answer: This is an AP with \( d = a \). Next three terms are: \( 5a, 6a, 7a \).

(xi) \( a, a^2, a^3, a^4, … \)

Solution:

The differences of the following sequence are:

• \( a_2 – a_1 = a^2 – a = a(a – 1) \)
• \( a_3 – a_2 = a^3 – a^2 = a^2(a – 1) \)
• \( a_4 – a_3 = a^4 – a^3 = a^3(a – 1) \)

Since \( a(a – 1) \neq a^2(a – 1) \neq a^3(a – 1) \) (unless \( a = 1 \), but generally not constant), the differences are not constant.

Answer: This is not an AP.

(xii) \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, … \)

Solution:

Simplifying the terms we get:

\( \sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2} \),

\( \sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2} \),

\( \sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2} \)

Sequence: \( \sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, … \)

The differences of the following sequence are:

• \( a_2 – a_1 = 2\sqrt{2} – \sqrt{2} = \sqrt{2} \)
• \( a_3 – a_2 = 3\sqrt{2} – 2\sqrt{2} = \sqrt{2} \)
• \( a_4 – a_3 = 4\sqrt{2} – 3\sqrt{2} = \sqrt{2} \)

The common difference is constant \(( d = \sqrt{2} )\). This is an AP.

Next three terms:

• \( a_5 = 4\sqrt{2} + \sqrt{2} = 5\sqrt{2} = \sqrt{50} \)
• \( a_6 = 5\sqrt{2} + \sqrt{2} = 6\sqrt{2} = \sqrt{72} \)
• \( a_7 = 6\sqrt{2} + \sqrt{2} = 7\sqrt{2} = \sqrt{98} \)

Answer: This is an AP with \( d = \sqrt{2} \). Next three terms are: \( \sqrt{50}, \sqrt{72}, \sqrt{98} \).

(xiii) \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, … \)

Solution:

The differences of the given sequence are:

• \( a_2 – a_1 = \sqrt{6} – \sqrt{3} = \sqrt{3}(\sqrt{2} – 1) \)
• \( a_3 – a_2 = \sqrt{9} – \sqrt{6} = 3 – \sqrt{6} = \sqrt{3}(\sqrt{3} – \sqrt{2}) \)
• \( a_4 – a_3 = \sqrt{12} – \sqrt{9} = 2\sqrt{3} – 3 = \sqrt{3}(2 – \sqrt{3}) \)

Since \( \sqrt{3}(\sqrt{2} – 1) \neq \sqrt{3}(\sqrt{3} – \sqrt{2}) \neq \sqrt{3}(2 – \sqrt{3}) \), the differences are not constant.

Answer: This is not an AP.

(xiv) \( 1^2, 3^2, 5^2, 7^2, … \) \(i.e., ( 1, 9, 25, 49, … )\)

Solution:

The differences are:

• \( a_2 – a_1 = 9 – 1 = 8 \)
• \( a_3 – a_2 = 25 – 9 = 16 \)
• \( a_4 – a_3 = 49 – 25 = 24 \)

Since \( 8 \neq 16 \neq 24 \), the differences are not constant.

Answer: This is not an AP.

(xv) \( 1^2, 5^2, 7^2, 73, … \) \(i.e., ( 1, 25, 49, 73, … )\)

Solution:

The differences are:

• \( a_2 – a_1 = 25 – 1 = 24 \)
• \( a_3 – a_2 = 49 – 25 = 24 \)
• \( a_4 – a_3 = 73 – 49 = 24 \)

The common difference is constant \(( d = 24 )\). This is an AP.

Next three terms are:

• \( a_5 = 73 + 24 = 97 \)
• \( a_6 = 97 + 24 = 121 \)
• \( a_7 = 121 + 24 = 145 \)

Answer: This is an AP with \( d = 24 \). Next three terms are: \( 97, 121, 145 \).