Table of Contents
Exercise 6.1 Solutions – Class X Mathematics
These solutions are based on the Telangana State Class X Mathematics textbook, focusing on arithmetic progressions (AP), identifying whether sequences form an AP, and finding terms, first terms, and common differences. Mathematical expressions are rendered using MathJax.
1. In which of the following situations, does the list of numbers involved form an arithmetic progression, and why?
(i) The minimum taxi fare is ₹20 for the first km and thereafter ₹8 for each additional km.
Fare for 1 km: ₹20, for 2 km: ₹20 + ₹8 = ₹28, for 3 km: ₹28 + ₹8 = ₹36, for 4 km: ₹36 + ₹8 = ₹44.
Sequence: 20, 28, 36, 44, …
Check differences: 28 – 20 = 8, 36 – 28 = 8, 44 – 36 = 8.
Since the difference is constant (8), this forms an arithmetic progression (AP).
Forms an AP: Yes, Reason: Constant difference of 8
(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.
Let initial air = \( V \). After 1st removal: \( V – \frac{1}{4}V = \frac{3}{4}V \).
After 2nd removal: \( \frac{3}{4}V – \frac{1}{4} \left( \frac{3}{4}V \right) = \frac{3}{4}V \cdot \frac{3}{4} = \left( \frac{3}{4} \right)^2 V \).
After 3rd removal: \( \left( \frac{3}{4} \right)^2 V \cdot \frac{3}{4} = \left( \frac{3}{4} \right)^3 V \).
Sequence: \( V, \frac{3}{4}V, \left( \frac{3}{4} \right)^2 V, \left( \frac{3}{4} \right)^3 V, \ldots \).
Ratios: \( \frac{\text{second term}}{\text{first term}} = \frac{3}{4} \), \( \frac{\text{third term}}{\text{second term}} = \frac{3}{4} \), constant ratio implies geometric progression (GP), not AP.
Forms an AP: No, Reason: Forms a geometric progression (GP) with common ratio \( \frac{3}{4} \)
(iii) The cost of digging a well, after every metre of digging, when it costs ₹150 for the first metre and rises by ₹50 for each subsequent metre.
Cost for 1st metre: ₹150, 2nd: ₹150 + ₹50 = ₹200, 3rd: ₹200 + ₹50 = ₹250, 4th: ₹250 + ₹50 = ₹300.
Sequence: 150, 200, 250, 300, …
Differences: 200 – 150 = 50, 250 – 200 = 50, 300 – 250 = 50.
Constant difference (50), so it forms an AP.
Forms an AP: Yes, Reason: Constant difference of 50
(iv) The amount of money in the account every year, when ₹10000 is deposited at compound interest at 8% per annum.
Principal = ₹10000, rate = 8%. Amount after 1 year: \( 10000 \left(1 + \frac{8}{100}\right) = 10000 \cdot 1.08 = 10800 \).
After 2 years: \( 10800 \cdot 1.08 = 10000 \cdot (1.08)^2 = 11664 \).
After 3 years: \( 10000 \cdot (1.08)^3 \approx 12597.12 \).
Sequence: 10000, 10800, 11664, 12597.12, …
Ratios: \( \frac{10800}{10000} = 1.08 \), \( \frac{11664}{10800} = 1.08 \), constant ratio implies GP, not AP.
Forms an AP: No, Reason: Forms a geometric progression (GP) with common ratio 1.08
2. Write first four terms of the AP, when the first term \( a \) and the common difference \( d \) are given as follows:
(i) \( a = 10, d = 10 \)
AP: \( a, a + d, a + 2d, a + 3d, \ldots \).
First term: 10.
Second term: \( 10 + 10 = 20 \).
Third term: \( 20 + 10 = 30 \).
Fourth term: \( 30 + 10 = 40 \).
First four terms: 10, 20, 30, 40
(ii) \( a = -2, d = 0 \)
First term: -2.
Second term: \( -2 + 0 = -2 \).
Third term: \( -2 + 0 = -2 \).
Fourth term: \( -2 + 0 = -2 \).
First four terms: -2, -2, -2, -2
(iii) \( a = 4, d = -3 \)
First term: 4.
Second term: \( 4 + (-3) = 1 \).
Third term: \( 1 + (-3) = -2 \).
Fourth term: \( -2 + (-3) = -5 \).
First four terms: 4, 1, -2, -5
(iv) \( a = -1, d = \frac{1}{2} \)
First term: -1.
Second term: \( -1 + \frac{1}{2} = -\frac{1}{2} \).
Third term: \( -\frac{1}{2} + \frac{1}{2} = 0 \).
Fourth term: \( 0 + \frac{1}{2} = \frac{1}{2} \).
First four terms: -1, -\frac{1}{2}, 0, \frac{1}{2}
(v) \( a = -1.25, d = -0.25 \)
First term: -1.25.
Second term: \( -1.25 + (-0.25) = -1.5 \).
Third term: \( -1.5 + (-0.25) = -1.75 \).
Fourth term: \( -1.75 + (-0.25) = -2 \).
First four terms: -1.25, -1.5, -1.75, -2
3. For the following APs, write the first term and the common difference:
(i) 3, 1, -1, -3, …
First term (\( a \)): 3.
Common difference (\( d \)): \( 1 – 3 = -2 \), \( -1 – 1 = -2 \), \( -3 – (-1) = -2 \).
First term: 3, Common difference: -2
(ii) -5, -1, 3, 7, …
First term: -5.
Common difference: \( -1 – (-5) = 4 \), \( 3 – (-1) = 4 \), \( 7 – 3 = 4 \).
First term: -5, Common difference: 4
(iii) \( \frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \ldots \)
Simplify: \( \frac{1}{3}, \frac{5}{3}, 3, \frac{13}{3}, \ldots \).
First term: \( \frac{1}{3} \).
Common difference: \( \frac{5}{3} – \frac{1}{3} = \frac{4}{3} \), \( 3 – \frac{5}{3} = \frac{4}{3} \), \( \frac{13}{3} – 3 = \frac{4}{3} \).
First term: \( \frac{1}{3} \), Common difference: \( \frac{4}{3} \)
(iv) 0.6, 1.7, 2.8, 3.9, …
First term: 0.6.
Common difference: \( 1.7 – 0.6 = 1.1 \), \( 2.8 – 1.7 = 1.1 \), \( 3.9 – 2.8 = 1.1 \).
First term: 0.6, Common difference: 1.1
4. Which of the following are APs? If they form an AP, find the common difference \( d \) and write the next three terms.
(i) 2, 4, 8, 16, …
Differences: \( 4 – 2 = 2 \), \( 8 – 4 = 4 \), \( 16 – 8 = 8 \).
Differences are not constant (2, 4, 8), so not an AP.
Is an AP: No
(ii) \( 2, \frac{5}{2}, 3, \frac{7}{2}, \ldots \)
Differences: \( \frac{5}{2} – 2 = \frac{1}{2} \), \( 3 – \frac{5}{2} = \frac{1}{2} \), \( \frac{7}{2} – 3 = \frac{1}{2} \).
Constant difference: \( \frac{1}{2} \), so it is an AP.
Next terms: \( \frac{7}{2} + \frac{1}{2} = 4 \), \( 4 + \frac{1}{2} = \frac{9}{2} \), \( \frac{9}{2} + \frac{1}{2} = 5 \).
Is an AP: Yes, Common difference: \( \frac{1}{2} \), Next three terms: 4, \frac{9}{2}, 5
(iii) -1.2, -3.2, -5.2, -7.2, …
Differences: \( -3.2 – (-1.2) = -2 \), \( -5.2 – (-3.2) = -2 \), \( -7.2 – (-5.2) = -2 \).
Constant difference: -2, so it is an AP.
Next terms: \( -7.2 + (-2) = -9.2 \), \( -9.2 + (-2) = -11.2 \), \( -11.2 + (-2) = -13.2 \).
Is an AP: Yes, Common difference: -2, Next three terms: -9.2, -11.2, -13.2
(iv) -10, -6, -2, 2, …
Differences: \( -6 – (-10) = 4 \), \( -2 – (-6) = 4 \), \( 2 – (-2) = 4 \).
Constant difference: 4, so it is an AP.
Next terms: \( 2 + 4 = 6 \), \( 6 + 4 = 10 \), \( 10 + 4 = 14 \).
Is an AP: Yes, Common difference: 4, Next three terms: 6, 10, 14
(v) \( 3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}, \ldots \)
Differences: \( 3 + \sqrt{2} – 3 = \sqrt{2} \), \( 3 + 2\sqrt{2} – (3 + \sqrt{2}) = \sqrt{2} \), \( 3 + 3\sqrt{2} – (3 + 2\sqrt{2}) = \sqrt{2} \).
Constant difference: \( \sqrt{2} \), so it is an AP.
Next terms: \( 3 + 4\sqrt{2} \), \( 3 + 5\sqrt{2} \), \( 3 + 6\sqrt{2} \).
Is an AP: Yes, Common difference: \( \sqrt{2} \), Next three terms: \( 3 + 4\sqrt{2}, 3 + 5\sqrt{2}, 3 + 6\sqrt{2} \)
(vi) 0.2, 0.22, 0.222, 0.2222, …
Differences: \( 0.22 – 0.2 = 0.02 \), \( 0.222 – 0.22 = 0.002 \), \( 0.2222 – 0.222 = 0.0002 \).
Differences are not constant, so not an AP.
Is an AP: No
(vii) 0, -4, -8, -12, …
Differences: \( -4 – 0 = -4 \), \( -8 – (-4) = -4 \), \( -12 – (-8) = -4 \).
Constant difference: -4, so it is an AP.
Next terms: \( -12 + (-4) = -16 \), \( -16 + (-4) = -20 \), \( -20 + (-4) = -24 \).
Is an AP: Yes, Common difference: -4, Next three terms: -16, -20, -24
(viii) \( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \ldots \)
Differences: \( -\frac{1}{2} – (-\frac{1}{2}) = 0 \), \( -\frac{1}{2} – (-\frac{1}{2}) = 0 \).
Constant difference: 0, so it is an AP.
Next terms: \( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2} \).
Is an AP: Yes, Common difference: 0, Next three terms: \( -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2} \)
(ix) 1, 3, 9, 27, …
Differences: \( 3 – 1 = 2 \), \( 9 – 3 = 6 \), \( 27 – 9 = 18 \).
Differences are not constant, so not an AP.
Is an AP: No
(x) \( a, 2a, 3a, 4a, \ldots \)
Differences: \( 2a – a = a \), \( 3a – 2a = a \), \( 4a – 3a = a \).
Constant difference: \( a \), so it is an AP.
Next terms: \( 5a, 6a, 7a \).
Is an AP: Yes, Common difference: \( a \), Next three terms: \( 5a, 6a, 7a \)
(xi) \( a, a^2, a^3, a^4, \ldots \)
Differences: \( a^2 – a = a(a – 1) \), \( a^3 – a^2 = a^2(a – 1) \), not constant unless \( a = 1 \), which is a special case.
Generally, not an AP unless specified otherwise.
Is an AP: No
(xii) \( \sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots \)
Simplify: \( \sqrt{2}, \sqrt{8} = 2\sqrt{2}, \sqrt{18} = 3\sqrt{2}, \sqrt{32} = 4\sqrt{2} \).
Sequence: \( \sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \ldots \).
Differences: \( 2\sqrt{2} – \sqrt{2} = \sqrt{2} \), \( 3\sqrt{2} – 2\sqrt{2} = \sqrt{2} \), \( 4\sqrt{2} – 3\sqrt{2} = \sqrt{2} \).
Constant difference: \( \sqrt{2} \), so it is an AP.
Next terms: \( 5\sqrt{2}, 6\sqrt{2}, 7\sqrt{2} \).
Is an AP: Yes, Common difference: \( \sqrt{2} \), Next three terms: \( 5\sqrt{2}, 6\sqrt{2}, 7\sqrt{2} \)
(xiii) \( \sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots \)
Simplify: \( \sqrt{3}, \sqrt{6}, \sqrt{9} = 3, \sqrt{12} = 2\sqrt{3} \).
Differences: \( \sqrt{6} – \sqrt{3} \), \( 3 – \sqrt{6} \), \( 2\sqrt{3} – 3 \), not constant (numerically different).
Not an AP.
Is an AP: No
