NCERT Solutions Class 10 Maths Chapter 5: Arithmetic Progressions
Detailed solutions for Exercise 5.1 (Introduction to Arithmetic Progressions).
#### **Q1. 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 ₹ 15 for the first km and ₹ 8 for each additional km.**
**(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.**
**(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.**
**(iv) The amount of money in the account every year, when ₹ 10000 is deposited at compound interest at 8 % per annum.**
#### **A1. Solution:**
**(i)**
Fare for 1st km = ₹ 15
Fare for 2nd km = $15 + 8 = 23$
Fare for 3rd km = $23 + 8 = 31$
List: $15, 23, 31, 39, \dots$
Since the difference between consecutive terms is constant ($d = 8$), **it is an AP.**
**(ii)**
Let initial volume = $V$.
After 1st removal: $V - \frac{1}{4}V = \frac{3}{4}V$
After 2nd removal: $\frac{3}{4}V - \frac{1}{4}(\frac{3}{4}V) = \frac{3}{4}V - \frac{3}{16}V = \frac{12V - 3V}{16} = \frac{9}{16}V = (\frac{3}{4})^2 V$
List: $V, \frac{3}{4}V, (\frac{3}{4})^2 V, \dots$
Difference 1: $\frac{3}{4}V - V = -\frac{1}{4}V$
Difference 2: $\frac{9}{16}V - \frac{3}{4}V = \frac{9V - 12V}{16} = -\frac{3}{16}V$
Differences are not constant. **It is NOT an AP.**
**(iii)**
Cost for 1st metre = ₹ 150
Cost for 2nd metre = $150 + 50 = 200$
Cost for 3rd metre = $200 + 50 = 250$
List: $150, 200, 250, \dots$
Difference is constant ($d = 50$). **It is an AP.**
**(iv)**
Principal = 10000, Rate = 8%.
Year 1 Amount: $10000(1 + \frac{8}{100})^1$
Year 2 Amount: $10000(1 + \frac{8}{100})^2$
Year 3 Amount: $10000(1 + \frac{8}{100})^3$
The terms increase by a factor (multiplication), not by a fixed difference. **It is NOT an AP.**
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#### **Q2. 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)** $a = -1, d = \frac{1}{2}$
**(v)** $a = -1.25, d = -0.25$
#### **A2. Solution:**
**(i) $a = 10, d = 10$**
$a_1 = 10$
$a_2 = 10 + 10 = 20$
$a_3 = 20 + 10 = 30$
$a_4 = 30 + 10 = 40$
**Terms: 10, 20, 30, 40**
**(ii) $a = -2, d = 0$**
$a_1 = -2$
$a_2 = -2 + 0 = -2$
$a_3 = -2$
$a_4 = -2$
**Terms: -2, -2, -2, -2**
**(iii) $a = 4, d = -3$**
$a_1 = 4$
$a_2 = 4 - 3 = 1$
$a_3 = 1 - 3 = -2$
$a_4 = -2 - 3 = -5$
**Terms: 4, 1, -2, -5**
**(iv) $a = -1, d = \frac{1}{2}$**
$a_1 = -1$
$a_2 = -1 + 0.5 = -0.5$ (or $-\frac{1}{2}$)
$a_3 = -0.5 + 0.5 = 0$
$a_4 = 0 + 0.5 = 0.5$ (or $\frac{1}{2}$)
**Terms: -1, $-\frac{1}{2}$, 0, $\frac{1}{2}$**
**(v) $a = -1.25, d = -0.25$**
$a_1 = -1.25$
$a_2 = -1.25 - 0.25 = -1.50$
$a_3 = -1.50 - 0.25 = -1.75$
$a_4 = -1.75 - 0.25 = -2.00$
**Terms: -1.25, -1.50, -1.75, -2.00**
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#### **Q3. For the following APs, write the first term and the common difference:**
**(i)** $3, 1, -1, -3, \dots$
**(ii)** $-5, -1, 3, 7, \dots$
**(iii)** $\frac{1}{3}, \frac{5}{3}, \frac{9}{3}, \frac{13}{3}, \dots$
**(iv)** $0.6, 1.7, 2.8, 3.9, \dots$
#### **A3. Solution:**
**(i)** $a = 3$, $d = 1 - 3 = -2$. **Ans: $a=3, d=-2$**
**(ii)** $a = -5$, $d = -1 - (-5) = -1 + 5 = 4$. **Ans: $a=-5, d=4$**
**(iii)** $a = \frac{1}{3}$, $d = \frac{5}{3} - \frac{1}{3} = \frac{4}{3}$. **Ans: $a=\frac{1}{3}, d=\frac{4}{3}$**
**(iv)** $a = 0.6$, $d = 1.7 - 0.6 = 1.1$. **Ans: $a=0.6, d=1.1$**
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#### **Q4. Which of the following are APs? If they form an AP, find the common difference $d$ and write three more terms.**
#### **A4. Solution:**
**(i) $2, 4, 8, 16, \dots$**
$4 - 2 = 2$
$8 - 4 = 4$
Difference is not constant. **Not an AP.**
**(ii) $2, \frac{5}{2}, 3, \frac{7}{2}, \dots$**
$\frac{5}{2} - 2 = 0.5$
$3 - \frac{5}{2} = 0.5$
Difference is constant. **It is an AP ($d=0.5$).**
Next three terms:
$3.5 + 0.5 = 4$
$4 + 0.5 = 4.5$ ($\frac{9}{2}$)
$4.5 + 0.5 = 5$
**(iii) $-1.2, -3.2, -5.2, -7.2, \dots$**
$-3.2 - (-1.2) = -2.0$
$-5.2 - (-3.2) = -2.0$
**It is an AP ($d=-2$).**
Next three terms: $-9.2, -11.2, -13.2$.
**(iv) $-10, -6, -2, 2, \dots$**
$-6 - (-10) = 4$
$-2 - (-6) = 4$
**It is an AP ($d=4$).**
Next three terms: $6, 10, 14$.
**(v) $3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \dots$**
$(3+\sqrt{2}) - 3 = \sqrt{2}$
$(3+2\sqrt{2}) - (3+\sqrt{2}) = \sqrt{2}$
**It is an AP ($d=\sqrt{2}$).**
Next three terms: $3+4\sqrt{2}, 3+5\sqrt{2}, 3+6\sqrt{2}$.
**(vi) $0.2, 0.22, 0.222, 0.2222, \dots$**
$0.22 - 0.2 = 0.02$
$0.222 - 0.22 = 0.002$
Difference is not constant. **Not an AP.**
**(vii) $0, -4, -8, -12, \dots$**
$-4 - 0 = -4$
$-8 - (-4) = -4$
**It is an AP ($d=-4$).**
Next three terms: $-16, -20, -24$.
**(viii) $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \dots$**
$-\frac{1}{2} - (-\frac{1}{2}) = 0$
**It is an AP ($d=0$).**
Next three terms: $-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}$.
**(ix) $1, 3, 9, 27, \dots$**
$3 - 1 = 2$
$9 - 3 = 6$
Difference is not constant. **Not an AP.**
**(x) $a, 2a, 3a, 4a, \dots$**
$2a - a = a$
$3a - 2a = a$
**It is an AP ($d=a$).**
Next three terms: $5a, 6a, 7a$.
**(xi) $a, a^2, a^3, a^4, \dots$**
$a^2 - a = a(a-1)$
$a^3 - a^2 = a^2(a-1)$
Difference is not constant. **Not an AP.**
**(xii) $\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots$**
Simplify terms: $\sqrt{2}, 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}$
$2\sqrt{2} - \sqrt{2} = \sqrt{2}$
$3\sqrt{2} - 2\sqrt{2} = \sqrt{2}$
**It is an AP ($d=\sqrt{2}$).**
Next three terms:
$5\sqrt{2} = \sqrt{50}$
$6\sqrt{2} = \sqrt{72}$
$7\sqrt{2} = \sqrt{98}$
**(xiii) $\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \dots$**
$\sqrt{6} - \sqrt{3} = \sqrt{3}(\sqrt{2}-1)$
$\sqrt{9} - \sqrt{6} = 3 - \sqrt{6}$
Difference is not constant. **Not an AP.**
**(xiv) $1^2, 3^2, 5^2, 7^2, \dots$**
Values: $1, 9, 25, 49$
$9 - 1 = 8$
$25 - 9 = 16$
Difference is not constant. **Not an AP.**
**(xv) $1^2, 5^2, 7^2, 73, \dots$**
Values: $1, 25, 49, 73$
$25 - 1 = 24$
$49 - 25 = 24$
$73 - 49 = 24$
**It is an AP ($d=24$).**
Next three terms:
$73 + 24 = 97$
$97 + 24 = 121$
$121 + 24 = 145$