Chapter 5 — ARITHMETIC PROGRESSIONS

Concept Before Solving

Solution Roadmap (Exam Strategy)

1. The taxi fare after each km when fare is ₹15 for first km and ₹8 for each additional km.

Solution (i)

Step 1: Write the values

1 km → ₹15
2 km → ₹15 + ₹8 = ₹23
3 km → ₹23 + ₹8 = ₹31
4 km → ₹31 + ₹8 = ₹39

Step 2: Form sequence

\[15, \ 23, \ 31, \ 39, \ldots\]

Step 3: Check difference

23 − 15 = 8
31 − 23 = 8
39 − 31 = 8

Conclusion: Difference is constant

\[a = 15, \quad d = 8\]

This is an Arithmetic Progression.

2. The amount of air when vacuum pump removes 1/4 each time.

Solution (ii)

Step 1: Assume initial air = V

After 1st pump → Remaining = (3/4)V
After 2nd pump → (3/4)² V
After 3rd pump → (3/4)³ V

Step 2: Sequence

\[V, \ \frac{3}{4}V, \ \left(\frac{3}{4}\right)^2 V, \ \left(\frac{3}{4}\right)^3 V, \ldots\]

Step 3: Check difference

Differences are NOT constant

Step 4: Check ratio

Each term multiplied by 3/4

Conclusion:

This is a Geometric Progression, not an AP.

3. Cost of digging well increases ₹50 per metre.

Solution (iii)

Step 1: Write costs

1st metre → ₹150
2nd metre → ₹150 + ₹50 = ₹200
3rd metre → ₹250
4th metre → ₹300

Step 2: Sequence

\[150, \ 200, \ 250, \ 300, \ldots\]

Step 3: Differences

200 − 150 = 50
250 − 200 = 50
300 − 250 = 50

Conclusion:

\[a = 150, \quad d = 50\]

This is an Arithmetic Progression.

4. Amount in account with compound interest.

Solution (iv)

Step 1: Write values

Year 1 → 10000 × 1.08 = 10800
Year 2 → 10800 × 1.08 = 11664
Year 3 → 11664 × 1.08 = 12597.12

Step 2: Sequence

\[10000, \ 10000(1.08), \ 10000(1.08)^2, \ 10000(1.08)^3, \ldots\]

Step 3: Check pattern

Each term multiplied by 1.08

Conclusion:

This is a Geometric Progression, not an AP.

Why This Question is Important

Concept Before Solving

Solution Roadmap

Q2. Write first four terms of the AP:

1. \(a = 10, d = 10\)

Solution (i)

\[\begin{aligned} a_1 &= 10 \\ a_2 &= 10 + 10 = 20 \\ a_3 &= 10 + 2 \times 10 = 30 \\ a_4 &= 10 + 3 \times 10 = 40 \end{aligned}\]

First four terms: 10, 20, 30, 40

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

Solution (ii)

\[\begin{aligned} a_1 &= -2 \\ a_2 &= -2 + 0 = -2 \\ a_3 &= -2 + 2 \times 0 = -2 \\ a_4 &= -2 + 3 \times 0 = -2 \end{aligned}\]

First four terms: -2, -2, -2, -2

Observation: When \(d=0\), all terms remain constant.

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

Solution (iii)

\[\begin{aligned} a_1 &= 4 \\ a_2 &= 4 + (-3) = 1 \\ a_3 &= 4 + 2(-3) = 4 - 6 = -2 \\ a_4 &= 4 + 3(-3) = 4 - 9 = -5 \end{aligned}\]

First four terms: 4, 1, -2, -5

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

Solution (iv)

\[\begin{aligned} a_1 &= -1 \\ a_2 &= -1 + \frac{1}{2} = -\frac{1}{2} \\ a_3 &= -1 + 2 \times \frac{1}{2} = -1 + 1 = 0 \\ a_4 &= -1 + 3 \times \frac{1}{2} = -1 + \frac{3}{2} = \frac{1}{2} \end{aligned}\]

First four terms: -1, -1/2, 0, 1/2

5. \(a = -1.25, d = -0.25\)

Solution (v)

\[\begin{aligned} a_1 &= -1.25 \\ a_2 &= -1.25 + (-0.25) = -1.50 \\ a_3 &= -1.25 + 2(-0.25) = -1.25 - 0.50 = -1.75 \\ a_4 &= -1.25 + 3(-0.25) = -1.25 - 0.75 = -2.00 \end{aligned}\]

First four terms: -1.25, -1.50, -1.75, -2.00

Why This Question Matters

Concept Before Solving

Solution Roadmap

Q3. Find first term and common difference:

1. \(3,\ 1,\ -1,\ -3,\ \ldots\)

Solution (i)

Step 1: Identify terms

\(a_1 = 3,\; a_2 = 1,\; a_3 = -1\)

$$\begin{aligned} d &= a_2 - a_1 \\ &= 1 - 3 \\ &= -2 \end{aligned}$$

Verification:

\(-1 - 1 = -2\), \(-3 - (-1) = -2\)

Answer: \(a = 3,\; d = -2\)

2. \(-5,\ -1,\ 3,\ 7,\ \ldots\)

Solution (ii)

\(a_1 = -5,\; a_2 = -1\)

$$\begin{aligned} d &= -1 - (-5) \\ &= -1 + 5 \\ &= 4 \end{aligned}$$

Verification:

\(3 - (-1) = 4\), \(7 - 3 = 4\)

Answer: \(a = -5,\; d = 4\)

3. \(\frac{1}{3},\ \frac{5}{3},\ \frac{9}{3},\ \frac{13}{3},\ \ldots\)

Solution (iii)

\(a_1 = \frac{1}{3},\; a_2 = \frac{5}{3}\)

$$\begin{aligned} d &= \frac{5}{3} - \frac{1}{3} \\ &= \frac{4}{3} \end{aligned}$$

Verification:

\(\frac{9}{3} - \frac{5}{3} = \frac{4}{3}\)

Answer: \(a = \frac{1}{3},\; d = \frac{4}{3}\)

4. \(0.6,\ 1.7,\ 2.8,\ 3.9,\ \ldots\)

Solution (iv)

\(a_1 = 0.6,\; a_2 = 1.7\)

$$\begin{aligned} d &= 1.7 - 0.6 \\ &= 1.1 \end{aligned}$$

Verification:

\(2.8 - 1.7 = 1.1\), \(3.9 - 2.8 = 1.1\)

Answer: \(a = 0.6,\; d = 1.1\)

Why This Question Matters

Concept Before Solving

Solution Roadmap

Q4. Check AP and find next three terms:

1. \(2, 4, 8, 16, \ldots\)
$$\begin{aligned} d_1 &= 4 - 2 = 2 \\ d_2 &= 8 - 4 = 4 \\ d_3 &= 16 - 8 = 8 \end{aligned}$$

Since \(d_1 \ne d_2 \ne d_3\), not an AP.

2. \(2, \frac{5}{2}, 3, \frac{7}{2}, \ldots\)
$$\begin{aligned} d_1 &= \frac{5}{2} - 2 = \frac{1}{2} \\ d_2 &= 3 - \frac{5}{2} = \frac{6-5}{2} = \frac{1}{2} \\ d_3 &= \frac{7}{2} - 3 = \frac{7-6}{2} = \frac{1}{2} \end{aligned}$$

All differences equal ⇒ AP with \(d = \frac{1}{2}\)

$$\begin{aligned} a_5 &= 2 + 4\cdot\frac{1}{2} = 4 \\ a_6 &= 2 + 5\cdot\frac{1}{2} = \frac{9}{2} \\ a_7 &= 2 + 6\cdot\frac{1}{2} = 5 \end{aligned}$$
3. \(-1.2, -3.2, -5.2, -7.2, \ldots\)
$$\begin{aligned} d_1 &= -3.2 - (-1.2) = -2 \\ d_2 &= -5.2 - (-3.2) = -2 \\ d_3 &= -7.2 - (-5.2) = -2 \end{aligned}$$

AP with \(d = -2\)

$$\begin{aligned} a_5 &= -1.2 + 4(-2) = -9.2 \\ a_6 &= -1.2 + 5(-2) = -11.2 \\ a_7 &= -1.2 + 6(-2) = -13.2 \end{aligned}$$
4. \(-10, -6, -2, 2, \ldots\)
$$\begin{aligned} d_1 &= -6 - (-10) = 4 \\ d_2 &= -2 - (-6) = 4 \\ d_3 &= 2 - (-2) = 4 \end{aligned}$$

AP with \(d = 4\)

$$\begin{aligned} a_5 &= -10 + 4\cdot4 = 6 \\ a_6 &= -10 + 5\cdot4 = 10 \\ a_7 &= -10 + 6\cdot4 = 14 \end{aligned}$$
5. \(3, 3+\sqrt{2}, 3+2\sqrt{2}, 3+3\sqrt{2}, \ldots\)
$$\begin{aligned} d_1 &= (3+\sqrt{2}) - 3 = \sqrt{2} \\ d_2 &= (3+2\sqrt{2}) - (3+\sqrt{2}) = \sqrt{2} \\ d_3 &= (3+3\sqrt{2}) - (3+2\sqrt{2}) = \sqrt{2} \end{aligned}$$

AP with \(d=\sqrt{2}\)

$$\begin{aligned} a_5 &= 3 + 4\sqrt{2} \\ a_6 &= 3 + 5\sqrt{2} \\ a_7 &= 3 + 6\sqrt{2} \end{aligned}$$
6. \(0.2, 0.22, 0.222, 0.2222, \ldots\)
$$\begin{aligned} d_1 &= 0.22 - 0.2 = 0.02 \\ d_2 &= 0.222 - 0.22 = 0.002 \end{aligned}$$

Since differences are not equal ⇒ Not an AP.

7. \(0, -4, -8, -12, \ldots\)
$$\begin{aligned} d_1 &= -4 - 0 = -4 \\ d_2 &= -8 - (-4) = -4 \\ d_3 &= -12 - (-8) = -4 \end{aligned}$$

AP with \(d = -4\)

$$\begin{aligned} a_5 &= 0 + 4(-4) = -16 \\ a_6 &= 0 + 5(-4) = -20 \\ a_7 &= 0 + 6(-4) = -24 \end{aligned}$$
8. \(-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}, \ldots\)
\[d = -\frac{1}{2} - (-\frac{1}{2}) = 0\]

AP with \(d = 0\)

Next three terms remain same:

\[-\frac{1}{2}, -\frac{1}{2}, -\frac{1}{2}\]
9. \(1, 3, 9, 27, \ldots\)
$$\begin{aligned} d_1 &= 3 - 1 = 2 \\ d_2 &= 9 - 3 = 6 \\ d_3 &= 27 - 9 = 18 \end{aligned}$$

Not an AP.

10. \(a, 2a, 3a, 4a, \ldots\)
$$\begin{aligned} d_1 &= 2a - a = a \\ d_2 &= 3a - 2a = a \\ d_3 &= 4a - 3a = a \end{aligned}$$

AP with \(d=a\)

\[a_5 = 5a,\quad a_6 = 6a,\quad a_7 = 7a\]
11. \(a, a^2, a^3, a^4, \ldots\)
$$\begin{aligned} d_1 &= a^2 - a \\ d_2 &= a^3 - a^2 \\ d_3 &= a^4 - a^3 \end{aligned}$$

Differences are not equal ⇒ Not an AP.

12. \(\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}, \ldots\)
$$\begin{aligned} \sqrt{8} &= 2\sqrt{2},\\ \sqrt{18} &= 3\sqrt{2},\\ \sqrt{32}&=4\sqrt{2} \end{aligned}$$ $$\begin{aligned} d_1 &= 2\sqrt{2} - \sqrt{2} = \sqrt{2} \\ d_2 &= 3\sqrt{2} - 2\sqrt{2} = \sqrt{2} \\ d_3 &= 4\sqrt{2} - 3\sqrt{2} = \sqrt{2} \end{aligned}$$

AP with \(d=\sqrt{2}\)

\[a_5=5\sqrt{2},\ a_6=6\sqrt{2},\ a_7=7\sqrt{2}\]
13. \(\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}, \ldots\)
$$\begin{aligned} d_1 &= \sqrt{6}-\sqrt{3} \\ d_2 &= 3-\sqrt{6} \end{aligned}$$

Not equal ⇒ Not AP

14. \(1^2, 3^2, 5^2, 7^2, \ldots\)
$$\begin{aligned} d_1 &= 9-1=8 \\ d_2 &= 25-9=16 \\ d_3 &= 49-25=24 \end{aligned}$$

Not AP

15. \(1^2, 5^2, 7^2, 73, \ldots\)
$$\begin{aligned} d_1 &= 25-1=24 \\ d_2 &= 49-25=24 \\ d_3 &= 73-49=24 \end{aligned}$$

AP with \(d=24\)

$$\begin{aligned} a_5 &= 1 + 4\times 24 = 97 \\ a_6 &= 1 + 5\times 24 = 121 \\ a_7 &= 1 + 6\times 24 = 145 \end{aligned}$$

Exam Significance