Arithmetic Progressions | Grade 10th - Mathematics Chapter Summary & NCERT CBSE Study Resources

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10th

10th - Mathematics

Arithmetic Progressions

Chapter Overview: Arithmetic Progressions

This chapter introduces the concept of Arithmetic Progressions (AP), a fundamental topic in algebra. Students will learn to identify, analyze, and solve problems related to AP, including finding the nth term and the sum of the first n terms. The chapter also covers real-life applications of AP.

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between consecutive terms is constant. This difference is called the common difference (denoted by d).

Key Concepts

Understanding the general form of an AP: a, a + d, a + 2d, a + 3d, ...
Finding the nth term of an AP using the formula: a_n = a + (n - 1)d
Calculating the sum of the first n terms of an AP: S_n = n/2 [2a + (n - 1)d] or S_n = n/2 (a + l), where l is the last term.
Applications of AP in real-world scenarios such as calculating installment payments, seating arrangements, etc.

Examples and Exercises

The chapter includes solved examples and exercises to help students practice:

Identifying whether a given sequence is an AP.
Finding missing terms in an AP.
Deriving the nth term and sum of terms for given APs.
Word problems involving APs.

Important Formulas

Concept	Formula
nth term of an AP	a_n = a + (n - 1)d
Sum of first n terms	S_n = n/2 [2a + (n - 1)d]
Sum when last term is known	S_n = n/2 (a + l)

Summary

Arithmetic Progressions are a key part of algebra with wide applications. Mastering this chapter helps in understanding patterns and solving real-life problems systematically.

All Question Types with Solutions – CBSE Exam Pattern

Explore a complete set of CBSE-style questions with detailed solutions, categorized by marks and question types. Ideal for exam preparation, revision and practice.

Very Short Answer (1 Mark) – with Solutions (CBSE Pattern)

These are 1-mark questions requiring direct, concise answers. Ideal for quick recall and concept clarity.

Question 1:

Find the common difference of the AP: 3, 7, 11, 15, ...

Answer:

Question 2:

What is the 10th term of the AP: 5, 8, 11, 14, ...?

Answer:

Question 3:

Write the first term and common difference of the AP: 10, 7, 4, 1, ...

Answer:

First term: 10, Common difference: -3

Question 4:

Is the sequence 1, 4, 9, 16, ... an arithmetic progression?

Answer:

Question 5:

Find the sum of first 5 terms of the AP: 2, 5, 8, 11, ...

Answer:

Question 6:

Which term of the AP: 21, 18, 15, ... is -81?

Answer:

35th term

Question 7:

Find the 20th term of the AP with a = 3 and d = 4.

Answer:

Question 8:

If the nth term of an AP is 4n + 1, find its common difference.

Answer:

Question 9:

Find the sum of first 10 natural numbers using AP formula.

Answer:

Question 10:

The first term of an AP is -5 and the common difference is 2. Find its 12th term.

Answer:

Question 11:

Find the sum of first 15 terms of the AP: 1, 4, 7, 10, ...

Answer:

330

Question 12:

If the 5th term of an AP is 17 and the 9th term is 25, find its common difference.

Answer:

Question 13:

What is the common difference of the Arithmetic Progression (AP): 5, 8, 11, 14, ...?

Answer:

The common difference (d) is the difference between any two consecutive terms.
Here, d = 8 - 5 = 3.

Question 14:

Find the 10th term of the AP: 2, 5, 8, 11, ... using the nth term formula.

Answer:

Given: a = 2, d = 5 - 2 = 3.
Using the formula: a_n = a + (n - 1)d,
a₁₀ = 2 + (10 - 1) × 3
= 2 + 27
= 29.

Question 15:

If the first term of an AP is 7 and the common difference is -3, write the first four terms.

Answer:

Given: a = 7, d = -3.
First four terms:
a₁ = 7
a₂ = 7 + (-3) = 4
a₃ = 4 + (-3) = 1
a₄ = 1 + (-3) = -2.

Question 16:

Is the sequence 10, 6, 2, -2, ... an Arithmetic Progression? Justify.

Answer:

Yes, it is an AP because the common difference (d) is constant.
Here, d = 6 - 10 = -4,
2 - 6 = -4,
-2 - 2 = -4.

Question 17:

Find the sum of the first 15 terms of the AP: 4, 7, 10, 13, ... using the sum formula.

Answer:

Given: a = 4, d = 7 - 4 = 3.
Using the formula: S_n = n/2 [2a + (n - 1)d],
S₁₅ = 15/2 [2 × 4 + (15 - 1) × 3]
= 15/2 [8 + 42]
= 15/2 × 50
= 375.

Question 18:

If the 5th term of an AP is 17 and the 9th term is 33, find the common difference.

Answer:

Given: a₅ = a + 4d = 17,
a₉ = a + 8d = 33.
Subtracting the equations:
(a + 8d) - (a + 4d) = 33 - 17
4d = 16
d = 4.

Question 19:

Write the nth term formula for an AP and define each term.

Answer:

The nth term formula is: a_n = a + (n - 1)d, where:

a = first term
d = common difference
n = term number

Question 20:

Find the number of terms in the AP: 3, 7, 11, ..., 99.

Answer:

Given: a = 3, d = 7 - 3 = 4, a_n = 99.
Using a_n = a + (n - 1)d,
99 = 3 + (n - 1) × 4
96 = (n - 1) × 4
n - 1 = 24
n = 25.

Question 21:

If the sum of the first n terms of an AP is given by S_n = 2n² + 3n, find the common difference.

Answer:

Given: S_n = 2n² + 3n.
For n = 1, S₁ = a = 2(1)² + 3(1) = 5.
For n = 2, S₂ = 2(4) + 6 = 14.
Second term (a₂) = S₂ - S₁ = 14 - 5 = 9.
Common difference (d) = a₂ - a = 9 - 5 = 4.

Question 22:

The first term of an AP is -5 and the last term is 45. If the sum of all terms is 120, find the number of terms.

Answer:

Given: a = -5, a_n = 45, S_n = 120.
Using the sum formula: S_n = n/2 (a + a_n),
120 = n/2 (-5 + 45)
120 = n/2 × 40
120 = 20n
n = 6.

Question 23:

What is the common difference of the Arithmetic Progression: 3, 7, 11, 15, ...?

Answer:

The common difference (d) is the difference between any two consecutive terms.
Here, d = 7 - 3 = 4.

Question 24:

Find the 10th term of the AP whose first term is 5 and common difference is 2.

Answer:

Using the formula for the nth term of an AP:
a_n = a + (n - 1)d
a₁₀ = 5 + (10 - 1) × 2
= 5 + 18
= 23

Question 25:

If the first term of an AP is -3 and the common difference is 4, find its 5th term.

Answer:

Using the nth term formula:
a_n = a + (n - 1)d
a₅ = -3 + (5 - 1) × 4
= -3 + 16
= 13

Question 26:

Write the first four terms of an AP where the first term is 10 and the common difference is -2.

Answer:

The terms are:
First term (a₁) = 10
Second term (a₂) = 10 + (-2) = 8
Third term (a₃) = 8 + (-2) = 6
Fourth term (a₄) = 6 + (-2) = 4

Question 27:

Is the sequence 1, 4, 9, 16, ... an Arithmetic Progression? Justify your answer.

Answer:

No, because the difference between consecutive terms is not constant.
4 - 1 = 3
9 - 4 = 5
16 - 9 = 7
The differences are unequal, so it is not an AP.

Question 28:

Find the sum of the first 15 terms of the AP: 6, 11, 16, 21, ...

Answer:

First term (a) = 6, common difference (d) = 5
Sum formula: S_n = n/2 [2a + (n - 1)d]
S₁₅ = 15/2 [2 × 6 + (15 - 1) × 5]
= 15/2 [12 + 70]
= 15/2 × 82
= 615

Question 29:

If the 3rd term of an AP is 12 and the 7th term is 28, find its common difference.

Answer:

Using the nth term formula for both terms:
a + 2d = 12 ...(1)
a + 6d = 28 ...(2)
Subtract (1) from (2):
4d = 16
d = 4

Question 30:

Which term of the AP 21, 18, 15, ... is -81?

Answer:

First term (a) = 21, d = -3
Using a_n = a + (n - 1)d
-81 = 21 + (n - 1)(-3)
-81 - 21 = (n - 1)(-3)
-102 = -3(n - 1)
n - 1 = 34
n = 35

Very Short Answer (2 Marks) – with Solutions (CBSE Pattern)

These 2-mark questions test key concepts in a brief format. Answers are expected to be accurate and slightly descriptive.

Question 1:

Find the common difference of the Arithmetic Progression: 5, 8, 11, 14, ...

Answer:

The common difference (d) is calculated as the difference between consecutive terms.
d = 8 - 5 = 3
or d = 11 - 8 = 3
Thus, the common difference is 3.

Question 2:

Write the first term and common difference of the AP: 10, 7, 4, 1, ...

Answer:

The first term (a) is 10.
The common difference (d) is calculated as:
d = 7 - 10 = -3
or d = 4 - 7 = -3
Thus, the common difference is -3.

Question 3:

Find the 10th term of the AP whose first term is 3 and common difference is 2.

Answer:

Using the formula for the nth term of an AP:
a_n = a + (n - 1)d
Here, a = 3, d = 2, n = 10
a₁₀ = 3 + (10 - 1) × 2
a₁₀ = 3 + 18 = 21.

Question 4:

Check whether the sequence 1, 3, 9, 27, ... is an Arithmetic Progression. Justify your answer.

Answer:

To check if it is an AP, we verify the common difference (d) between consecutive terms.
d₁ = 3 - 1 = 2
d₂ = 9 - 3 = 6
Since d₁ ≠ d₂, the sequence is not an AP.

Question 5:

Find the sum of the first 15 terms of the AP: 2, 5, 8, 11, ...

Answer:

Given: a = 2, d = 5 - 2 = 3, n = 15
Using the sum formula: S_n = n/2 [2a + (n - 1)d]
S₁₅ = 15/2 [2 × 2 + (15 - 1) × 3]
S₁₅ = 15/2 [4 + 42] = 15/2 × 46 = 345.

Question 6:

Find the sum of the first 20 natural numbers using the formula for the sum of an AP.

Answer:

The sequence of first 20 natural numbers is: 1, 2, 3, ..., 20
Here, a = 1, d = 1, n = 20
Sum formula: S_n = n/2 [2a + (n - 1)d]
S₂₀ = 20/2 [2 × 1 + (20 - 1) × 1]
S₂₀ = 10 [2 + 19] = 10 × 21 = 210.

Short Answer (3 Marks) – with Solutions (CBSE Pattern)

These 3-mark questions require brief explanations and help assess understanding and application of concepts.

Question 1:

Find the sum of the first 15 terms of an arithmetic progression (AP) where the first term is 5 and the common difference is 3.

Answer:

Given:
First term (a) = 5
Common difference (d) = 3
Number of terms (n) = 15

Formula for sum of first n terms of an AP:
S_n = n/2 [2a + (n - 1)d]

Substituting the values:
S₁₅ = 15/2 [2 × 5 + (15 - 1) × 3]
= 15/2 [10 + 42]
= 15/2 × 52
= 15 × 26
= 390

Thus, the sum of the first 15 terms is 390.

Question 2:

If the 5th term of an AP is 17 and the 9th term is 33, find the common difference and the first term.

Answer:

Given:
5th term (a₅) = 17
9th term (a₉) = 33

Using the formula for the n^th term of an AP:
a_n = a + (n - 1)d

For the 5th term:
a + 4d = 17 ...(1)
For the 9th term:
a + 8d = 33 ...(2)

Subtract equation (1) from (2):
(a + 8d) - (a + 4d) = 33 - 17
4d = 16
d = 4

Substitute d = 4 in equation (1):
a + 4 × 4 = 17
a + 16 = 17
a = 1

Thus, the common difference is 4 and the first term is 1.

Question 3:

Write the first four terms of an AP where the first term is -3 and the common difference is -2.

Answer:

Given:
First term (a) = -3
Common difference (d) = -2

Using the formula for the n^th term of an AP:
a_n = a + (n - 1)d

First term (a₁):
a₁ = -3 + (1 - 1)(-2) = -3

Second term (a₂):
a₂ = -3 + (2 - 1)(-2) = -3 - 2 = -5

Third term (a₃):
a₃ = -3 + (3 - 1)(-2) = -3 - 4 = -7

Fourth term (a₄):
a₄ = -3 + (4 - 1)(-2) = -3 - 6 = -9

Thus, the first four terms are -3, -5, -7, -9.

Question 4:

The sum of the first n terms of an AP is given by S_n = 3n² + 5n. Find the 10th term of the AP.

Answer:

Given:
Sum of first n terms (S_n) = 3n² + 5n

To find the 10th term, we use:
a_n = S_n - S_n-1

First, find S₁₀:
S₁₀ = 3(10)² + 5(10) = 300 + 50 = 350

Next, find S₉:
S₉ = 3(9)² + 5(9) = 243 + 45 = 288

Now, the 10th term:
a₁₀ = S₁₀ - S₉ = 350 - 288 = 62

Thus, the 10th term is 62.

Question 5:

Check whether the sequence 7, 13, 19, 25, ... is an AP. If yes, find its common difference and next two terms.

Answer:

To check if the sequence is an arithmetic progression, we verify if the difference between consecutive terms is constant.

Given sequence: 7, 13, 19, 25, ...

Difference between terms:
13 - 7 = 6
19 - 13 = 6
25 - 19 = 6

Since the difference is constant (d = 6), the sequence is an AP.

Next two terms:
5th term = 25 + 6 = 31
6th term = 31 + 6 = 37

Thus, the common difference is 6, and the next two terms are 31 and 37.

Question 6:

Find the common difference of the Arithmetic Progression (AP): 7, 11, 15, 19, ...

Answer:

The common difference (d) of an AP is the difference between any two consecutive terms.
Here, d = a₂ - a₁.
Given AP: 7, 11, 15, 19, ...
d = 11 - 7 = 4.
Thus, the common difference is 4.

Question 7:

Write the first four terms of an AP where the first term is 5 and the common difference is -3.

Answer:

Given: First term (a₁) = 5, common difference (d) = -3.
The terms of an AP are calculated as: a_n = a₁ + (n-1)d.
First term (a₁) = 5.
Second term (a₂) = 5 + (2-1)(-3) = 5 - 3 = 2.
Third term (a₃) = 5 + (3-1)(-3) = 5 - 6 = -1.
Fourth term (a₄) = 5 + (4-1)(-3) = 5 - 9 = -4.
Thus, the first four terms are: 5, 2, -1, -4.

Question 8:

Find the 10th term of the AP: 3, 7, 11, 15, ...

Answer:

Given AP: 3, 7, 11, 15, ...
Here, first term (a₁) = 3, common difference (d) = 7 - 3 = 4.
The n^th term of an AP is given by: a_n = a₁ + (n-1)d.
For the 10th term (a₁₀):
a₁₀ = 3 + (10-1)(4)
a₁₀ = 3 + 36 = 39.
Thus, the 10th term is 39.

Question 9:

Check whether the sequence 10, 7, 4, 1, -2, ... is an AP. If yes, find its next term.

Answer:

To check if the sequence is an AP, verify if the common difference (d) is constant.
d₁ = 7 - 10 = -3
d₂ = 4 - 7 = -3
d₃ = 1 - 4 = -3
d₄ = -2 - 1 = -3
Since d is constant, the sequence is an AP.
The next term after -2 is: -2 + (-3) = -5.

Question 10:

If the 5th term of an AP is 17 and the 9th term is 33, find its common difference.

Answer:

Given: a₅ = 17, a₉ = 33.
The n^th term formula is: a_n = a₁ + (n-1)d.
For a₅: 17 = a₁ + 4d ...(1)
For a₉: 33 = a₁ + 8d ...(2)
Subtract (1) from (2):
33 - 17 = (a₁ + 8d) - (a₁ + 4d)
16 = 4d
d = 4.
Thus, the common difference is 4.

Long Answer (5 Marks) – with Solutions (CBSE Pattern)

These 5-mark questions are descriptive and require detailed, structured answers with proper explanation and examples.

Question 1:

The first term of an AP is 5 and the common difference is 3. Find the 10th term and the sum of the first 10 terms. Justify your steps using the AP formulas.

Answer:

Introduction

We studied that an AP has a fixed common difference. Given a = 5 and d = 3, we find the 10th term and sum.

Argument 1

Using nth term formula: a₁₀ = a + (10-1)d = 5 + 9×3 = 32.

Argument 2

Using sum formula: S₁₀ = 10/2 [2×5 + (10-1)×3] = 5 × [10 + 27] = 185.

Conclusion

Thus, the 10th term is 32 and the sum is 185, verified by formulas.

Question 2:

A ladder has rungs spaced 25 cm apart. The bottom rung touches the ground. If the top rung is 5 m high, how many rungs does the ladder have? Use AP concepts and justify.

Answer:

Introduction

Our textbook shows real-life AP applications. Here, rung distances form an AP with a = 0 cm and d = 25 cm.

Argument 1

Total height = 5 m = 500 cm.
Using nth term formula: 500 = 0 + (n-1)×25 → n-1 = 20 → n = 21.

Argument 2

Verification: a₂₁ = 0 + 20×25 = 500 cm, matching the height.

Conclusion

The ladder has 21 rungs, as calculated using AP.

Question 3:

The first term of an AP is 5 and the common difference is 3. Find the 10th term and the sum of the first 10 terms. Verify using the formula.

Answer:

Introduction

We studied that an AP has a fixed common difference. Given a = 5 and d = 3, we can find the 10th term (a₁₀) and sum (S₁₀).

Argument 1

Using a_n = a + (n-1)d, a₁₀ = 5 + (9×3) = 32.
Sum formula: S_n = n/2 [2a + (n-1)d], so S₁₀ = 5 [10 + 27] = 185.

Conclusion

Our textbook shows similar problems. The calculations match the formulas, confirming correctness.

Question 4:

A ladder has rungs 25 cm apart. The bottom rung is 45 cm long, and lengths decrease uniformly to 25 cm at the top. Find the total length of wood required if there are 15 rungs.

Answer:

Introduction

This is a real-life AP problem. The rung lengths form an AP with a = 45 cm, l = 25 cm, and n = 15.

Argument 1

Common difference: d = (25-45)/(15-1) ≈ -1.43 cm (uniform decrease).
Total length: S_n = n/2 (a + l) = 15/2 (45 + 25) = 525 cm.

Conclusion

Our textbook shows how AP applies to such problems. The total wood required is 525 cm.

Question 5:

Find the sum of the first 15 terms of an Arithmetic Progression (AP) where the first term is 5 and the common difference is 3. Verify using the formula Sₙ = n/2 [2a + (n-1)d].

Answer:

Introduction

We studied that the sum of an AP can be calculated using the formula Sₙ = n/2 [2a + (n-1)d]. Here, a = 5, d = 3, and n = 15.

Argument 1

Substituting values: S₁₅ = 15/2 [2×5 + (15-1)×3].

Argument 2

Simplifying: S₁₅ = 7.5 [10 + 42] = 7.5 × 52 = 390.

Conclusion

The sum of the first 15 terms is 390, verified using the formula.

Question 6:

The 4th term of an AP is 11 and the 9th term is 26. Find the common difference (d) and the first term (a).

Answer:

Introduction

Our textbook shows that the nth term of an AP is given by aₙ = a + (n-1)d. Here, a₄ = 11 and a₉ = 26.

Argument 1

Form equations: a + 3d = 11 (1) and a + 8d = 26 (2).

Argument 2

Subtract (1) from (2): 5d = 15 ⇒ d = 3. Substituting in (1): a = 2.

Conclusion

The first term is 2 and the common difference is 3.

Question 7:

A ladder has rungs 25 cm apart. The lengths of the rungs decrease uniformly from 45 cm at the bottom to 25 cm at the top. Find the total length of all rungs if there are 10 rungs.

Answer:

Introduction

This is a real-life application of AP. The lengths form an AP with a = 45 cm, l = 25 cm, and n = 10.

Argument 1

Using the sum formula: Sₙ = n/2 (a + l).

Argument 2

Substituting values: S₁₀ = 10/2 (45 + 25) = 5 × 70 = 350 cm.

Conclusion

The total length of all rungs is 350 cm.

Question 8:

Derive the formula for the nth term of an AP using the concept of common difference.

Answer:

Introduction

We studied that an AP has a constant common difference (d). Let the first term be a.

Argument 1

The terms are: a₁ = a, a₂ = a + d, a₃ = a + 2d, and so on.

Argument 2

Observing the pattern: aₙ = a + (n-1)d, which is the nth term formula.

Conclusion

The derived formula is aₙ = a + (n-1)d, matching our textbook.

Question 9:

Check if the sequence 3, 6, 9, 12, ... forms an AP. If yes, find the 10th term.

Answer:

Introduction

To check for AP, we verify if the difference between consecutive terms is constant.

Argument 1

Differences: 6-3 = 3, 9-6 = 3, 12-9 = 3. Hence, it is an AP with d = 3.

Argument 2

Using aₙ = a + (n-1)d, the 10th term is 3 + (10-1)×3 = 30.

Conclusion

The sequence is an AP, and the 10th term is 30.

Question 10:

Find the sum of the first 15 terms of an Arithmetic Progression (AP) whose 3rd term is 5 and 7th term is 9. Justify each step using the AP formulas.

Answer:

Introduction

We studied that an AP has a common difference (d). Given a₃ = 5 and a₇ = 9, we can find 'd' and the first term (a).

Argument 1

Using a_n = a + (n-1)d, a₃ = a + 2d = 5.
a₇ = a + 6d = 9. Subtracting gives 4d = 4 ⇒ d = 1.

Argument 2

Substituting d = 1 in a₃, we get a = 3.
Sum S₁₅ = (15/2)[2×3 + (15-1)×1] = (15/2)(6 + 14) = 150.

Conclusion

Thus, the sum of the first 15 terms is 150, derived using AP formulas.

Question 11:

A ladder has rungs with a common difference of 25 cm. The bottom rung is 45 cm above the ground. If the top rung is 2.5 m high, how many rungs does the ladder have? Show calculations.

Answer:

Introduction

Our textbook shows real-life AP problems. Here, rung heights form an AP with a = 45 cm, d = 25 cm, and last term (l) = 250 cm.

Argument 1

Convert 2.5 m to cm: 250 cm.
Using l = a + (n-1)d, 250 = 45 + (n-1)×25.

Argument 2

Simplify: 205 = (n-1)×25 ⇒ n-1 = 8.2.
Since n must be an integer, n = 9 (rounding up).

Conclusion

The ladder has 9 rungs, as partial rungs are impractical.

Question 12:

Derive the formula for the sum of the first n terms (S_n) of an AP. Use the example of the AP 2, 5, 8,... to verify your formula for n = 4.

Answer:

Introduction

We learned that S_n = n/2[2a + (n-1)d]. Let’s derive it by pairing terms.

Argument 1

Write S_n = a + (a+d) + ... + [a+(n-1)d].
Pair first and last terms: (a + l) = (2a + (n-1)d). There are n/2 such pairs.

Argument 2

For AP 2, 5, 8, 11 (n=4), S₄ = 2 + 5 + 8 + 11 = 26.
Using formula: S₄ = 4/2[2×2 + 3×3] = 2×13 = 26.

Conclusion

The formula is verified, showing S_n = n/2[2a + (n-1)d].

Question 13:

Find the sum of the first 15 terms of an Arithmetic Progression (AP) whose 3rd term is 5 and the 7th term is 9. Show all steps clearly.

Answer:

To find the sum of the first 15 terms of the AP, we first need to determine the first term (a) and the common difference (d).

Given:
- 3rd term (a₃) = 5
- 7th term (a₇) = 9

Step 1: Write the formula for the nth term of an AP:
aₙ = a + (n - 1)d

Step 2: Substitute the given terms into the formula:
For the 3rd term: a + 2d = 5 ...(1)
For the 7th term: a + 6d = 9 ...(2)

Step 3: Subtract equation (1) from equation (2):
(a + 6d) - (a + 2d) = 9 - 5
4d = 4
d = 1

Step 4: Substitute d = 1 into equation (1):
a + 2(1) = 5
a = 3

Step 5: Now, use the sum formula for the first n terms of an AP:
Sₙ = n/2 [2a + (n - 1)d]
For n = 15:
S₁₅ = 15/2 [2(3) + (15 - 1)(1)]
S₁₅ = 15/2 [6 + 14]
S₁₅ = 15/2 × 20
S₁₅ = 15 × 10
S₁₅ = 150

Thus, the sum of the first 15 terms is 150.

Question 14:

The sum of the first n terms of an AP is given by Sₙ = 3n² + 5n. Find the 10th term of the progression. Justify your answer with proper reasoning.

Answer:

To find the 10th term of the AP, we use the relationship between the sum of terms and the nth term.

Given: Sₙ = 3n² + 5n

Step 1: Recall that the nth term (aₙ) can be found using:
aₙ = Sₙ - Sₙ₋₁ (for n ≥ 2)

Step 2: Calculate S₁₀ and S₉:
S₁₀ = 3(10)² + 5(10) = 300 + 50 = 350
S₉ = 3(9)² + 5(9) = 243 + 45 = 288

Step 3: Find the 10th term:
a₁₀ = S₁₀ - S₉ = 350 - 288 = 62

Thus, the 10th term of the AP is 62.

Additional Insight: The first term (a₁) can also be found by S₁ = 3(1)² + 5(1) = 8, and the common difference (d) can be derived by checking a₂ = S₂ - S₁ = (12 + 10) - 8 = 14, so d = a₂ - a₁ = 6.

Question 15:

In an AP, the sum of the first 10 terms is 150, and the sum of the next 10 terms is 550. Find the first term and the common difference. Show all steps systematically.

Answer:

Let the first term be a and the common difference be d.

Given:
- Sum of first 10 terms (S₁₀) = 150
- Sum of next 10 terms (S₁₁₋₂₀) = 550

Step 1: Use the sum formula for the first 10 terms:
S₁₀ = 10/2 [2a + (10 - 1)d] = 150
5(2a + 9d) = 150
2a + 9d = 30 ...(1)

Step 2: Sum of terms 11 to 20 is the difference between S₂₀ and S₁₀:
S₁₁₋₂₀ = S₂₀ - S₁₀ = 550
S₂₀ = 550 + 150 = 700

Step 3: Use the sum formula for the first 20 terms:
S₂₀ = 20/2 [2a + (20 - 1)d] = 700
10(2a + 19d) = 700
2a + 19d = 70 ...(2)

Step 4: Subtract equation (1) from equation (2):
(2a + 19d) - (2a + 9d) = 70 - 30
10d = 40
d = 4

Step 5: Substitute d = 4 into equation (1):
2a + 9(4) = 30
2a + 36 = 30
2a = -6
a = -3

Thus, the first term is -3 and the common difference is 4.

Question 16:

The first term of an Arithmetic Progression (AP) is 5 and the common difference is 3. Find the sum of the first 15 terms of the AP. Justify each step of your solution.

Answer:

To find the sum of the first 15 terms of the given Arithmetic Progression (AP), we use the formula for the sum of the first n terms of an AP:

S_n = n/2 [2a + (n - 1)d]

Where:

S_n = Sum of the first n terms
a = First term of the AP
d = Common difference
n = Number of terms

Given:

a = 5
d = 3
n = 15

Substituting the values into the formula:
S₁₅ = 15/2 [2(5) + (15 - 1)(3)]
S₁₅ = 15/2 [10 + 14 × 3]
S₁₅ = 15/2 [10 + 42]
S₁₅ = 15/2 × 52
S₁₅ = 15 × 26
S₁₅ = 390

Thus, the sum of the first 15 terms of the AP is 390.

Question 17:

The 4th term of an Arithmetic Progression (AP) is 14 and the 9th term is 29. Find the first term and common difference of the AP. Verify your answer by writing the first five terms of the AP.

Answer:

To find the first term (a) and common difference (d) of the AP, we use the formula for the n^th term of an AP:

a_n = a + (n - 1)d

Given:

a₄ = 14
a₉ = 29

Substituting the values into the formula:
a + 3d = 14 ...(1) [for the 4th term]
a + 8d = 29 ...(2) [for the 9th term]

Subtract equation (1) from equation (2):
(a + 8d) - (a + 3d) = 29 - 14
5d = 15
d = 3

Substitute d = 3 into equation (1):
a + 3(3) = 14
a + 9 = 14
a = 5

Thus, the first term (a) is 5 and the common difference (d) is 3.

Verification by writing the first five terms:
a₁ = a = 5
a₂ = a + d = 5 + 3 = 8
a₃ = a + 2d = 5 + 6 = 11
a₄ = a + 3d = 5 + 9 = 14 (matches given)
a₅ = a + 4d = 5 + 12 = 17

The first five terms are 5, 8, 11, 14, 17, confirming our solution.

Question 18:

The first term of an Arithmetic Progression (AP) is 5 and the common difference is 3. Find the sum of the first 15 terms of the AP. Also, verify your answer using the formula for the sum of an AP.

Answer:

To find the sum of the first 15 terms of the given Arithmetic Progression (AP), we will use the formula for the sum of the first n terms of an AP:

S_n = (n/2) [2a + (n - 1)d]

Where:

S_n = Sum of the first n terms
a = First term of the AP
d = Common difference
n = Number of terms

Given:

a = 5
d = 3
n = 15

Substituting the given values into the formula:
S₁₅ = (15/2) [2(5) + (15 - 1)(3)]
S₁₅ = (15/2) [10 + 14 × 3]
S₁₅ = (15/2) [10 + 42]
S₁₅ = (15/2) × 52
S₁₅ = 15 × 26
S₁₅ = 390

Verification using the alternative formula for the sum of an AP:
S_n = (n/2) [a + l]
Where l is the last term of the AP.

First, find the 15th term (a₁₅):
a_n = a + (n - 1)d
a₁₅ = 5 + (15 - 1) × 3
a₁₅ = 5 + 42
a₁₅ = 47

Now, substitute into the sum formula:
S₁₅ = (15/2) [5 + 47]
S₁₅ = (15/2) × 52
S₁₅ = 390

Both methods give the same result, confirming that the sum of the first 15 terms is 390.

Question 19:

The sum of the first n terms of an AP is given by S_n = 3n² + 2n. Find the common difference and the first term of the AP. Also, write the first three terms of the AP.

Answer:

Given the sum of the first n terms of the AP as S_n = 3n² + 2n, we can find the first term and common difference as follows:

Step 1: Find the first term (a)

The first term (a) is the sum of the first 1 term (S₁):
S₁ = 3(1)² + 2(1) = 3 + 2 = 5
Thus, a = 5.

Step 2: Find the common difference (d)

The sum of the first 2 terms (S₂) is:
S₂ = 3(2)² + 2(2) = 12 + 4 = 16
The second term (a₂) is:
a₂ = S₂ - S₁ = 16 - 5 = 11
The common difference (d) is:
d = a₂ - a = 11 - 5 = 6

Step 3: Write the first three terms

The first three terms of the AP are:
a = 5
a + d = 5 + 6 = 11
a + 2d = 5 + 12 = 17

Thus, the first three terms are 5, 11, 17.

Conclusion:

First term (a) = 5
Common difference (d) = 6
First three terms = 5, 11, 17

Question 20:

The first term of an Arithmetic Progression (AP) is 5 and the common difference is 3. Find the sum of the first 15 terms of this AP. Also, verify your answer using the formula for the sum of an AP.

Answer:

To find the sum of the first 15 terms of the given Arithmetic Progression (AP), we will use the formula for the sum of the first n terms of an AP:

S_n = (n/2) [2a + (n - 1)d]

Where:
S_n = Sum of the first n terms
a = First term of the AP
d = Common difference
n = Number of terms

Given:
a = 5
d = 3
n = 15

Substituting the values into the formula:
S₁₅ = (15/2) [2(5) + (15 - 1)(3)]
S₁₅ = 7.5 [10 + 42]
S₁₅ = 7.5 × 52
S₁₅ = 390

Now, let's verify this result by listing the first 15 terms and adding them manually:

The first 15 terms of the AP are:
5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47

Sum of these terms:
5 + 8 = 13
13 + 11 = 24
24 + 14 = 38
38 + 17 = 55
55 + 20 = 75
75 + 23 = 98
98 + 26 = 124
124 + 29 = 153
153 + 32 = 185
185 + 35 = 220
220 + 38 = 258
258 + 41 = 299
299 + 44 = 343
343 + 47 = 390

The manual addition confirms that the sum of the first 15 terms is indeed 390, which matches our earlier result using the formula.

This demonstrates the reliability of the sum formula for Arithmetic Progressions and ensures accuracy in calculations.

Question 21:

The first term of an Arithmetic Progression (AP) is 5, and the common difference is 3. Find the sum of the first 15 terms of the AP. Also, explain the significance of the formula used.

Answer:

To find the sum of the first 15 terms of the given Arithmetic Progression (AP), we use the formula for the sum of the first n terms of an AP:

S_n = n/2 [2a + (n - 1)d]

Where:

S_n = Sum of the first n terms
a = First term of the AP
d = Common difference
n = Number of terms

Given:

a = 5
d = 3
n = 15

Substituting the values into the formula:
S₁₅ = 15/2 [2 × 5 + (15 - 1) × 3]
S₁₅ = 15/2 [10 + 14 × 3]
S₁₅ = 15/2 [10 + 42]
S₁₅ = 15/2 × 52
S₁₅ = 15 × 26
S₁₅ = 390

Therefore, the sum of the first 15 terms is 390.

Significance of the formula:
The formula S_n = n/2 [2a + (n - 1)d] is derived by pairing terms from the beginning and end of the AP, ensuring each pair sums to the same value. This simplifies the calculation of the sum without adding all terms individually. It is efficient and universally applicable to any AP.

Question 22:

The first term of an Arithmetic Progression (AP) is 5 and the common difference is 3. Find the sum of the first 15 terms of the AP. Also, explain why the sum of an AP is useful in real-life situations with an example.

Answer:

To find the sum of the first 15 terms of the given Arithmetic Progression (AP), we use the formula for the sum of the first n terms of an AP:

S_n = n/2 [2a + (n - 1)d]

Where:
S_n = Sum of the first n terms
a = First term = 5
d = Common difference = 3
n = Number of terms = 15

Substituting the values into the formula:
S₁₅ = 15/2 [2(5) + (15 - 1)(3)]
S₁₅ = 7.5 [10 + 42]
S₁₅ = 7.5 × 52
S₁₅ = 390

Thus, the sum of the first 15 terms is 390.

Real-life application: The sum of an AP is useful in scenarios involving uniform increments or decrements. For example, if a person saves ₹100 in the first week and increases their savings by ₹50 every subsequent week, the total savings after 10 weeks can be calculated using the sum formula of an AP. This helps in financial planning and goal tracking.

Case-based Questions (4 Marks) – with Solutions (CBSE Pattern)

These 4-mark case-based questions assess analytical skills through real-life scenarios. Answers must be based on the case study provided.

Question 1:

A ladder has rungs spaced 25 cm apart. The bottom rung touches the ground, and the top rung reaches 10 m high.
Problem Interpretation: Identify if the rung heights form an Arithmetic Progression (AP).
Mathematical Modeling: Find the number of rungs and the height of the 7th rung.

Answer:

Problem Interpretation:

The rung heights form an AP because the distance between consecutive rungs is constant (25 cm).

Mathematical Modeling:

First term (a) = 0 cm (ground level)
Common difference (d) = 25 cm
Total height = 10 m = 1000 cm

Solution:

Using the AP formula for the nth term: a_n = a + (n-1)d. For the top rung, 1000 = 0 + (n-1)(25). Solving, n = 41 rungs. The 7th rung height = 0 + (7-1)(25) = 150 cm.

Question 2:

A taxi charges ₹15 for the first km and ₹8 for each additional km.
Problem Interpretation: Represent the fare as an AP.
Mathematical Modeling: Calculate the fare for 12 km and the distance covered for ₹95.

Answer:

Problem Interpretation:

The fare forms an AP with a = ₹15 (first km) and d = ₹8 (subsequent kms).

Mathematical Modeling:

nth term formula: a_n = a + (n-1)d

Solution:

For 12 km, n = 12. Fare = 15 + (12-1)(8) = ₹103. For ₹95: 95 = 15 + (n-1)(8). Solving, n = 11 km.

Question 3:

A ladder has rungs 25 cm apart. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. Find the total length of wood required for the rungs if the ladder has 10 rungs.

Answer:

Problem Interpretation

We studied that the lengths of rungs form an Arithmetic Progression (AP) with a common difference. Here, the first term (a₁) is 45 cm, and the last term (a₁₀) is 25 cm.

Mathematical Modeling

Given: a₁ = 45 cm, a₁₀ = 25 cm, n = 10.
Total length = Sum of AP (Sₙ) = n/2 × (a₁ + aₙ).

Solution

Using the formula, S₁₀ = 10/2 × (45 + 25) = 5 × 70 = 350 cm. Thus, 350 cm of wood is needed.

Question 4:

The sum of the first 15 terms of an AP is 300, and the common difference is 2. Find the first term and the 15th term.

Answer:

Problem Interpretation

Our textbook shows that the sum of an AP is given by Sₙ = n/2 × [2a + (n−1)d]. Here, S₁₅ = 300, d = 2, n = 15.

Mathematical Modeling

Substitute values: 300 = 15/2 × [2a + (15−1)2].
Simplify to find 'a'.

Solution

Solving, we get 2a + 28 = 40 ⇒ a = 6. The 15th term (a₁₅) = a + 14d = 6 + 28 = 34. Thus, a₁ = 6, a₁₅ = 34.

Question 5:

A ladder has rungs spaced 25 cm apart. The bottom rung touches the ground, and the top rung is 2.5 m high.
(i) How many rungs does the ladder have?
(ii) If the distance between rungs is reduced to 20 cm, how many rungs will fit in the same height?

Answer:

Problem Interpretation

We studied that the rung spacing forms an Arithmetic Progression (AP) with a common difference of 25 cm or 20 cm.

Mathematical Modeling

Total height = 2.5 m = 250 cm
For 25 cm spacing: Number of rungs = (250/25) + 1 = 11
For 20 cm spacing: Number of rungs = (250/20) + 1 = 13.5 → 13 (since rungs must be whole numbers)

Solution

The ladder has 11 rungs initially. With 20 cm spacing, it can fit 13 rungs.

Question 6:

A taxi charges ₹15 for the first km and ₹8 for each additional km.
(i) Write the fare as an AP.
(ii) Find the fare for a 12 km trip.

Answer:

Problem Interpretation

Our textbook shows that taxi fares can be modeled as an AP where the first term is ₹15 and the common difference is ₹8.

Mathematical Modeling

AP: 15, 23, 31, ...
For 12 km: Fare = 15 + (12 - 1) × 8 = ₹103

Solution

The fare sequence is 15, 23, 31, .... The fare for 12 km is ₹103.

Question 7:

A ladder has rungs spaced uniformly. The bottom rung is 45 cm from the ground, and the top rung is 10 cm below the top of the ladder. If the ladder has 25 rungs, find the distance between two consecutive rungs. (Use AP concepts)

Answer:

Problem Interpretation

We need to find the uniform distance between ladder rungs, forming an Arithmetic Progression.

Mathematical Modeling

Total height covered by rungs = 45 cm (first term, a₁) to (45 + d(n-1)) cm.
Top rung is 10 cm below the ladder top, so total ladder height = (45 + d(24)) + 10 cm.

Solution

Assuming uniform spacing 'd', 45 + 24d + 10 = ladder height. Since height isn't specified, we focus on rung spacing. From NCERT examples, d = (last term - first term)/(n-1). Here, d = (45 + 24d - 45)/24 → d = 25/24 ≈ 1.04 cm.

Question 8:

A taxi charges ₹15 for the first km and ₹8 for each additional km. Write the fare as an AP and find the fare for 12 km.

Answer:

Problem Interpretation

Taxi fare forms an AP where the first term (a₁) is ₹15 and common difference (d) is ₹8 for extra km.

Mathematical Modeling

AP terms: 15, 23, 31, ... (since 15 + 8(n-1)).
For 12 km, n = 12.

Solution

Using the AP formula aₙ = a₁ + (n-1)d, a₁₂ = 15 + (12-1)×8 = 15 + 88 = ₹103. Our textbook shows similar problems with step notation, confirming the calculation.

Question 9:

A ladder has rungs spaced uniformly. The bottom rung is 20 cm from the ground, and the top rung is 100 cm. If there are 9 rungs, find the distance between two consecutive rungs. (Uniform spacing, arithmetic progression)

Answer:

Problem Interpretation

We need to find the equal distance between ladder rungs, forming an arithmetic progression.

Mathematical Modeling

First term (a₁) = 20 cm (bottom rung)
Last term (a₉) = 100 cm (top rung)
Number of terms (n) = 9

Solution

Using the AP formula: aₙ = a₁ + (n-1)d. Substituting, 100 = 20 + 8d ⇒ d = 10 cm. Thus, the distance between rungs is 10 cm.

Question 10:

A taxi charges ₹15 for the first km and ₹8 for each additional km. Write the fare as an arithmetic progression and find the fare for 12 km.

Answer:

Problem Interpretation

Taxi fare increases uniformly per km, forming an arithmetic progression.

Mathematical Modeling

First term (a₁) = ₹15 (first km)
Common difference (d) = ₹8 (additional km)

Solution

AP: 15, 23, 31, ... For 12 km (n=12), fare = a₁ + (n-1)d = 15 + 11×8 = ₹103. The total fare is ₹103.

Question 11:

A farmer plants saplings in rows such that the number of saplings in each row forms an arithmetic progression. The first row has 5 saplings, and the common difference is 3.

(a) Find the number of saplings in the 10th row.
(b) If there are 15 rows, how many total saplings are planted?

Answer:

(a) Given: First term (a) = 5, Common difference (d) = 3.
Using n^th term formula: a_n = a + (n - 1)d.
For 10th row: a₁₀ = 5 + (10 - 1) × 3 = 5 + 27 = 32 saplings.

(b) Total rows (n) = 15.
Using sum formula: S_n = n/2 [2a + (n - 1)d].
S₁₅ = 15/2 [10 + 42] = 15 × 26 = 390 saplings.

Question 12:

A staircase has steps with heights in arithmetic progression. The 3rd step is 18 cm high, and the 7th step is 30 cm high.

(a) Determine the common difference.
(b) Calculate the height of the 1st step.

Answer:

(a) Given: a₃ = 18, a₇ = 30.
Using n^th term formula: a_n = a + (n - 1)d.
Equations: a + 2d = 18 and a + 6d = 30.
Subtracting: 4d = 12 → d = 3 cm.

(b) Substituting d = 3 in first equation: a + 6 = 18 → a = 12 cm.

Question 13:

A taxi fare follows an arithmetic progression where the fare for the first km is ₹15 and increases by ₹8 per subsequent km.

(a) Find the fare for the 12th km.
(b) If a passenger travels 20 km, what is the total fare?

Answer:

(a) Given: a = 15, d = 8.
Using n^th term formula: a₁₂ = 15 + (12 - 1) × 8 = 15 + 88 = ₹103.

(b) Total distance (n) = 20 km.
Using sum formula: S₂₀ = 20/2 [30 + 152] = 10 × 182 = ₹1820.

Question 14:

A farmer plants saplings in rows such that the number of saplings in each row forms an arithmetic progression. The first row has 5 saplings, and each subsequent row has 3 more saplings than the previous one. If there are 15 rows in total:

(a) Find the number of saplings in the 10th row.
(b) Calculate the total number of saplings planted by the farmer.

Answer:

(a) Number of saplings in the 10th row:

Given: First term (a) = 5, Common difference (d) = 3.
The nth term of an arithmetic progression is given by: aₙ = a + (n − 1)d.
For the 10th row (n = 10):
a₁₀ = 5 + (10 − 1) × 3
a₁₀ = 5 + 27
a₁₀ = 32
Thus, the 10th row has 32 saplings.

(b) Total number of saplings:

Total rows (n) = 15.
Sum of first n terms of an AP: Sₙ = n/2 [2a + (n − 1)d].
S₁₅ = 15/2 [2 × 5 + (15 − 1) × 3]
S₁₅ = 7.5 [10 + 42]
S₁₅ = 7.5 × 52
S₁₅ = 390
The farmer planted a total of 390 saplings.

Question 15:

The seating arrangement of an auditorium follows an arithmetic progression. The first row has 20 seats, and each subsequent row has 2 more seats than the previous one. The last row has 50 seats.

(a) Determine the number of rows in the auditorium.
(b) If the ticket price per seat is ₹150, calculate the total revenue if all seats are occupied.

Answer:

(a) Number of rows in the auditorium:

Given: First term (a) = 20, Common difference (d) = 2, Last term (aₙ) = 50.
The nth term of an AP is: aₙ = a + (n − 1)d.
Substituting values:
50 = 20 + (n − 1) × 2
30 = (n − 1) × 2
n − 1 = 15
n = 16
The auditorium has 16 rows.

(b) Total revenue:

Total seats (Sₙ) = Sum of all rows.
Sum formula: Sₙ = n/2 [a + aₙ].
S₁₆ = 16/2 [20 + 50]
S₁₆ = 8 × 70
S₁₆ = 560 seats.
Revenue = Total seats × Ticket price.
Revenue = 560 × 150
Revenue = ₹84,000
The total revenue is ₹84,000.

Question 16:

A farmer plants trees in rows to form an Arithmetic Progression. The number of trees in the 5th row is 18, and in the 8th row, it is 27. The farmer wants to know:

How many trees are there in the first row?
What is the common difference between consecutive rows?

Justify your answer step-by-step.

Answer:

Given, the number of trees in the 5th row (a₅) = 18
Number of trees in the 8th row (a₈) = 27

We know the general form of an Arithmetic Progression is: a_n = a + (n - 1)d, where:
a = first term, d = common difference

For the 5th row:
a + 4d = 18 → (Equation 1)

For the 8th row:
a + 7d = 27 → (Equation 2)

Subtract Equation 1 from Equation 2:
(a + 7d) - (a + 4d) = 27 - 18
3d = 9
d = 3

Substitute d = 3 in Equation 1:
a + 4(3) = 18
a = 18 - 12
a = 6

Thus:
Number of trees in the first row = 6
Common difference (d) = 3

Question 17:

Riya saves money every month in an Arithmetic Progression. Her total savings in the first 5 months is ₹1,250, and in the first 10 months, it is ₹3,500. Help her find:

Her monthly savings pattern (first term and common difference).
How much she will save in the 15th month?

Answer:

Given:
Sum of first 5 months (S₅) = ₹1,250
Sum of first 10 months (S₁₀) = ₹3,500

We know the sum of first n terms of an AP is: S_n = n/2 [2a + (n - 1)d]

For S₅:
5/2 [2a + 4d] = 1,250
[2a + 4d] = 500 → (Equation 1)

For S₁₀:
10/2 [2a + 9d] = 3,500
[2a + 9d] = 700 → (Equation 2)

Subtract Equation 1 from Equation 2:
(2a + 9d) - (2a + 4d) = 700 - 500
5d = 200
d = 40

Substitute d = 40 in Equation 1:
2a + 4(40) = 500
2a = 500 - 160
a = 170

Thus:
First term (a) = ₹170
Common difference (d) = ₹40

Savings in the 15th month (a₁₅):
a₁₅ = a + 14d
= 170 + 14(40)
= 170 + 560
= ₹730

Question 18:

Answer:

Given:
First term (a) = 5 saplings
Common difference (d) = 3 saplings
Number of terms (n) = 15 rows

We need to find the total number of saplings, which is the sum of the arithmetic progression (S_n).

Formula for sum of an AP:
S_n = n/2 [2a + (n-1)d]

Substituting the values:
S₁₅ = 15/2 [2(5) + (15-1)(3)]
= 15/2 [10 + 42]
= 15/2 × 52
= 15 × 26
= 390 saplings

Thus, the farmer planted a total of 390 saplings.

Question 19:

The monthly savings of Riya form an arithmetic progression. She saves ₹500 in the first month and increases her savings by ₹50 every subsequent month. Calculate her total savings after 2 years. Also, determine in which month her savings will first exceed ₹1000.

Answer:

Given:
First term (a) = ₹500
Common difference (d) = ₹50
Number of terms (n) = 24 months (2 years)

Part 1: Total savings after 2 years
Formula for sum of an AP:
S_n = n/2 [2a + (n-1)d]

Substituting the values:
S₂₄ = 24/2 [2(500) + (24-1)(50)]
= 12 [1000 + 1150]
= 12 × 2150
= ₹25,800

Part 2: Month when savings exceed ₹1000
We need to find the smallest n such that a_n > 1000.
Formula for nth term of an AP:
a_n = a + (n-1)d

Substituting the values:
500 + (n-1)(50) > 1000
(n-1)(50) > 500
n-1 > 10
n > 11

Thus, Riya's savings will first exceed ₹1000 in the 12th month.

Final answers:
Total savings after 2 years = ₹25,800
Month when savings exceed ₹1000 = 12th month

Question 20:

Answer:

The problem describes an arithmetic progression (AP) where:

First term (a) = 5 saplings
Common difference (d) = 3 saplings
Number of terms (n) = 15 rows

To find the total number of saplings, we calculate the sum of the AP using the formula:

S_n = n/2 [2a + (n - 1)d]

Substitute the given values:
S₁₅ = 15/2 [2 × 5 + (15 - 1) × 3]

Simplify the expression step-by-step:
= 15/2 [10 + 14 × 3]
= 15/2 [10 + 42]
= 15/2 × 52
= 15 × 26
= 390

Thus, the farmer planted a total of 390 saplings. This method ensures accuracy by applying the standard AP sum formula.

Question 21:

The seating arrangement of an auditorium follows an arithmetic progression. The first row has 20 seats, and each subsequent row has 2 more seats than the previous one. If the last row has 50 seats, how many rows are there in the auditorium? Show your calculations clearly.

Answer:

This scenario involves an arithmetic progression (AP) with:

First term (a) = 20 seats
Common difference (d) = 2 seats
Last term (l) = 50 seats

To find the number of rows (n), use the formula for the n^th term of an AP:

l = a + (n - 1)d

Substitute the known values:
50 = 20 + (n - 1) × 2

Solve for n step-by-step:
50 - 20 = (n - 1) × 2
30 = (n - 1) × 2
n - 1 = 15
n = 16

Therefore, the auditorium has 16 rows. This solution correctly applies the AP term formula to determine the total number of rows.

Question 22:

A farmer plants trees in rows to form an arithmetic progression. The number of trees in the 5th row is 22, and in the 8th row, it is 31.

(a) Find the number of trees in the 1st row.

(b) How many trees are there in the 15th row?

Answer:

Given: The number of trees in rows follows an arithmetic progression (AP).

Let: First term = a, Common difference = d.

Step 1: Write the equations for the 5th and 8th terms of the AP.
5th term: a + 4d = 22
8th term: a + 7d = 31

Step 2: Subtract the 5th term equation from the 8th term equation to find d.
(a + 7d) - (a + 4d) = 31 - 22
3d = 9
d = 3

Step 3: Substitute d = 3 into the 5th term equation to find a.
a + 4(3) = 22
a = 22 - 12
a = 10

(a) Answer: The number of trees in the 1st row is 10.

Step 4: Find the 15th term using the AP formula.
15th term = a + 14d
= 10 + 14(3)
= 10 + 42
= 52

(b) Answer: The number of trees in the 15th row is 52.

Question 23:

The sum of the first n terms of an AP is given by S_n = 3n² + 5n.

(a) Find the first term and common difference of the AP.

(b) What is the sum of the first 10 terms?

Answer:

Given: Sum of first n terms, S_n = 3n² + 5n.

(a) Step 1: Find the first term (a) by substituting n = 1 in S_n.
S₁ = 3(1)² + 5(1) = 3 + 5 = 8
First term (a) = S₁ = 8.

Step 2: Find the sum of the first 2 terms (S₂) and subtract S₁ to get the second term.
S₂ = 3(2)² + 5(2) = 12 + 10 = 22
Second term = S₂ - S₁ = 22 - 8 = 14

Step 3: Calculate the common difference (d).
d = Second term - First term = 14 - 8 = 6.

(a) Answer: First term = 8, Common difference = 6.

(b) Step 4: Substitute n = 10 in S_n to find the sum of the first 10 terms.
S₁₀ = 3(10)² + 5(10) = 300 + 50 = 350

(b) Answer: The sum of the first 10 terms is 350.

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