202-NCERT New Syllabus Grade 10 Triangles Ex-6.1

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NCERT New Syllabus Mathematics

Class: 10

Exercise 6.1

Topic: Triangles

Understanding Triangles: A Key Concept in Class 10 Mathematics

Triangles form one of the foundational blocks in geometry, and in the Class 10 NCERT syllabus, this topic holds significant importance. From learning the basic properties of triangles to diving deeper into congruence, similarity, and the Pythagoras theorem, the chapter on triangles equips students with essential tools to solve geometric problems.

In this blog, we will systematically explore each concept, providing clear explanations and practical solutions to the exercises. By the end of this discussion, you'll have a solid understanding of triangles, enabling you to confidently approach simple and complex problems.

Let's begin our journey into the world of triangles and uncover the logic that shapes this fascinating topic!

EXERCISE 6.1

Q1. Fill in the blanks using the correct word given in brackets :

(i) All circles are _________. (congruent, similar)

Ans: similar, "All circles are similar".

 

(ii) All squares are_________ . (similar, congruent)

 Ans: similar, "All squares are similar".

(iii) All _________ triangles are similar. (isosceles, equilateral)

Ans: similar, "All equilateral tringles are similar".

 

(iv) Two polygons of the same number of sides are similar, if (a) their

corresponding angles are _________  and (b) their corresponding sides are_________ . (equal, proportional)

 

Ans: a) equal, b) proportional, 

Two polygons of the same number of sides are similar, if

(a) their corresponding angles are equal and

(b) their corresponding sides are proportional.

 

2. Give two different examples of a pair of

(i) similar figures. (ii) non-similar figures.

Solution:

(i) similar figures:

a) All equilateral tringles are similar.

b) All squares are similar.

(ii) non-similar figures:

a) Equilateral tringle and isosceles traingle are non-similar.

b) Square and rectangle are non-similar. 

 

3. State whether the following quadrilaterals are similar or not:

Ans: No, these quadrilaterals are not similar.

Conclusion

In conclusion, mastering the topic of triangles is essential for building a strong foundation in geometry. By understanding concepts like congruence, similarity, and the Pythagoras theorem, you can easily solve a wide range of problems. As you continue practicing, these geometric principles will become second nature, helping you excel in exams and real-life applications.

Keep exploring, stay curious, and remember—the beauty of mathematics lies in its logical simplicity!

Stay tuned for more insights and solutions to other important topics from the Class 10 NCERT syllabus.

#Class10Maths #NCERTSolutions #Triangles #GeometryBasics #MathForStudents #MathMadeEasy #PythagorasTheorem #MathBlog #StudentLife #LearningIsFun #MathConcepts

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Anil Satpute

201-NCERT New Syllabus Grade 10 Arithmetic Progressions Ex-5.4

Click here to explore the previous ⇨ NCERT New Syllabus Class 10 - Arithmetic Progressions Exercise 5.3

NCERT New Syllabus Mathematics

Class: 10

Exercise 5.4

Topic: Arithmetic Progressions

Introduction to Arithmetic Progressions

Arithmetic Progressions (AP) are a fundamental mathematical concept that forms the basis for understanding various real-world applications, ranging from financial planning to engineering design. In this chapter of the 10th-grade NCERT syllabus, we delve into sequences where the difference between consecutive terms remains constant. This simple yet powerful concept helps students recognize patterns and solve problems involving future predictions, geometric designs, and even natural phenomena.

An arithmetic progression represented as a sequence of numbers like a, a+d, a+2d, and so on, is defined by the first term (a) and the common difference (d). This progression opens up a range of possibilities for solving complex problems by understanding the behavior of numbers in a linear format.

In this blog, we’ll explore the key concepts of AP, including its general form, the formula for the nth term, and the sum of n terms. We’ll also solve various problems from the new NCERT syllabus, making it easier for students to grasp the concept and excel in their exams.

1) The nth term a_n of the AP with first term a and common difference d is given by 

a_n = a + (n – 1) d.

2) a_n is also called the general term of the AP. If there are m terms in the AP, then

a_m represents the last term which is sometimes also denoted by l. 

EXERCISE 5.4

1. Which term of the AP: 121, 117, 113, . . ., is its first negative term?

[Hint: Find n for a_n < 0]

Solution:

1) According to the problem, a₁ = 121_, a₂ = 117_, a₃ = 113. . . so d = – 4,

2) We will have to find n using the above information.

a_n = a + (n – 1) d

a_n = 121 + (n – 1) (– 4)

a_n = 121 + (– 4n + 4)

a_n = 121 – 4n + 4
a_n = 125 – 4n

3) We have to find which first term is negative. 

a_n < 0

125 – 4n < 0

125 < 4n

4n > 125

n > 125/4

n > 31.25

4) Therefore, the first negative term of this AP is the 32nd term. 

2. The sum of the third and the seventh terms of an AP is 6 and their product

is 8. Find the sum of first sixteen terms of the AP.

Solution:

1) Let the first term be "a" and the common difference be "d".

2) Here a₃ + a₇ = 6 ---------- equation 1

3) Here a₃ x a₇ = 8 ---------- equation 2

4) Now we will find a₃ and a₇,

a) First we will find a₃ 

a_n = a + (n – 1) d

a₃ = a + (3 – 1) d

a₃ = a + 2d ---------- equation 3

b) Now we will find a₇ 

a_n = a + (n – 1) d

a₇ = a + (7 – 1) d

a₇ = a + 6d ---------- equation 4

5) From equations 1, 3, and 4, we have,

a₃ + a₇ = 6

a + 2d + a + 6d = 6
2a + 8d = 6
2(a + 4d) = 6
(a + 4d) = 6/2
(a + 4d) = 3
a = 3 – 4d ---------- equation 5

6) From equations 2, 3, 4, and 5, we have,

a₃ x a₇ = 8

(a + 2d) x (a + 6d) = 8

(3 – 4d + 2d) x (3 – 4d + 6d) = 8

(3 – 2d) x (3 + 2d) = 8

(3² – 4d²) = 8

(9 – 4d²) = 8

4d² = 9 – 8
4d² = 1
d² = 1/4
d² = 1/4

d = ± 1/2
d = 1/2 or – 1/2 ---------- equation 6

7) Put d = 1/2 and d = - 1/2 from equation 6 in equation 5,

a) First we take d = 1/2, we get

a = 3 – 4d

a = 3 – 4(1/2)

a = 3 – 2

a = 1 ---------- equation 7

b) First we take d = – 1/2, we get

a = 3 – 4d

a = 3 – 4(– 1/2)

a = 3 – (– 2)

a = 3 + 2

a = 5 ---------- equation 8

8) We know that the sum of the first n terms of an AP is given by:

S_n = (n/2)[2a + (n – 1) d]

a) First we will find sum of first 16 terms with a = 1 and d = 1/2:

S_n = (n/2)[2a + (n – 1) d]

S₁₆ = (16/2)[2(1) + (16 – 1) (1/2)]
S₁₆ = 8[2 + (15) (1/2)]

S₁₆ = 8[2 + (15/2)] 

S₁₆ = 8[(4 + 15)]/2
S₁₆ = 4(19)
S₁₆ = 76 ---------- equation 9

b) First we will find sum of first 16 terms with a = 5 and d = (– 1/2):

S_n = (n/2)[2a + (n – 1) d]

S₁₆ = (16/2)[2(5) + (16 – 1) (– 1/2)]
S₁₆ = 8[10 + (15) (– 1/2)]

S₁₆ = 8[10 – (15/2)] 

S₁₆ = 8[(20 – 15)]/2
S₁₆ = 4(5)
S₁₆ = 20 ---------- equation 10

 9) From equations 9 and 10, we have,

a) S₁₆ = 76 when a = 1 and d = 1/2

b) S₁₆ = 20 when a = 5 and d = – 1/2.

3. A ladder has rungs 25 cm apart. (see the following fig.). The rungs decrease

uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are 2½ m apart, what is the length of the wood required for the rungs?

[Hint : Number of rungs = (250/25) + 1]

Solution:

1) The distance between the rungs is 25 cm.

2) Distance between the top rung and the bottom

     rung is 2½ m. i.e. 5/2 m = 2.5 m = 250 cm.

3) So total number of rungs = [(250/25) + 1] = 11.

4) The length of rungs is decreasing uniformly from bottom to top, so they are in AP.

5) Here, a₁ = 45, l = 25 and n = 11, so,

        S_n = (n/2)[a + l]

        S₁₁ = (11/2)[45 + 25]

        S₁₁ = (11/2)[70]

        S₁₁ = 11(35)

        S₁₁ = 385

6) Therefore, the length of the wood required for the rungs is 385 cm.

4. The houses of a row are numbered consecutively from 1 to 49. Show that

there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x. [Hint : s_{x - 1} = S₄₉ – S_x]

Solution:

1) Row houses are numbered 1, 2, 3, . . 49.

2) So, a₁ = 1_, a₂ = 2_, a₃ = 3. . . so d = 1,

3) The sum of the number of houses preceding the xth house will be S_(x-1).

S_n = (n/2)[2a + (n – 1) d]

S_(x-1) = ((x – 1)/2)[2(1) + (x – 1 – 1) (1)]

S_(x-1) = ((x – 1)/2)[2 + (x – 2)] 

S_(x-1) = ((x – 1)/2)[x]

S_(x-1) = x(x – 1)/2 ---------- equation 1

4) Now we will find S₄₉

S_n = (n/2)[2a + (n – 1) d]

S₄₉ = (49/2)[2(1) + (49 – 1) (1)]

S₄₉ = (49/2)[2 + 48] 

S₄₉ = (49/2)(2)[1 + 24]

S₄₉ = (49)[25]

S₄₉ = 1225 ---------- equation 2

5) Now we will find S_x

S_n = (n/2)[2a + (n – 1) d]

S_x = (x/2)[2(1) + (x – 1) (1)]

S_x = (x/2)[2 + (x – 1)] 

S_x = (x/2)[x + 1]

S_x = x(x + 1)/2 ---------- equation 3

6) Now we will find S₄₉ – S_x using equations 2 and 3.

S₄₉ – S_x = 1225 – [x(x + 1)/2] ---------- equation 4

7) According to the problem, S_(x-1) =  S₄₉ – S_x  ---------- equation 5

8) From equations 1, 4, and 5, we have,

S_(x-1) =  S₄₉ – S_x

x(x – 1)/2 =  1225 – [x(x + 1)/2]

x(x – 1)/2 =  [2450 – (x(x + 1))]/2

x(x – 1) =  [2450 – (x(x + 1))]
x² – x = 2450 – x² – x
2x² = 2450
x² = 1225
x = ± 35

x = 35 or – 35

9) Therefore, the number of houses can't be negative, the value of 

x will be 35.

 

5. A small terrace at a football ground comprises of 15 steps each of which is

50 m long and built of solid concrete. Each step has a rise of 1/4 m and a tread of 1/2 m. (see the following fig.). Calculate the total volume of concrete required to build the terrace. [Hint: Volume of concrete required to build the first step = (1/4) x (1/2) x 50 m³]

 

1) According to the problem and the figure, we have,

i) The heights of the steps are given bellow:

a) The height of the first step is (1/4) m

b) The height of the second step is (1/4) + (1/4) = (1/2) m

c) The height of the 3rd step is (1/2) + (1/4) = (3/4) m

d) The height of the 4th step is (3/4) + (1/4) = (1) m

2) If we consider the height in the changing form, the width will be the same for all

the steps. i.e. the width will be 1/2 m for all the steps.

3) Here the length of the terrace is 50 m.

4) Here we can find:

The volume of the steps = volume of the cuboid

The volume of the steps = Length x Breadth x Height

5) Now we will find the volumes of the steps:

a) The volume of the first step

= (1/4) x (1/2) x (50)

= (1/8) x (50)
= (50/8)

= (25/4) ---------- equation 1

b) The volume of the second step

= (1/2) x (1/2) x (50)

= (1/4) x (50)
= (50/4) ---------- equation 2 

c) The volume of the 3rd step

= (3/4) x (1/2) x (50)

= (3/4) x (25)
= (75/4) ---------- equation 3 

d) The volume of the 4th step is (3/4) + (1/4) = (1) m 

= (1) x (1/2) x (50)

= (1) x (25)

= (25) ---------- equation 4

6) From equations 1, 2, 3, and 4, volumes of the steps are in AP.

7) So, here, a = 25/4, d = 50/4 – 25/4 = 25/4, and n = 15,

8) so the sum of these 15 steps will be,

S_n = (n/2)[2a + (n – 1) d]
S₁₅ = (15/2)[2(25/4) + (15 – 1) (25/4)]
S₁₅ = (15/2)[2(25/4) + (14) (25/4)]
S₁₅ = (15/2) x (2(25/4)) x [1 + (7)]
S₁₅ = (15/2) x (2(25/4)) x [8]
S₁₅ = [(15 x 2 x 25)/8] x [8]
S₁₅ = (15 x 2 x 25)
S₁₅ = (30 x 25)
S₁₅ = (750)

9) The concrete required to build the terrace is 750 m³.

Conclusion: The Power of Arithmetic Progressions

As we wrap up our journey through Arithmetic Progressions, we've seen how sequences unfold and reveal patterns in everyday life—from predicting future events to organizing data. The concept of common differences and the sum of terms allows us to model various scenarios in both math and real-world situations. Whether you’re solving problems in exams or exploring the magic of number sequences, AP knowledge is a powerful tool. Keep practicing, and you’ll see how every sequence leads to new possibilities!

Related Hashtags: 

#ArithmeticProgressions #Class10Math #MathematicsNCERT #APFormulas #MathHelp #LearnMath #MathConcepts #CBSEMath #ProgressionPatterns #MathSequences #MathInLife #NCERTClass10 #SequenceAndPattern #MathWizards #LearningWithNumbers #MathMastery #EndlessPossibilities

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Anil Satpute

200-NCERT New Syllabus Grade 10 Arithmetic Progressions Ex-5.3

Click here to explore the previous ⇨ NCERT New Syllabus Class 10 - Arithmetic Progressions Exercise 5.2

NCERT New Syllabus Mathematics

Class: 10

Exercise 5.3

Topic: Arithmetic Progressions

Introduction to Arithmetic Progressions

Arithmetic Progressions (AP) are a fundamental mathematical concept that forms the basis for understanding various real-world applications, ranging from financial planning to engineering design. In this chapter of the 10th-grade NCERT syllabus, we delve into sequences where the difference between consecutive terms remains constant. This simple yet powerful concept helps students recognize patterns and solve problems involving future predictions, geometric designs, and even natural phenomena.

An arithmetic progression represented as a sequence of numbers like a, a+d, a+2d, and so on, is defined by the first term (a) and the common difference (d). This progression opens up a range of possibilities for solving complex problems by understanding the behavior of numbers in a linear format.

In this blog, we’ll explore the key concepts of AP, including its general form, the formula for the nth term, and the sum of n terms. We’ll also solve various problems from the new NCERT syllabus, making it easier for students to grasp the concept and excel in their exams.

1) The nth term a_n of the AP with first term a and common difference d is given by 

a_n = a + (n – 1) d.

2) a_n is also called the general term of the AP. If there are m terms in the AP, then

a_m represents the last term which is sometimes also denoted by l. 

EXERCISE 5.3

Q1. Find the sum of the following APs:

(i) 2, 7, 12, . . ., to 10 terms.     (ii) –37, –33, –29, . . ., to 12 terms.

(iii) 0.6, 1.7, 2.8, . . ., to 100 terms.     (iv) 1/15, 1/12, 1/10,  . . ., to 11 terms.

Explanation:

1) The nth term a_n of an AP with the first term 'a' and common difference 'd' is given

by a_n = a + (n – 1) d.

2) The sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

3) If a_n = l is the last term of an AP, then

S = (n/2)[2a + (n – 1) d]

S = (n/2)[a + (a + (n – 1) d)]
S = (n/2)[a + a_n]
S = (n/2)[a + l].

Solution:

(i) 2, 7, 12, . . ., to 10 terms.

1) Here, a = 2, d = 7 – 2 = 5, and n = 10

2) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

S = (10/2)[2(2) + (10 – 1) (5)]

S = (5)[4 + 5(9)]
S = (5)[4 + 45]
S = (5)(49)
S = 245

3) Therefore, the sum of the given AP is 245. 

 

(ii) –37, –33, –29, . . ., to 12 terms.

1) Here, a = – 37, d = – 33 + 37 = 4, and n = 12

2) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

S = (12/2)[2(– 37) + (12 – 1) (4)]

S = (6)[– 74 + 4(11)]
S = (6)[– 74 + 44]
S = (6)(– 30)
S = – 180

3) Therefore, the sum of the given AP is – 180. 

(iii) 0.6, 1.7, 2.8, . . ., to 100 terms.

1) Here, a = 0.6, d = 1.7 – 0.6 = 1.1, and n = 100

2) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

S = (100/2)[2(0.6) + (100 – 1) (1.1)]

S = (50)[1.2 + 1.1(99)]
S = (50)[1.2 + 108.9]
S = (50)(110.1)
S = 5505

3) Therefore, the sum of the given AP is 5505. 

(iv) 1/15, 1/12, 1/10,  . . ., to 11 terms.

1) Here, a = 1/15, 

d = 1/12 – 1/15

d = (15 – 12)/(12x15)

d = (3)/(12x15)
d = 1/(4x15)
d = 1/60 and n = 11.

2) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

S = (11/2)[2(1/15) + (11 – 1) (1/60)]

S = (11/2)[2/15 + 10/60]
S = (11/2)[2/15 + 1/6]
S = (11/2)/[(4/30) + (5/30)]

S = (11/2)/(4 + 5)/30 

S = (11/2)/(9/30)
S = (11/2)/(3/10)

S = 33/20 

3) Therefore, the sum of the given AP is 33/20.

Q2. Find the sums given below :

(i) 7 + 10½ + 14 + . . . + 84

(ii) 34 + 32 + 30 + . . . + 10

(iii) –5 + (–8) + (–11) + . . . + (–230)

Solution:

(i) 7 + 10½ + 14 + . . . + 84

1) Here, a = 7, d = 10½ – 7 = 3½ = 7/2, nth term is a_n = l = 84

2) We know that,

a_n = a + (n – 1) d

84 = 7 + (n – 1) (7/2)

(n – 1) (7/2) = 84 – 7
(n – 1) (7/2) = 77
(n – 1) = 77(2/7)
(n – 1) = 11(2)
(n – 1) = 22
n = 22 + 1
n = 23 

3) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[a + l]

S = (23/2)[7 + 84]

S = (23/2)[91]
S = 2093/2
S = 1046.5

4) Therefore, the sum of the given AP is 1046.5.

(ii) 34 + 32 + 30 + . . . + 10

1) Here, a = 34, d = 32 – 34 = – 2, nth term is a_n = 10

2) We know that,

a_n = a + (n – 1) d

10 = 34 + (n – 1) (– 2)

(n – 1) (– 2) = 10 – 34
(n – 1) (– 2) = – 24
(n – 1) = (– 24)/(– 2)
(n – 1) = 12
n = 12 + 1
n = 13 

3) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[a + l]

S = (13/2)[34 + 10]

S = (13/2)[44]

S = (13)[22]
S = 286 

4) Therefore, the sum of the given AP is 286.

(iii) –5 + (–8) + (–11) + . . . + (–230)

1) Here, a = – 5, d = (– 8) – (– 5) = – 3, nth term is a_n = – 230

2) We know that,

a_n = a + (n – 1) d

– 230 = – 5 + (n – 1) (– 3)

(n – 1) (– 3) = 5 – 230
(n – 1) (– 3) = – 225
(n – 1) = (– 225)/(– 3)
(n – 1) = 75
n = 75 + 1
n = 76

3) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[a + l]

S = (76/2)[– 5 + (– 230)]

S = (38)[– 235]

S = – 8930 

4) Therefore, the sum of the given AP is – 8930.

Q3. In an AP:

(i) given a = 5, d = 3, a_n = 50, find n and S_n.

(ii) given a = 7, a₁₃ = 35, find d and S₁₃.

(iii) given a₁₂ = 37, d = 3, find a and S₁₂.

(iv) given a₃ = 15, S₁₀ = 125, find d and a₁₀.

(v) given d = 5, S_n = 75, find a and a₉.

(vi) given a = 2, d = 8, S_n = 90, find n and a_n.

(vii) given a = 8, a_n = 62, S_n = 210, find n and d.

(viii) given a_n = 4, d = 2, S_n = – 14, find n and a.

(ix) given a = 3, n = 8, S = 192, find d.

(x) given l = 28, S = 144, and there are total 9 terms. Find a.

Solution:

(i) given a = 5, d = 3, a_n = 50, find n and S_n.

1) Here, a = 5, d = 3, nth term is a_n = 50.

2) We know that,

a_n = a + (n – 1) d

50 = 5 + (n – 1) (3)

(n – 1) (3) = 50 – 5
(n – 1) (3) = 45
(n – 1) = 45/3
(n – 1) = 15
n = 15 + 1
n = 16

3) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[a + l]

S = (16/2)[5 + 50]

S = (8)[55]

S = 440 

4) Therefore, S_n = 440, n = 16.

(ii) given a = 7, a₁₃ = 35, find d and S₁₃.

1) Here, a = 7, d = ?, 13th term is a₁₃ = 35. Find d and S₁₃

2) We know that,

a_n = a + (n – 1) d

a₁₃ = 7 + (13 – 1) (d)

a₁₃ = 7 + 12d

35 = 7 + 12d
12d = 35 – 7
12d = 28

d = 28/12 

d = 7/3 --------- equation 1

3) We know that the sum of the first n terms of an AP is given by:

S_n = (n/2)[a + l]

S₁₃ = (13/2)[7 + 35]

S₁₃ = (13/2)[42]

S₁₃ = (13)[21]
S₁₃ = 273

4) Therefore, S₁₃ = 273, d = 7/3.

(iii) given a₁₂ = 37, d = 3, find a and S₁₂.

1) Here, a = ?, d = 3, 12th term is a₁₂ = 37. Find "a" and S₁₂"

2) We know that,

a_n = a + (n – 1) d

a₁₂ = a + (12 – 1) (3)

37 = a + 3(11)

37 = a + 33
a = 37 – 33
a = 4 --------- equation 1

3) We know that the sum of the first n terms of an AP is given by:

S_n = (n/2)[a + l]
S₁₂ = (12/2)[4 + 37]

S₁₂ = (6)[41]
S₁₂ = 246 

4) Therefore, S₁₂ = 246, a = 4.  

 

(iv) given a₃ = 15, S₁₀ = 125, find d and a₁₀.

1) Here, a₃ = 15, S₁₀ = 125. Find d and a₁₀.

2) We know that,

a_n = a + (n – 1) d
a₃ = a + (3 – 1) (d)

15 = a + 2d

a + 2d = 15 --------- equation 1 

3) We know that the sum of the first n terms of an AP is given by:

S_n = (n/2)[2a + (n – 1) d]
S₁₀ = (10/2)[2(a) + (10 – 1) d]

125 = (5)[2(a) + 9d] 

[2(a) + 9d] = 125/5

[2(a) + 9d] = 25 

2a + 9d = 25 --------- equation 2

4) Subtract double of equation 1 from equation 2

2a + 9d = 25

2a + 4d = 30

  ( – )  ( – )     ( – )  

  --------------------------- 

      5d = –  5

d = – 5/5

d = – 1 --------- equation 3

5) Put d = – 1 from equation 3 in equation 1, and we get,

a + 2d = 15

a + 2(– 1) = 15
a – 2 = 15
a = 15 + 2
a = 17 --------- equation 4 

6) We know that,

a_n = a + (n – 1) d

a₁₀ = 17 + (10 – 1) (– 1) 

a₁₀ = 17 + 9 (– 1)

a₁₀ = 17 – 9
a₁₀ = 8

7) Therefore, a₁₀ = 8, d = – 1. 

 

(v) given d = 5, S₉ = 75, find a and a₉.

1) Here, d = 5, S₉ = 75. Find a and a_9.

2) We know that,

S_n = (n/2)[2a + (n – 1) d]

S₉ = (9/2)[2a + (9 – 1) (5)]

75 = (9/2)[2a + 8 (5)]
75 = (9/2)[2a + 40]
75 = 9[a + 20]
[a + 20] = 75/9
[a + 20] = 25/3
a = (25/3) – 20
a = (25 – 60)/3
a = – 35/3

3) We know that the 

a_n = a + (n – 1) d

a₉ = – 35/3 + (9 – 1) (5)

a₉ = – 35/3 + (8)(5)

a₉ = – 35/3 + 40
a₉ = (– 35 + 120)/3
a₉ = 85/3

4) Therefore, a₉ = 85/3, a = – 35/3.

(vi) given a = 2, d = 8, S_n = 90, find n and a_n.

1) Here, a = 2, d = 8, S_n = 90. Find n and a_n.

2) We know that,

S_n = (n/2)[2a + (n – 1) d]

S_n = (n/2)[2(2) + (n – 1)(8)]
90 = (n/2)[4 + 8(n – 1)]
90 = n[2 + 4(n – 1)]
90 = n(2 + 4n – 4)
90 = n(4n – 2)
90 = 4n² – 2n

45 = 2n² – n
2n² – n – 45 = 0

2n² – 10n + 9n – 45 = 0

2n(n – 5) + 9(n – 5) = 0
(n – 5)(2n + 9) = 0
(n – 5) = 0 or (2n + 9) = 0
n = 5 or 2n = – 9
n = 5 or n = – 9/2 

3) As n is always a positive integer, ignore n = – 9/2. So we have n = 5.

4) We know that,

a_n = a + (n – 1) d
a₅ = 2 + (5 – 1) (8)

a₅ = 2 + (8)(4)a₅ = 2 + 32

a₅ = 34

5) Therefore, a₅ = 34, n = 5. 

(vii) given a = 8, a_n = 62, S_n = 210, find n and d.

1) Here, a = 8, a_n = 62, S_n = 210. Find n and d.

2) We know that,

S_n = (n/2)[a + l]

210 = (n/2)[8 + 62]
210 = (n/2)[70]
210 = 35n
35n = 210
n = 210/35
n = 6

 3) Now we will find d using the formula:

a_n = a + (n – 1) d
62 = 8 + (6 – 1) (d)

62 = 8 + 5d

5d = 62 – 8

5d = 54

d = 54/5

5) Therefore, d = 54/5, n = 6. 

(viii) given a_n = 4, d = 2, S_n = – 14, find n and a.

1) Here, d = 2, a_n = 4, S_n = – 14. Find n and a.

2) We know that,

S_n = (n/2)[a + l]

– 14 = (n/2)[a + 4] 

(n/2)[a + 4] = – 14
n[a + 4] = – 14(2)
n[a + 4] = – 28 --------- equation 1

 3) We know that,

a_n = a + (n – 1) d
4 = a + (n – 1) (2)

4 = a + 2n – 2a + 2n = 4 + 2

a + 2n = 6

a = 6 – 2n --------- equation 2

4) Put a = 6 – 2n from equation 2 in equation 1, we get, 

n(a + 4) = – 28
n((6 – 2n) + 4) = – 28
n(6 – 2n + 4) = – 28

n(10 – 2n) = – 28 

2n(5 – n) = – 28

n(5 – n) = – 14

5n – n² = – 14

n² – 5n – 14 = 0

n² – 7n + 2n – 14 = 0

n(n – 7) + 2(n – 7) = 0

(n – 7)(n + 2) = 0
(n – 7) = 0 or (n + 2) = 0
n = 7 or n = – 2

5) As n is always a positive integer, ignore n = – 2. So we have n = 7.

6) Put n = 7 in equation 2, and we get,

a = 6 – 2na = 6 – 2(7)

a = 6 – 14

a = – 8

7) Therefore, a = – 8, n = 7.

(ix) given a = 3, n = 8, S = 192, find d.

1) Here, a = 3, n = 8, sum of first n terms is S_n = 192, find d.

2) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

192 = (8/2)[2(3) + (8 – 1) d]

192 = 4[6 + 7d]

4[6 + 7d] = 192

[6 + 7d] = 192/4

[6 + 7d] = 48

7d = 48 – 6

7d = 42
d = 42/7
d = 6

3) Therefore, d = 6.

 

(x) given l = 28, S = 144, and there are total 9 terms. Find a.

1) Here, l = 28, S_n = 144, n = 9, Find a.

2) We know that,

S_n = (n/2)[a + l]

144 = (9/2)[a + 28] 

(9/2)[a + 28] = 144
[a + 28] = 144(2/9)
[a + 28] = 16 (2)

a + 28 = 32

a = 32 – 28

a = 4

3) Therefore, a = 4.

Q4. How many terms of the AP: 9, 17, 25, . . . must be taken to give a sum of

636?

Solution:

1) Here, a₁ = a = 9, a₂ = 17, a₃ = 25, sum of first n terms is S_n = 636, find n.

2) According to the problem,

d = a₂ – a₁

d = 17 – 9
d = 8

3) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[2a + (n – 1) d]

636 = (n/2)[2(9) + (n – 1) (8)]

636 = (n/2)[18 + 8(n – 1)]

636 = n[9 + 4(n – 1)]

636 = n[9 + 4n – 4]

636 = n[4n + 5]

636 = 4n² + 5n

4n² + 5n – 636 = 0

4n² + 53n – 48n – 636 = 0

n(4n + 53) – 12(4n + 53) = 0

(4n + 53)(n – 12) = 0
(4n + 53) = 0 or (n – 12) = 0
n = – 53/4 or n = 12

4) As n is always a positive integer, ignore n = – 53/4. So we have n = 12.

Q5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Solution:

1) Here, a₁ = a = 5, a_n = 45, S_n = 400, find n and d.

2) We know that the sum of the first n terms of an AP is given by:

S = (n/2)[a + l]

400 = (n/2)[(5) + 45]

400 = (n/2)[50]

400 = n[25]

n[25] = 400

n = 400/25

n = 16

3) Now we will find the value of d. 

a_n = a + (n – 1) d 

45 = 5 + d (16 – 1)

15d = 45 – 5

15d = 40

d = 40/15

d = 8/3 

4) Therefore, d = 8/3, n = 16.

Q6. The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there, and what is their sum?

Solution:

1) Here, a₁ = a = 17, a_n = 350, d = 9, find n and S_n.

2) Now we will find the value of d. 

a_n = a + (n – 1) d 

350 = 17 + 9(n – 1)

9(n – 1) = 350 – 17

9(n – 1) = 333

(n – 1) = 333/9

(n – 1) = 37

n = 37 + 1

n = 38 

3) We know that the sum of the first n terms of an AP is given by:

S_n = (n/2)[a + l]

S_n = (38/2)[(17) + 350]

S_n = 19[17 + 350]

S_n = 19(367)

S_n = 6973

4) Therefore, n = 38, S_n = 6973.

Q7. Find the sum of the first 22 terms of an AP in which d = 7 and the 22nd term is 149.

Solution:

1) Here, a₂₂ = 149, d = 7, find and S₂₂.

2) Now we will find the value of d. 

a_n = a + (n – 1) d

a₂₂ = a + (22 – 1) (7) 

149 = a + 7(22 – 1)

a = 149 – 7(21)

a = 149 – 147

a = 2 

3) We know that the sum of the first n terms of an AP is given by:

S_n = (n/2)[a + l]

S₂₂ = (22/2)[(2) + 149]

S₂₂ = (11)[151]

S₂₂ = 1661

4) Therefore, a = 2, S_n = 1661.

Q8. Find the sum of the first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Solution:

1) Here, a₂ = 14, a₃ = 18, find and S₅₁.

2) According to the problem,

d = a₃ – a₂

d = 18 – 14
d = 4

3) Now we will find the value of d. 

a_n = a + (n – 1) d

a₂ = a + (2 – 1) (4) 

14 = a + 4(1)

a = 14 – 4

a = 10

4) We know that the sum of the first n terms of an AP is given by:

S_n = (n/2)[2a + (n – 1)d]

S₅₁ = (51/2)[2(10) + (51 – 1)(4)]

S₅₁ = (51/2)[20 + (50)(4)]

S₅₁ = (51/2)[20 + 200]

S₅₁ = (51/2)[220]
S₅₁ = 51[110]
S₅₁ = 5610

5) Therefore, S₅₁ = 5610.

Q9. If the sum of the first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of the first n terms.

Solution:

1) Here, S₇ = 49, S₁₇ = 289, find and S_n. Let the first term be 'a' and the common

difference be 'd'. 

2) We know that the sum of the first n terms of an AP is given by:

S_n = (n/2)[2a + (n – 1)d]

S₇ = (7/2)[2a + (7 – 1)(d)]

49 = (7/2)[2a + (6)(d)]

49 = 7[a + 3d]

a + 3d = 49/7
a + 3d = 7 --------- equation 1

3) Now we will use S₁₇ = 289 to get another equation

S_n = (n/2)[2a + (n – 1)d]

S₁₇ = (17/2)[2a + (17 – 1)d]

S₁₇ = (17/2)[2a + (16)d]

S₁₇ = 17[a + 8d]
289 = 17[a + 8d]
17[a + 8d] = 289
a + 8d = 289/17
a + 8d = 17 --------- equation 2

4) Subtract equation 1 from equation 2

a + 8d = 17

a + 3d =   7

     ( – )  ( – )     ( – )

--------------------------

    5d = 10

d = 10/5

d = 2 --------- equation 3

5) Put d = 2 from equation 3 in equation 1, and we get,

a + 3d = 7

a + 3(2) = 7
a + 6 = 7

a = 7 – 6

a = 1

2) We know that the sum of the first n terms of an AP is given by:

S_n = (n/2)[2a + (n – 1)d] 

S_n = (n/2)[2(1) + (n – 1)(2)]

S_n = n[1 + (n – 1)]

S_n = n[n]

S_n = n² 

5) Therefore, S_n = n².

Q10. Show that a_1, a_2, . . ., a_n, . . . form an AP where a_n is defined as below :

(i) a_n = 3 + 4n     (ii) a_n = 9 – 5n. 

Also, find the sum of the first 15 terms in each case.

Solution:

(i) a_n = 3 + 4n.

1) Put n = 1, 2, 3 in equation a_n = 3 + 4n to get a₁, a₂, a₃  and so on.

a) Put n = 1,

a_n = 3 + 4n

a₁ = 3 + 4(1)

a₁ = 3 + 4

a₁ = 7

b) Put n = 2,

a_n = 3 + 4n

a₂ = 3 + 4(2)

a₂ = 3 + 8

a₂ = 11

c) Put n = 3,

a_n = 3 + 4n

a₃ = 3 + 4(3)

a₃ = 3 + 12

a₃ = 15 

2) Here, a₁ = 7, a₂ = 11, a₃ = 15.

a) First difference:

d = a₂ – a₁

d = 11 – 7

d = 4 --------- equation 1

b) Second difference:

d = a₃ – a₂

d = 15 – 11

d = 4 --------- equation 2

3) Here the common difference d = a₂ – a₁ = a₃ – a₂ = 4, 

so the equation a_n = 3 + 4n form an AP and their terms are 7, 11, 15 . . .

4) Now we will find the sum of the first 15 terms.

5) We know that the sum of the first n terms of an AP is given by:

S_n = (n/2)[2a + (n – 1) d]

S₁₅ = (15/2)[2(7) + (15 – 1) (4)]

S₁₅ = (15/2)[2(7) + 4(14)]
S₁₅ = (15)[7 + 2(14)]

S₁₅ = (15)[7 + 28] 

S₁₅ = (15)(35)

S₁₅ = 525

 6) Therefore, here S_n = 525.

(ii) a_n = 9 – 5n.

1) Put n = 1, 2, 3 in equation a_n = 9 – 5n to get a₁, a₂, a₃  and so on.

a) Put n = 1,

a_n = 9 – 5n

a₁ = 9 – 5(1)

a₁ = 9 – 5

a₁ = 4

b) Put n = 2,

a_n = 9 – 5n

a₂ = 9 – 5(2)

a₂ = 9 – 10

a₂ = – 1

c) Put n = 3,

a_n = 9 – 5n

a₃ = 9 – 5(3)

a₃ = 9 – 15

a₃ = – 6

2) Here, a₁ = 4, a₂ = – 1, a₃ = – 6.

a) First difference:

d = a₂ – a₁

d = – 1 – 4

d = – 5 --------- equation 1

b) Second difference:

d = a₃ – a₂

d = –  6 – (– 1)

d = –  6 + 1 

d = – 5 --------- equation 2

3) Here the common difference d = a₂ – a₁ = a₃ – a₂ = – 5, 

so the equation a_n = 9 – 5n form an AP and their terms are 4, – 1, – 6 . . .

4) Now we will find the sum of the first 15 terms.

5) We know that the sum of the first n terms of an AP is given by:

S_n = (n/2)[2a + (n – 1) d]

S₁₅ = (15/2)[2(4) + (15 – 1) (– 5)]

S₁₅ = (15/2)[2(4) – 5(14)]
S₁₅ = (15)[4 – 5(7)]

S₁₅ = (15)[4 – 35] 

S₁₅ = (15)[– 31] 

S₁₅ = – 465

 6) Therefore, here S_n = – 465.

 

Q11. If the sum of the first n terms of an AP is 4n – n², what is the first term

(that is S₁)? What is the sum of the first two terms? What is the second term? Similarly, find the 3rd, the 10th, and the nth terms.

Solution:

1) Here S_n = 4n – n² so,

a) So, first term  S₁ will be,

S_n = 4n – n²

S₁ = 4(1) – (1)²

S₁ = 4 – 1

S₁ = 3 = a₁ = a

b) To get sum of first two terms, put n = 2, in S_n = 4n – n²

S_n = 4n – n²

S₂ = 4(2) – (2)²

S₂ = 8 – 4

S₂ = 4

c) To get sum of first three terms, put n = 3, in S_n = 4n – n²

S_n = 4n – n²

S₃ = 4(3) – (3)²

S₃ = 12 – 9

S₃ = 3 

2) We know that the second term a₂ = S₂ – S₁.

a₂ = S₂ – S₁

a₂ = 4 – 3

a₂ = 1

3) The common difference will be:

d = a₂ – a₁

d = 1 – 3

d = – 2

4) Now we will find 3rd term:

a_n = a + (n – 1) d

a₃ = 3 + (3 – 1) (– 2)

a₃ = 3 + (2) (– 2)
a₃ = 3 – 4
a₃ = – 1

5) Now we will find the 10th term:

a_n = a + (n – 1) d

a₁₀ = 3 + (10 – 1) (– 2)

a₁₀ = 3 + (9) (– 2)
a₁₀ = 3 – 18
a₁₀ = – 15

6) Now we will find the nth term:

a_n = a + (n – 1) d

a_n = 3 + (n – 1) (– 2)

a_n = 3 – 2n + 2
a_n = 5 – 2n

7) Therefore,

a) First term  s₁ = a = 3.

b) Sum of first two terms s₂ = 4. 

c) The second term a₂ = s₂ – s₁ = 1.

d) Sum of first three terms s₃ = 3. 

e) The third term a₃ = – 1.
f) The 10th term a₁₀ = – 15.
g) The nth term a_n = 5 – 2n.

 

Q12. Find the sum of the first 40 positive integers divisible by 6.

Solution:

1) Here a₁ will be 6, and a₂ will be 12, so here d = 6.

So, now we will find sum of first 40 integers,

S_n = (n/2)[2a + (n – 1) d

S₄₀ = (40/2)[2(6) + (40 – 1) (6)]

S₄₀ = 20[12 + 6(39)]

S₄₀ = 20[12 + 234]

S₄₀ = 20[246]

S₄₀ = 4920

2) The sum of the first 40 terms will be 4920.

Q13. Find the sum of the first 15 multiples of 8.

Solution:

1) Here a₁ will be 8, and a₂ will be 16, so here d = 8.

So, now we will find sum of first 15 multiples of 8,

S_n = (n/2)[2a + (n – 1) d

S₁₅ = (15/2)[2(8) + (15 – 1) (8)]

S₁₅ = 15[8 + 4(14)]

S₁₅ = 15[8 + 56]

S₁₅ = 15[64]

S₁₅ = 960

2) The sum of the first 15 multiples of 8 is 960.

Q14. Find the sum of the odd numbers between 0 and 50.

Solution:

1) Here, our odd numbers are 1, 3, 5, . . 47, 49.

2) We have a₁ = 1, a₂ = 3, d = 2 and a_n = 49,

a_n = a + (n – 1) d

49 = 1 + 2(n – 1)

2(n – 1) = 49 – 1

2(n – 1) = 48

(n – 1) = 48/2

(n – 1) = 24 

n = 25

 3) According to the problem,

So, now we will find sum of odd numbers from 0 to 50.

S_n = (n/2)[a + l]

S₂₅ = (25/2)[1 + 49]

S₂₅ = (25/2)[50]

S₂₅ = 25[25]

S₂₅ = 625

4) The sum of all odd numbers between 0 and 50 is 625.

Q15. A contract on a construction job specifies a penalty for delay of

completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money does the contractor have to pay as a penalty if he has delayed the work by 30 days?

Solution:

1) Here a₁ will be 200, and a₂ will be 250, so here d = 50.

So, now we will find total penalty for the delayed work by 30 days, so we will have to find S_n.

S_n = (n/2)[2a + (n – 1) d

S₃₀ = (30/2)[2(200) + (30 – 1) (50)]

S₃₀ = 15[400 + 50(29)]

S₃₀ = 15[400 + 1450]

S₃₀ = 15[1850]

S₃₀ = 27750

2) The total penalty for the delayed work by 30 days is 27750.

Q16. A sum of Rs 700 is to be used to give seven cash prizes to students of a

school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Solution:

1) Let the cost of the first prize be Rs x.

2) According to the problem, the cost of the second prize is (x – 20).

3) The cost of the third prize is (x – 40). So the common difference is d = – 20.

4) As the sum of Rs 700 is to be used give 7 cash prizes, S₇ = 700.

So, now we will find total penalty for the delayed work by 30 days, so we will have to find S_n.

S_n = (n/2)[2a + (n – 1) d

S₇ = (7/2)[2(x) + (7 – 1) (– 20)]

S₇ = (7)[(x) + (6) (– 10)]

700 = (7)[(x) – 60]

100 = x – 60

x = 100 + 60

x = 160 

5) The value of each of the prizes was (i) Rs 160, (ii) Rs 140, 

(iii) Rs 120, (iv) Rs 100, (v) Rs 80, (vi) Rs 60, and (vii) Rs 40.

Q17. In a school, students thought of planting trees in and around the school

to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted by the students?

Solution:

1) Here a₁ will be 1, and a₂ will be 2, a_n will be 12, so here d = 1.

So, now we will find total penalty for the delayed work by 30 days, so we will have to find S_n.

S_n = (n/2)[2a + (n – 1) d

S₁₂ = (12/2)[2(1) + (12 – 1) (1)]

S₁₂ = 6[2 + 1(11)]

S₁₂ = 6[2 + 11]

S₁₂ = 6[13]

S₁₂ = 78

2) Therefore, one section of a school will plant 78 trees. So 3 sections will

plant 78 x 3 = 234 plants.

Q18. A spiral is made up of successive semicircles, with centers alternately at

A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, . . . as shown in Fig. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take p = 22/7)

Solution:

1) We know that the perimeter of the first semicircle is 𝞹r.

2) According to the problem, the perimeters of the semicircle are:

a) r = 0.5 so  a₁ = 0.5 𝞹 = 𝞹/2

b) r = 1 so  a₂ = 𝞹 = 𝞹

c) r = 1.5 so  a₃ = 1.5 𝞹 = 3𝞹/2

3) Here a_1, a_2, and a_3, are the semicircle's perimeter and are in AP.

4) Here d = a₂ – a₁ = 𝞹 – (𝞹/2) = (𝞹/2)

5) We will have to find S₁₃, so we have,

S_n = (n/2)[2a + (n – 1) d

S₁₃ = (13/2)[2(𝞹/2) + (13 – 1) (𝞹/2)]

S₁₃ = (13/2)(𝞹/2)[2 + 12]

S₁₃ = (13/2)(𝞹/2)[14]

S₁₃ = (13/2)(𝞹)[7]

S₁₃ = (13/2)(22/7)[7]

S₁₃ = (13/2)(22)

S₁₃ = (13)(11)

S₁₃ = 143

6) The length of such a spiral of thirteen consecutive semicircles is 143 cm.

Q19. 200 logs are stacked in the following manner: 20 logs in the bottom row,

19 in the next row, 18 in the row next to it, and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

Solution:

1) The number of logs in each row forms an arithmetic progression (AP).

2) According to the problem, the first term a₁ = 20_, a₂ = 19_, a₃ = 18. . . and 

S_n = 200, so d = – 1,

3) We will have to find n using the above information.

S_n = (n/2)[2a + (n – 1) d

S_n = (n/2)[2(20) + (n – 1) (– 1)]

200 = (n/2)[40 – (n – 1)]

200 = (n/2)[40 – n + 1)] 

400 = n(41 – n)

400 = 41n –  n²

n² – 41n + 400 = 0

n² – 16n – 25n + 400 = 0

n(n – 16) – 25(n – 25) = 0

(n – 16)(n – 25) = 0

(n – 16) = 0 or (n – 25) = 0
n = 16 or n = 25

4) Now we will find a₁₆ and a₂₅ using the formula a_n = a + (n – 1) d. 

a) First we will find a₁₆ 

a_n = a + (n – 1) d 

a₁₆ = 20 + (– 1) (16 – 1)

a₁₆ = 20 – (15)

a₁₆ = 5

b) First we will find a₂₅ 

a_n = a + (n – 1) d 

a₂₅ = 20 + (– 1) (25 – 1)

a₂₅ = 20 – (24)

a₂₅ = – 4

5) The number of logs can't be negative, a₂₅ = – 4 is impossible.

So 200 logs can be placed in 16 rows and there are 5 logs in the 16th log.

Q20. In a potato race, a bucket is placed at the starting point, which is 5 m

from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.

A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? 

[Hint: To pick up the first potato and the second potato, the total distance (in meters) run by a competitor is 2 × 5 + 2 × (5 + 3)]

Solution:

1) The potatoes are placed at 5, 8, 11, 14, 17 . . . distances from the bucket.

2) These distances form an AP.

3) According to the problem, a competitor travels distances as 10, 16, 22, 28, . . .

4) So a₁ = 10_, a₂ = 16_, a₃ = 22, a₄ = 28. . . , so d = 6,

5) Now we will have to find S₁₀ using the above information.

S_n = (n/2)[2a + (n – 1) d

S₁₀ = (10/2)[2(10) + (10 – 1) (6)]

S₁₀ = (10/2)(2)[10 + (9) (3)]

S₁₀ = 10[10 + 27] 

S₁₀ = 10[37]

S₁₀ = 370

6) Therefore, a competitor will have to run a total distance of 370 m.

Conclusion: The Power of Arithmetic Progressions

As we wrap up our journey through Arithmetic Progressions, we've seen how sequences unfold and reveal patterns in everyday life—from predicting future events to organizing data. The concept of common differences and the sum of terms gives us the ability to model various scenarios in both math and real-world situations. Whether you’re solving problems in exams or exploring the magic of number sequences, the knowledge of AP serves as a powerful tool. Keep practicing, and you’ll see how every sequence leads to new possibilities!

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Anil Satpute