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Arithmetic Progression

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Class 10 Maths Chapter 9
Arithmetic Progression
Important Questions

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Class 10 Mathematics chapter 9, ‘Arithmetic Progression’ is a crucial topic that necessitates a comprehensive grasp and ample practice. To assist students in their exam preparation, we have assembled a collection of essential questions pertaining to Class 10 Arithmetic Progressions. An arithmetic progression (AP) is a sequence of numbers where each term is obtained by adding a fixed value (common difference, ‘d’) to the previous term. It’s a fundamental concept in mathematics. Key formulas include finding the nth term (an = a + (n – 1) * d) and the sum of the first ‘n’ terms (Sn = n/2 * [2a + (n – 1) * d]). APs are useful in solving various mathematical and real-world problems, helping to understand patterns and relationships between numbers. Remember to carefully read the problem of arithmetic progression class 10 ICSE questions apply the relevant formula, and double-check your answers for accuracy when working with APs.

Introduction

An arithmetic progression (AP), in Arithmetic Progression class 10 ICSE Board is also known as an arithmetic sequence, is a fundamental concept in mathematics that you’ll encounter in ICSE Class 10 mathematics. It’s a sequence of numbers in which each term after the first is obtained by adding a fixed constant value to the previous term. This constant value is called the “common difference.” The general form of an arithmetic progression is represented as: , a + d, a + 2d, a + 3d, … Where:’a’ is the first term in the sequence. ‘d’ is the common difference between consecutive terms. Key points to understand about arithmetic progressions: Common Difference (d): The difference between any two consecutive terms in an AP is always the same. It can be positive, negative, or zero. Formula for nth Term (an): The nth term of an arithmetic progression can be calculated using the formula: an = a + (n – 1) * d Sum of First ‘n’ Terms (Sn): You can find the sum of the first ‘n’ terms of an AP using the formula: Sn = (n/2) * [2a + (n – 1) * d] nth Term from the End: To find the nth term from the end of an AP, you can use the formula:an’ = a + (N – n) * d Where ‘an” is the nth term from the end, ‘N’ is the total number of terms, ‘n’ is the position from the end, ‘a’ is the first term, and ‘d’ is the common difference. Finding ‘d’ If you know the first term (a), the nth term (an), and the total number of terms (n), you can find the common difference ‘d’ using the formula:d = (an – a) / (n – 1) Finding ‘n’: If you know the first term (a), the common difference (d), and a term (an) in the sequence, you can find the position ‘n’ using the formula: n = (an – a) / d + 1

What is Arithmetic Progression?

An arithmetic progression (AP), Arithmetic Progression class 10 ICSE Board also known as an arithmetic sequence, is a fundamental concept in ICSE Class 10 mathematics. It’s a sequence of numbers in which each term after the first is obtained by adding a fixed constant value to the previous term. This constant value is called the “common difference.” The general form of an arithmetic progression is represented as: a, a + d, a + 2d, a + 3d, … Where: ‘a’ is the first term in the sequence. ‘d’ is the common difference between consecutive terms. Key points to understand about arithmetic progressions: Common Difference (d): The difference between any two consecutive terms in an AP is always the same. It can be positive, negative, or zero. Formula for nth Term (an): The nth term of an arithmetic progression can be calculated using the formula: an = a + (n – 1) * d Sum of First ‘n’ Terms (Sn): You can find the sum of the first ‘n’ terms of an AP using the formula: Sn = (n/2) * [2a + (n – 1) * d] nth Term from the End: To find the nth term from the end of an AP, you can use the formula: an’ = a + (N – n) * d Where ‘an” is the nth term from the end, ‘N’ is the total number of terms, ‘n’ is the position from the end, ‘a’ is the first term, and ‘d’ is the common difference. Finding ‘d’: If you know the first term (a), the nth term (an), and the total number of terms (n), you can find the common difference ‘d’ using the formula: d = (an – a) / (n – 1) Finding ‘n’: If you know the first term (a), the common difference (d), and a term (an) in the sequence, you can find the position ‘n’ using the formula: n = (an – a) / d + 1
arithmetic progression class 10 icse board questions

Class 10 Arithmetic Progression Important Questions and Answers

Q1. Which term of the A.P. 1, 4, 7, 10, .... is 58?
Options
(a) 18
(b) 19
(c) 20
(d) 21

Ans. (c) 20
Explanation:
A.P. = 1,4,7,10………
Here, first term (a) = 1
And common difference (d) = 4 - 1 = 3
Let nth  term of the given A.P. is 58.
⇒ 58 = a + (n-1)d
⇒ 58 = 1 + (n-1)× 3
⇒ n - 1 = \dfrac{57}{3} = 19 \\ ⇒ n = 20
So, 58 is 20th term of the A.P.

Q2. The sum of 25 terms of an A.P.
- \dfrac{2}{3}, - \dfrac{2}{3}, - \dfrac{2}{3} \space is :
Options
(a) 0
(b) -\dfrac{2}{3} \\[4.5 bp] (c) -\dfrac{50}{3}\\ (d) -50

Ans. (c)
Explanation:
\text{sum of 25 terms of an A.P.} \\[4.5 bp] \dfrac{-2}{3}, \dfrac{-2}{3},\dfrac{-2}{3} is \\[4.5 bp] = \dfrac{n}{2} [2a+(n-1)d] = \dfrac{25}{2} \\[4.5 bp] \begin{bmatrix} 2 × \left(\dfrac{-2}{3}\right) + (25 - 1 ) × 0\end{bmatrix} \\[4.5 bp] = \dfrac{}{}\begin{bmatrix} -\dfrac{4}{3}\end{bmatrix} = \dfrac{-50}{3}

Q3. In an Arithmetic Progression (A.P.) the fourth and sixth terms are 8 and 14 respectively. Find the :
(a) First term
(b) Common difference
(c) Sum of the first 20 terms.

Explanation:
Let, a and d be the first term and common difference of the given A.P.
Then, a_4 = 8 \space and \space a_6 = 14 \space (Given)\\ ⇒ a + 3d = 8 ...(i)
and a + 5d = 14 ...(ii)
On subtracting equation (i) from (ii),
we get,
2d = 6
⇒ d = 3On putting d = 3 in equation (i),
we get,
a + 3 × 3 = 8
⇒ a = 8 – 9 = – 1
(a) First term (a) = – 1. Ans.
(b) Common difference (c) Sum of first 20 terms (S20) (d) = 3. Ans.
∵ S_n = \dfrac{n}{2} [2a+(n–1)d] \\[4.5 bp] ∴ S_{20} = \dfrac{20}{2} [2× (-1) +(20–1)×3] \\ = 10(-2 + 57 )
= 550.

Q4.  Determine the 10th term from the last term (towards the first term) of the A.P. 10, 7, 4, … , – 62.

Explanation:
Here a = 10, d = 7 – 10 = – 3
and last term l = – 62
10th term from the last term
i.e., n = 10
Required term = l – (n – 1) d
= – 62 – (10 – 1) (– 3)
= – 62 + 27
= – 35
Therefore, the 10th term from the last term is – 35.

Q5. The 10th term of an A.P. is – 15 and 31st term is – 57, find the 15th term.

Explanation:
Let a be the first term and d be the common difference of the A.P. Then from the formula :
t_n = a + (n–1)d,\\[4.5 bp] \text{we have,} \\[4.5 bp] t_{10} ​ =a + (10–1)d = a + 9d \\[4.5 bp] t_{31} ​ = a + (31–1)d = a + 30d \\[4.5 bp] \text{we have,}\\[4.5 bp] a + 9d = – 15 …(i) \\[4.5 bp] a + 30d = –57 …(ii)\\[4.5 bp] Solve equations (i) and (ii) to get the values of a and d.
Subtracting (i) from (ii), we have

21d = –57 + 15 = –42 \\[4.5 bp] ∴ d = \dfrac{–42}{21} = –2 \\[4.5 bp] \text{Again from (i),} \space a = – 15 – 9d \\[4.5 bp] = –15 – 9(–2) \\[4.5 bp] = –15 + 18 = 3 \\[4.5 bp] \text{Now,} \space t_{15} = a + (15 – 1) d \\[4.5 bp] = 3 + 14(–2) = –25
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ICSE Class 10 Maths Chapter wise Important Questions

Chapter No.Chapter Name
Chapter 1Goods and Service Tax (GST)
Chapter 2Banking
Chapter 3Shares and Dividends
Chapter 4Linear inequations
Chapter 5Quadratic Equations in one variable
Chapter 6Ratio and proportion
Chapter 7Factorization
Chapter 8Matrices
Chapter 9Arithmetic Progression
Chapter 10Geometric Progression
Chapter 11Coordinate Geometry
Chapter 12Reflection
Chapter 13Similarity
Chapter 14Loci
Chapter 15Circles
Chapter 16Constructions
Chapter 17Mensuration
Chapter 18Trigonometry
Chapter 19Statistics
Chapter 20Probability

Conclusion

The chapter factorization class 10 ICSE is a foundational and pivotal topic that serves as a cornerstone for various mathematical concepts and practical applications. Mastering these factorisation class 10 ICSE important questions equips you with valuable problem-solving skills for everyday situations, making it a crucial part of your math education.

Frequently Asked Questions

Ans: An AP is a sequence of numbers where the difference between any two consecutive terms is constant.
Ans: The common difference is the fixed value by which consecutive terms in an AP increase or decrease.
Ans: Use the formula: a_n = a_1+ ( n-1 )d, where an is the nth term, a1 is the first term, and d is the common difference.
Ans: You can calculate it with the formula: S_n = \dfrac{n}{2} [2a_1+(n–1)d]
Ans: Use the formula: n = \frac{a – a}{d}+1
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