Arithmetic Progression

Important Questions

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.

Table of Contents

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

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

(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.

(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}

(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.

**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.

**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]
\text{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

Chapter No. | Chapter Name |
---|---|

Chapter 1 | Goods and Service Tax (GST) |

Chapter 2 | Banking |

Chapter 3 | Shares and Dividends |

Chapter 4 | Linear inequations |

Chapter 5 | Quadratic Equations in one variable |

Chapter 6 | Ratio and proportion |

Chapter 7 | Factorization |

Chapter 8 | Matrices |

Chapter 9 | Arithmetic Progression |

Chapter 10 | Geometric Progression |

Chapter 11 | Coordinate Geometry |

Chapter 12 | Reflection |

Chapter 13 | Similarity |

Chapter 14 | Loci |

Chapter 15 | Circles |

Chapter 16 | Constructions |

Chapter 17 | Mensuration |

Chapter 18 | Trigonometry |

Chapter 19 | Statistics |

Chapter 20 | Probability |

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.

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

Chapter Wise Important Questions for ICSE Board Class 10 Mathematics |
---|

Goods and Service Tax (GST) |

Banking |

Shares and Dividends |

Linear inequations |

Quadratic Equations in one variable |

Ratio and proportion |

Factorization |

Matrices |

Arithmetic Progression |

Geometric Progression |

Coordinate Geometry |

Reflection |

Similarity |

Loci |

Circles |

Constructions |

Mensuration |

Trigonometry |

Statistics |

Probability |

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