Arithmetic Progressions

Important Questions

Here are some important questions for Class 10 Mathematics Chapter 5, Arithmetic Progressions, carefully selected to help students prepare for the CBSE Class 10 Mathematics Examination in 2024-25. These questions cover various types of problems and are designed to assist students in understanding Arithmetic Progressions better. By practicing these diverse question types, students can clarify any doubts they may have and improve their problem-solving skills, leading to better performance in the chapter on Arithmetic Progressions.

Table of Contents

In Chapter 5 of Class 10 Mathematics, Arithmetic Progressions, you will study the motivation for studying Arithmetic Progression, the derivation of the nth term and sum of the first n terms of an A.P., and there are applications in solving real-life problems.

An Arithmetic Progression is a sequence of numbers in which the difference between consecutive terms remains constant. This constant difference is known as the common difference (d).

(b) 210

(c) 400

(d) 420

**Ans.** (c)

**Explanation:**

The given A.P. is

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39

Thus, a = 1, d = 2 and n = 20

We know that,

S_n = \dfrac{n}{a}\{2a + (n - 1)d \} \\[4.5 bp]
= \dfrac{20}{2}\{2(1) + (20 - 1)2 \} \\[4.5 bp]
= 10\{2+38\} = 400

(b) 60

(c) 59

(d) 55

**Ans.** (b)

**Explanation:**

Multiples of 4 lying between 10 and 250 are 12 , 16, 20, 24……, 248

Here, a = 12, d = 16 - 12 = 4, l = 248

l = a_n = a + (n - 1)d

248 = 12 + (n - 1)× 4

⇒ 248 - 12 = 4(n - 1)

⇒ n - 1 = 59

∴ n = 59 + 1 = 60

**Ans.** common difference = 6

**Explanation:**

We have,

S_n = 3n^2 +5n \\[4.5 bp]
⇒ S_(n - 1) = 3(n - 1)^2 + 5(n - 1) \\[4.5 bp]
= 3(n^2 + 1 - 2n) + 5n - 5 \\[4.5 bp]
= 3n^2 + 3 - 6n + 5n - 5 \\[4.5 bp]
= 3n^2 - n - 2 \\[4.5 bp]
\text{General term of A.P. is given as,} \\[4.5 bp]
T_n = S_n - S_(n-1) \\[4.5 bp]
= 3n^2 + 5n - 3n^2+ n + 2 \\[4.5 bp]
= 6n + 2 \\[4.5 bp]
\text{ Putting n = 1,2,3, we get }\\[4.5 bp]
T_1 = 6 × 1 + 2 = 8 \\[4.5 bp]
T_2= 6 × 2 + 2 = 14 \\[4.5 bp]
T_3 = 6 × 3 + 2 = 20 \\[4.5 bp]
\text{Now,} \space T_2 - T_1 = 14 - 8 = 6 \\[4.5 bp]
\text{and} \space T_3 - T_2 = 20 - 14 = 6

∴ common difference = 6

**Ans.** 28th term will be the first negative term of given A.P.

**Explanation:**

Given, A.P. is 20, 19\dfrac{1}{4}, 18\dfrac{1}{2}, 17\frac{3}{4},….. \\[4.5 bp]
= 20, \dfrac{77}{4}, \dfrac{37}{2}, \dfrac{71}{4}, \\[4.5 bp]
\text{Here,} \space a = 20, d = \dfrac{77}{4} – 20 = \dfrac{77 – 80}{2} = \dfrac{–3}{4} \\[4.5 bp]
\text{Let} \space a_n \space \text{ be its first negative term
an} \space < 0 \\[4.5 bp]
\text{Then} \space a_n + (n – 1)d < 0. \\[4.5 bp]
\Longrightarrow 20+(n – 1)\left(–\dfrac{3}{4}\right) < 0 \\[4.5 bp]
\Longrightarrow 20 – \dfrac{3}{4}n + \dfrac{3}{4} < 0 \\[4.5 bp]
\Longrightarrow 20 + \dfrac{3}{4} < \dfrac{3}{4}n \\[4.5 bp]
\Longrightarrow \dfrac{83}{4} < \dfrac{3}{4}n \\[4.5 bp]
\Longrightarrow n > \dfrac{83}{4} × \dfrac{4}{3} \\[4.5 bp]
\Longrightarrow n > \dfrac{83}{3} = 27.66...

28th term will be the first negative term of given A.P.

(ii) intersecting lines

**Ans.** X = 35.

**Explanation:**

Given, the houses in a row numbered consecutively from 1 to 49.

Now, sum of numbers preceding the number X

= \dfrac{X(X – 1)}{2} \\[4.5 bp]
\text{and sum of numbers following the number X } \\[4.5 bp]
=\dfrac{49(50)}{2} – \frac{X(X – 1)}{2} – X \\[4.5 bp]
= \dfrac{2450 – X^2 + X – 2X}{2} \\[4.5 bp]
= \dfrac{2450 – X^2 – X}{2} \\[4.5 bp]
\text{According to the question,
Sum of no’s preceding X = Sum of no’s following X } \\[4.5 bp]
\dfrac{X(X – 1)}{2} = \dfrac{2450 – X^2 – X}{2} \\[4.5 bp]
\Longrightarrow X^2 – X = 2450 – X^2 – X \\[4.5 bp]
\Longrightarrow 2X^2 = 2450 \\[4.5 bp]
\Longrightarrow 2X^2 = 1225 \\[4.5 bp]
\Longrightarrow X = 35

Hence, at X = 35, sum of number of houses preceding the house no. X is equal to the sum of the number of houses following X.

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

Chapter 1 | Real Numbers |

Chapter 2 | Polynomials |

Chapter 3 | Pair of Linear Equations in Two Variable |

Chapter 4 | Quadratic Equations |

Chapter 5 | Arithmetic Progressions |

Chapter 6 | Triangles |

Chapter 7 | Coordinate Geometry |

Chapter 8 | Introduction to Trigonometry |

Chapter 9 | Some Applications of Trigonometry |

Chapter 10 | Circles |

Chapter 11 | Areas Related to Circle |

Chapter 12 | Surface Areas and Volumes |

Chapter 13 | Statistics |

Chapter 14 | Probability |

If you want to improve your understanding of the concepts in this chapter, you can visit oswal.io. They have a wide range of questions that will help you practice and reinforce what you’ve learned. By solving these questions, you can strengthen your knowledge and become better at solving problems.

Ans: To identify an Arithmetic Progression, check if the difference between consecutive terms is constant. If the difference between any two consecutive terms is the same throughout the sequence, then it is an A.P.

The formula for the nth term of an A.P. is given as: a_n = a + (n – 1)d, where ‘a’ is the first term and ‘d’ is the common difference.

Ans: The sum of the first n terms of an A.P. is denoted as Sn and can be calculated using the formula: Sn = (n/2) [2a + (n – 1) d], where ‘a’ is the first term and ‘d’ is the common difference.

Ans: Yes, the common difference in an A.P. can be negative. It means that the sequence is decreasing, and each term is obtained by subtracting the absolute value of ‘d’ from the previous term.

Ans: To find the nth term (a_n) or the sum of the first n terms (Sn) directly, the common difference (d) must be known, and the first term (a) should be given or easily determined.

Chapter Wise Important Questions for CBSE Board Class 10 Maths |
---|

Real Numbers |

Polynomials |

Pair of Linear Equations in Two Variables |

Quadratic Equations |

Arithmetic Progressions |

Triangles |

Coordinate Geometry |

Introduction to Trigonometry |

Some Applications of Trigonometry |

Circles |

Areas Related to Circles |

Surface Areas and Volumes |

Statistics |

Probability |

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