Arithmetic Progressions

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Class 10 Math Chapter 5
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 2023-24. 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.

Introduction

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.

What is an Arithmetic Progression (A.P.)?

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).

Class 10 Arithmetic Progressions Important Questions

Q 1. The sum of first 20 odd natural numbers is :
Options
(a) 100
(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

Q 2. The number of multiples of 4 that lie between 10 and 250 is
Options
(a) 62
(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)

\Longrightarrow = \dfrac{236}{4} = n – 1

⇒ n - 1 = 59
∴ n = 59 + 1 = 60

Q 3. If the sum of n terms of an A.P. is S_n = 3n^2 + 5n, then write its common difference.

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

Q 4. Which term of the progression
20, 19 \dfrac{1}{4}, 18\dfrac{1}{2}, 17\dfrac{3}{4}, ….. is the first negative term?

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.

Q 5. The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of X such that the sum of numbers of houses proceeding the house numbered X is equal to sum of the numbers of houses following X.
(i) coincident lines
(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.

CBSE Class 10 Maths Chapter wise Important Questions

Conclusion

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.

Frequently Asked Questions

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.