Class 10 RD Sharma Solutions - Chapter 2 Polynomials - Exercise 2.3
Last Updated : 28 Aug, 2024
Chapter 2 of RD Sharma's Class 10 Mathematics book delves into the polynomials a fundamental algebraic concept. This chapter introduces polynomials, their types and their properties laying the groundwork for the solving polynomial equations. Exercise 2.3 focuses on the solving polynomial equations using the various techniques.
Polynomials
A polynomial is an algebraic expression consisting of the variables and coefficients combined using the addition, subtraction and multiplication, and non-negative integer exponents of the variables. For example: 3x2+2x−5 is a polynomial. The degree of a polynomial is the highest power of the variable in the expression. The Polynomials are classified based on their degree and the number of terms they contain such as the linear, quadratic and cubic polynomials.
Question 1. Apply division algorithm to find the quotient q(x) and remainder r(x) on dividing f(x) by g(x) in each of the following:
(i) f(x) = x3 – 6x2 + 11x – 6, g(x) = x2 + x + 1
(ii) f(x) = 10x4 + 17x3 – 62x2 + 30x – 105, g(x) = 2x2 + 7x + 1
(iii) f(x) = 4x3 + 8x2 + 8x + 7, g(x) = 2x2 – x + 1
(iv) f(x) = 15x3 – 20x2 + 13x – 12, g(x) = x2 – 2x + 2
Solution:
(i) Here we have to divide f(x) = x3 – 6x2 + 11x – 6 by g(x) = x2 + x + 1
So, to get quotient q(x) and remainder r(x), we use division algorithm
Therefore,
Remainder r(x) = 17x - 1
Quotient q(x) = x - 7
(ii) Here we have to divide f(x) = 10x4 + 17x3 – 62x2 + 30x – 105 by g(x) = 2x2 + 7x + 1
So, to get quotient q(x) and remainder r(x), we use division algorithm
Therefore,
Remainder r(x) = 53x - 1
Quotient q(x) = 5x2 - 9x - 2
(iii) Here we have to divide f(x) = 4x3 + 8x2 + 8x + 7 by g(x) =2x2 – x + 1
So, to get quotient q(x) and remainder r(x), we use division algorithm
Therefore,
Remainder r(x) = 11x + 2
Quotient q(x) = 2x - 5
(iv) f(x) = 15x3 – 20x2 + 13x – 12, g(x) = x2 – 2x + 2
Here we have to divide f(x) = 15x3 – 20x2 + 13x – 12 by g(x) = x2 – 2x + 2
So, to get quotient q(x) and remainder r(x), we use division algorithm
Therefore,
Remainder r(x) = 3x + 32
Quotient q(x) = 15x + 10
Question 2. Check whether the first polynomial is a factor of the second polynomial by applying the division algorithm:
(i) g(t) = t2 – 3; f(t) = 2t4 + 3t3 – 2t2 – 9t – 12
(ii) g(x) = x2 – 3x + 1; f(x) = x5 – 4x3 + x2 + 3x + 1
(iii) g(x) = 2x2 – x + 3; f(x) = 6x5 − x4 + 4x3 – 5x2 – x – 15
Solution:
(i) Here, we have to check whether g(t) = t2 – 3 is a factor of f(t) = 2t4 + 3t3 – 2t2 – 9t – 12
So, by using division algorithm, we get
As the remainder left is 0.
Therefore,
g(t) = t2 – 3 is a factor of f(t) = 2t4 + 3t3 – 2t2 – 9t – 12
(ii) Here, we have to check whether g(x) = x2 – 3x + 1 is a factor off(x) = x5 – 4x3 + x2 + 3x + 1
So, by using division algorithm, we get
As the remainder left is 2.
Therefore,
g(x) = x2 – 3x + 1 is not a factor of f(x) = x5 – 4x3 + x2 + 3x + 1
(iii) Here, we have to check whether g(x) = 2x2 – x + 3 is a factor of f(x) = 6x5 − x4 + 4x3 – 5x2 – x – 15
So, by using division algorithm, we get
As the remainder left is 0.
Therefore,
g(x) = 2x2 – x + 3 is a factor of f(x) = 6x5 − x4 + 4x3 – 5x2 – x – 15
Question 3. Obtain all zeroes of the polynomial f(x) = f(x) = 2x4 + x3 – 14x2 – 19x – 6, if two of its zeroes are -2 and -1.
Solution:
Given: f(x) = 2x4 + x3 – 14x2 – 19x – 6
Here we have given the two zeroes of the polynomial that are -2 and -1,
Hence, its factors will be (x + 2) and (x + 1)
Further,
(x + 2)(x + 1) = x2 + x + 2x + 2 = x2 + 3x + 2
So, by using division algorithm, we get
f(x) = 2x4 + x3 – 14x2 – 19x – 6 = (2x2 – 5x – 3)(x2 + 3x + 2)
= (2x + 1)(x – 3)(x + 2)(x + 1)
Hence, the factors of f(x) = 2x4 + x3 – 14x2 – 19x – 6 are (2x + 1), (x – 3), (x + 2), (x + 1)
Therefore, the zeroes of the polynomial are -1/2, 3, -2, -1
Question 4. Obtain all zeroes of f(x) = x3 + 13x2 + 32x + 20, if one of its zeroes is -2.
Solution:
We have been given the zero of the polynomial f(x) = x3 + 13x2 + 32x + 20 is -2.
Hence, its factor is (x + 2).
So, by using division algorithm, we get
Thus,
f(x) = x3 + 13x2 + 32x + 20
= (x2 + 11x + 10)(x + 2)
= (x2 + 10x + x + 10)(x + 2)
= (x + 10)(x + 1)(x + 2)
Hence, the factors of f(x) = x3 + 13x2 + 32x + 20 are (x + 10), (x + 1), (x + 2)
Thus, the zeroes of the polynomial are -1, -10, -2.
Question 5. Obtain all zeroes of the polynomial f(x) = x4 – 3x3 – x2 + 9x – 6, if the two of its zeroes are -√3 and √3.
Solution:
Here, we are given two zeros of the polynomial f(x) = x4 – 3x3 – x2 + 9x – 6 that are -√3 and √3.
Thus, the factors are (x + √3)(x − √3) ⇒ x2 - 3.
So, by using division algorithm, we get
Hence,
f(x) = x4 – 3x2 – x2 + 9x – 6 = (x2 – 3)(x2 – 3x + 2)
(x + √3)(x – √3)(x2 – 2x – 2 + 2)
= (x + √3)(x – √3)(x – 1)(x – 2)
Thus, the factors of f(x) = x4 – 3x3 – x2 + 9x – 6 are (x + √3)(x – √3)(x – 1)(x – 2).
Therefore, the zeroes of the polynomial are -√3, √3, 1, 2.
Question 6. Obtain all zeroes of the polynomial f(x) = 2x4 – 2x3 – 7x2 + x – 1, if the two of its zeroes are -√3/2 and √3/2.
Solution:
Here, we are given two zeros of the polynomial f(x) = 2x4 – 2x3 – 7x2 + x – 1 that are -√3/2 and √3/2.
Thus, the factors are \left(x-\sqrt\frac{3}{2}\right) \left(x+\sqrt\frac{3}{2}\right) ⇒ x2 - 3/2.
So, by using division algorithm, we get
Hence,
f(x) = 2x^4– 2x^3–7x^2+ x – 1=\left(x-\sqrt\frac{3}{2}\right) \left(x+\sqrt\frac{3}{2}\right)(2x^2-2x-4)\\ \left(x-\sqrt\frac{3}{2}\right)\left(x+\sqrt\frac{3}{2}\right)(2x^2-4x+2x-4)\\ \left(x-\sqrt\frac{3}{2}\right)\left(x+\sqrt\frac{3}{2}\right)(x+2)(x-2)
Factors of f(x) = 2x4 – 2x3 – 7x2 + x – 1 are \left(x-\sqrt\frac{3}{2}\right),\ \left(x+\sqrt\frac{3}{2}\right),\ (x+2),\ (x-2) .
Thus, the zeroes of the polynomial are -1, 2, -√3/2 and √3/2.
Question 7. Find all the zeroes of the polynomial x4 + x3 – 34x2 – 4x + 120, if the two of its zeroes are 2 and – 2.
Solution:
Here, we are given two zeros of the polynomial x4 + x3 – 34x2 – 4x + 120 that are 2 and -2.
Thus, the factors are (x + 2)(x - 2)⇒ x2 - 4.
So, by using division algorithm, we get
Hence,
x4 + x3 – 34x2 – 4x + 120 = (x2 – 4)(x2 + x – 30)
= (x – 2)(x + 2)(x2 + 6x – 5x – 30)
= (x – 2)(x + 2)(x + 6)(x – 5)
So, the factors of x4 + x3 – 34x2 – 4x + 120 are (x – 2), (x + 2), (x + 6), (x – 5)
Thus, the zeroes of the polynomial = x = 2, – 2, – 6, 5
Question 8. Find all the zeroes of the polynomial 2x4 + 7x3 – 19x2 – 14x + 30, if the two of its zeroes are √2 and -√2.
Solution:
Here, we are given two zeros of the polynomial 2x4 + 7x3 – 19x2 – 14x + 302 that are √2 and -√2.
Thus, the factors are (x + √2)(x - √2) ⇒ x2 - 2.
So, by using division algorithm, we get
Hence,
2x4 + 7x3 – 19x2 – 14x + 30 = (x2 – 2)(2x2 + 7x – 15)
= (2x2 + 10x – 3x – 15)(x + √2)(x – √2)
= (2x – 3)(x + 5)(x + √2)(x – √2)
So, the factors of 2x4 + 7x3 – 19x2 – 14x + 30 are (2x – 3), (x + 5), (x + √2), (x – √2)
Thus, the zeroes of the polynomial is √2, -√2, -5, 3/2.
Question 9. Find all the zeroes of the polynomial f(x) = 2x3 + x2 – 6x – 3, if two of its zeroes are -√3 and √3.
Solution:
Here, we are given two zeros of the polynomial f(x) = 2x3 + x2 – 6x – 3 that are -√3 and √3.
Thus, the factors are (x + √3)(x - √3) ⇒ x2 - 3.
So, by using division algorithm, we get
Hence,
f(x) = 2x3 + x2 – 6x – 3
= (x2 – 3)(2x + 1)
= (x + √3)(x - √3)(2x + 1)
Factors of f(x) = 2x3 + x2 – 6x – 3 are (x + √3), (x - √3), 2x + 1
Thus, the zeroes for the given polynomial are √3, -√3, -1/2
Question 10. Find all the zeroes of the polynomial f(x) = x3 + 3x2 – 2x – 6, if the two of its zeroes are √2 and -√2.
Solution:
Here, we are given two zeros of the polynomial f(x) = x3 + 3x2 – 2x – 6 that are √2 and -√2.
Thus, the factors are (x + √2)(x - √2)⇒ x2 - 2.
So, by using division algorithm, we get
Hence,
f(x) = x3 + 3x2 – 2x – 6
= (x2 – 2)(x + 3)
= (x + √2)(x - √2)(x + 3)
Factors of f(x) = x3 + 3x2 – 2x – 6 are (x + √2), (x - √2), (x + 3)
Thus, the zeroes of the given polynomial is -√2, √2, and – 3.
Question 11. What must be added to the polynomial f(x) = x4 + 2x3 – 2x2 + x − 1 so that the resulting polynomial is exactly divisible by g(x) = x2 + 2x − 3.
Solution:
Here we have to add to the polynomial f(x) = x4 + 2x3 – 2x2 + x − 1 so that the
resulting polynomial is exactly divisible by g(x) = x2 + 2x − 3.
So, divide f(x) = x4 + 2x3 – 2x2 + x − 1 by g(x) = x2 + 2x − 3 to get the answer.
As the remainder left is (x – 2) to get the resulting polynomial exactly divisible by
g(x) = x2 + 2x − 3 we must add (x - 2) to f(x) = x4 + 2x3 – 2x2 + x − 1.
Question 12. What must be subtracted from the polynomial f(x) = x4 + 2x3 – 13x2 –12x + 21 so that the resulting polynomial is exactly divisible by g(x) = x2 – 4x + 3.
Solution:
Here we have to subtract to the polynomial f(x) = x4 + 2x3 – 13x2 – 12x + 21
so that the resulting polynomial is exactly divisible by g(x) = x2 - 4x + 3.
So, divide f(x) = x4 + 2x3 – 13x2 – 12x + 21 by g(x) = x2 - 4x + 3 to get the answer.
As the remainder left is (2x - 3) to get the resulting polynomial exactly divisible by
g(x) = x2 - 4x + 3 we must add (2x - 3) to f(x) = x4 + 2x3 – 13x2 – 12x + 21.
Question 13. Given that √2 is a zero of the cubic polynomial f(x) = 6x3 + √2x2– 10x – 4√2, find its other two zeroes.
Solution:
Here, we are given that √2 is the zero of the cubic polynomial
f(x) = 6x3 + √2x2– 10x – 4√2, thus, factor of the polynomial is (x - √2)
So, by using division algorithm, we get
Hence,
f(x) = 6x3 + √2x2 – 10x – 4√2
= (x - √2)(6x2 + 7√2x + 4)
= (x - √2)(6x2 + 4√2x + 3√2x + 4)
= (x - √2)(3x + 2√2)(2x + √2)
The factors of f(x) = 6x3 + √2x2– 10x – 4√2 are (x - √2), (3x + 2√2), (2x + √2)
Therefore, the zeros of the polynomial are -2√2/3, -√2/2, √2
Question 14. Given that x - √5 is a factor of the cubic polynomial x3 - 3√5x2 + 13x - 3√5, find all the zeroes of the polynomial.
Solution:
Here, we have x - √5 as factor of the cubic polynomial x3 - 3√5x2 + 13x - 3√5
To find all the zeros of the polynomial, we have to divide the polynomial x3 - 3√5x2 + 13x - 3√5 by the factor x - √5
Hence,
x3 - 3√5x2 + 13x - 3√5
= (x - √5)(x2 - 2√5 + 3)
= (x - √5)(x - (√5 + √2))(x - (√5 - √2))
So, the factors of the cubic polynomial x3 - 3√5x2 + 13x - 3√5 are (x - √5), (x - (√5 + √2)), (x - (√5 - √2))
Therefore, the zero of the polynomial are √5, (√5 - √2), (√5 + √2)
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Chapter 5: Trigonometric Ratios
Chapter 6: Trigonometric Identities
Chapter 7: Statistics
Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.1 | Set 1Problem 1: Calculate the mean for the following distribution:x:56789f:4 814113 Solution: x ffx542068487 1498811889327 N = 40 ∑ fx = 281 We know that, Mean = ∑fx/ N = 281/40 = 7.025 Problem 2: Find the mean of the following data :x:19212325272931f:13151618161513 Solution: x ffx19132472115315231636825
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.1 | Set 2Problem 11: Five coins were simultaneously tossed 1000 times and at each toss, the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4, and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.No. of heads per tossNo. of tosses 03811442
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.2Question1: The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:Number of calls (x) 0 1 2 3 4 5 6 Number of intervals (f) 15 24 29 46 54 43 39 Compute the mean number of calls per interval. Solution: Let
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.3 | Set 1Chapter 7 of RD Sharma’s Class 10 textbook focuses on Statistics a fundamental area in mathematics that deals with data collection, analysis, interpretation, and presentation. Exercise 7.3 | Set 1 of this chapter covers the various problems designed to enhance students' understanding and application
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.3 | Set 2Question 14. Find the mean of the following frequency distribution:Class interval:25 – 2930 – 3435 – 3940 – 4445 – 4950 – 5455 – 59Frequency:1422166534 Solution: Let’s consider the assumed mean (A) = 42 Class intervalMid – value xidi = xi – 42ui = (xi – 42)/5fifiui25 – 2927-15-314-4230 – 3432-10-222
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Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 1In this article, we will explore the solutions to Exercise 7.4 from Chapter 7 of RD Sharma's Class 10 Mathematics textbook which focuses on Statistics. This chapter is crucial as it lays the foundation for understanding how data is collected, organized, analyzed, and interpreted in the various field
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Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 2Question 11. An incomplete distribution is given below :Variable10-2020-3030-4040-5050-6060-7070-80Frequency1230-65-2518You are given that the median value is 46 and the total number of items is 230. (i) Using the median formula fill up missing frequencies. (ii) Calculate the AM of the completed dis
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 1Statistics is a branch of mathematics dealing with data collection, analysis, interpretation, and presentation. It provides methods for summarizing and drawing conclusions from the data which is essential in various fields such as economics, sociology, and science. Chapter 7 of RD Sharma's Class 10
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 2Question 11. Find the mean, median, and mode of the following data:Classes0-5050-100100-150150-200200-250250-300300-350Frequency2356531 Solution: Let mean (A) = 175 Classes Class Marks (x) Frequency (f) c.f. di = x - A A = 175 fi * di0-502522-150-30050-1007535-100-300100-150125510-50-250150-200175-A
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.6Chapter 7 of RD Sharma's Class 10 textbook focuses on Statistics a branch of mathematics that deals with the collection, analysis, interpretation, and presentation of data. This chapter is crucial for understanding how to summarize and make sense of numerical data in various contexts. The aim is to
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Chapter 8: Quadratic Equations
Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.1Question 1. Which of the following are quadratic equations?(i) x2 + 6x - 4 = 0 Solution: As we know that, quadratic equation is in form of: ax2 + bx + c = 0 Here, given equation is x2 + 6x - 4 =0 in quadratic equation. So, It is a quadratic equations. (ii) √3x2 − 2x + ½= 0 Solution: As we know that,
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.2Quadratic equations are a fundamental concept in algebra that involves a polynomial of degree two. They play a crucial role in various fields of mathematics, physics, engineering, and many other disciplines. Understanding quadratic equations is essential for solving problems related to motion, area,
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 1Solve the following quadratic equations by factorizationQuestion 1. (x - 4) (x + 2) = 0 Solution: We have equation, (x - 4) (x + 2) = 0 Implies that either x - 4 = 0 or x + 2 = 0 Therefore, roots of the equation are 4 or -2. Question 2. (2x + 3) (3x - 7) = 0 Solution: We have equation, (2x + 3) (3x
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 2The Quadratic equations are fundamental to algebra and appear frequently in various mathematical contexts. They play a crucial role in understanding polynomial functions and solving real-world problems. In this article, we will delve into Exercise 8.3 | Set 2 from RD Sharma’s Class 10 Mathematics bo
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.4Question 1: Find the roots of the following quadratic (if they exist) by the method of completing the square: x^2-4\sqrt{2x}+6=0 . Solution: Given: x^2-4\sqrt{2x}+6=0 We have to make the equation a perfect square. => x^2-4\sqrt{2x} = -6 => x^2 - 2\times2\sqrt{2x} + (2\sqrt{2})^2 = -6 + (2\sqrt
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.5Question 1. Find the discriminant of the following quadratic equations:(i) 2x2 - 5x + 3 = 0 Solution: Given quadratic equation: 2x2 - 5x + 3 = 0 ....(1) As we know that the general form of quadratic equation is ax2 + bx + c = 0 ....(2) On comparing eq(1) and (2), we get Here, a = 2, b = -5 and c = 3
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.6 | Set 1Question 1. Determine the nature of the roots of following quadratic equations : (i) 2x² – 3x + 5 = 0 (ii) 2x² – 6x + 3 = 0 (iii) 3/5 x² – 2/3 x + 1 = 0 (iv) 3x² – 4√3 x + 4 = 0 (v) 3x² – 2√6 x + 2 = 0 Solution: (i) 2x² – 3x + 5 = 0 Here a=2, b=-3, c=5 D=b2-4ac=(-3)2-4*2*5 =-9-40=-31 D<0 Roots ar
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.6 | Set 2Question 11. If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal-roots, find the value of k. Solution: 2x² + px – 15 = 0 -5 is its one root It will satisfy it 2(-5)2+p(-5)-15=0 2*25-5p-15=0 50-15-5p=0 -5p+35=0 -5p=-35 p=-35/-5=7 Now in
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Class 10 RD Sharma Solutions- Chapter 8 Quadratic Equations - Exercise 8.7 | Set 1Question 1: Find two consecutive numbers whose squares have the sum 85. Solution: Let first number is = x ⇒ Second number = (x+1) Now according to given condition— ⇒ Sum of squares of the numbers = 85 ⇒ x2 + (x+1)2 = 85 ⇒ x2 + x2 + 2x + 1 = 85 [ because (a+b)2 = a2 + 2ab + b2] ⇒ 2x2 + 2x + 1 - 85 =
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Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.7 | Set 2Question 11: The sum of a number and its square is 63/4 , find the numbers. Solution: Let the number is = x so it's square is = x2 Now according to condition- ⇒(number)+(number)2=63/4 ⇒ x+x2 = 63/4 ⇒ x2+x-63/4=0 Multiplying by 4- ⇒ 4x2+4x-63=0 ⇒ 4x2 +(18-14)x - 63 =0 [because 63*4=252 so 18*14=252
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.8Chapter 8 of RD Sharma's Class 10 textbook focuses on the Quadratic Equations an essential topic in mathematics that lays the foundation for understanding more advanced algebraic concepts. In this chapter, students will learn how to solve quadratic equations using various methods such as factorizati
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.9The Quadratic equations are a fundamental topic in algebra often introduced at the Class 10 level. They are equations of the form ax2 +bx+c=0 where a, b, and c are constants and x is the variable. Understanding quadratic equations is crucial as they form the basis for many mathematical concepts and
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.10Question 1. The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides. Solution: Let the length of other two sides of the triangle be x and y. Therefore, according to the question, x - y = 5 or, x =
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.11Question 1: The perimeter of the rectangular field is 82 m and its area is 400 m2. Find the breadth of the rectangle? Solution: Given: Perimeter = 82 m and its area = 400 m2 Let the breadth of the rectangle be 'b' m. As we know, Perimeter of a rectangle = 2×(length + breadth) 82 = 2×(length + b) 41
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.12Question 1: A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work. Solution: Let us assume that B takes 'x' days to complete a piece of work. So, B’s 1 day work = 1/x Now, A takes
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.13Question 1. A piece of cloth costs ₹ 35. If the piece were 4 m longer and each meter costs ₹ 1 less, the cost would remain unchanged. How long is the piece? Solution: Let us considered the length of piece of cloth = x m Given: The total cost = ₹ 35 So, the cost of 1 m cloth = ₹ 35/x According to the
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Chapter 9: Arithmetic Progressions
Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progression Exercise 9.1Problem 1: Write the first terms of each of the following sequences whose nth term are:(i) an = 3n + 2 Solution: Given: an = 3n + 2 By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence a1 = (3 × 1) + 2 = 3 + 2 = 5 a2 = (3 × 2) + 2 = 6 + 2 = 8 a3 = (3 × 3) + 2 = 9 + 2 = 11 a4 = (3 × 4)
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.2Problem 1: Show that the sequence defined by an = 5n – 7 is an A.P., find its common difference. Solution: Given: an = 5n – 7 Now putting n = 1, 2, 3, 4,5 we get, a1 = 5.1 – 7 = 5 – 7 = -2 a2 = 5.2 – 7 = 10 – 7 = 3 a3 = 5.3 – 7 = 15 – 7 = 8 a4 = 5.4 – 7 = 20 – 7 = 13 We can see that, a2 – a1 = 3 – (
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.3Problem 1: For the following arithmetic progressions write the first term a and the common difference d:(i) -5, -1, 3, 7, ………… Solution: Given sequence is -5, -1, 3, 7, ………… ∴ First term is -5 And, common difference = a2 - a1 = -1 - (-5) = 4 ∴ Common difference is 4 (ii) 1/5, 3/5, 5/5, 7/5, …… Solut
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.4 | Set 1In arithmetic progressions (AP), the terms follow a consistent pattern where each term is derived from the previous one by adding a fixed value, known as the common difference. To analyze various aspects of APs, such as finding specific terms or determining the number of terms, certain formulas and
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.4 | Set 2Question 14. The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference. Solution: Let’s assume that the first term and the common difference of the A.P to be a and d respectively. Given that, 4th term of an A.P.
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Class 10 RD Sharma Solutions- Chapter 9 Arithmetic Progressions - Exercise 9.5Question 1: Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P. Solution: Since (8x + 4), (6x – 2) and (2x + 7) are in A.P. Now we know that condition for three number being in A.P.- ⇒2*(Middle Term)=(First Term)+(Last Term) ⇒2(6x-2)=(8x+4)+(2x+7) ⇒12x-4=10x+11 ⇒(12x-10x)=11+4
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 1Class 10 RD Sharma Solutions are comprehensive guides to the exercises and problems presented in the RD Sharma Mathematics textbook, which is a popular reference for students studying mathematics at the 10th class level. These solutions provide step-by-step explanations and answers to the questions
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 2Question 25. In an A.P. the first term is 22, nth term is –11 and the sum of first n term is 66. Find n and the d, the common difference. Solution: Given A.P. has first term(a) = 22, nth term(an) = –11 and sum(Sn) = 123. Now by using the formula of sum of n terms of an A.P. Sn = n[a + an] / 2 =>
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 3Question 49. Find the sum of first n odd natural numbers. Solution: First odd natural numbers are 1, 3, 5, 7, . . .2n – 1. First term(a) = 1, common difference(d) = 3 – 1 = 2 and nth term(an) = 2n – 1. Now by using the formula of sum of n terms of an A.P. Sn = n[a + an] / 2 So, = n[1 + 2n – 1] / 2 =
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