Class 10 Exam > Class 10 Notes > RD Sharma Solutions for Class 10 Mathematics > Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-8)
Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-8) | RD Sharma Solutions for Class 10 Mathematics PDF Download
Page No 5.30
Ques.1. Find the value of x for which (8x + 4), (6x − 2) and (2x + 7) are in A.P.
Ans. Here, we are given three terms,
First-term (a1) = 8x + 4
Second term (a2) = 6x - 2
Third term (a3) = 2x + 7
We need to find the value of x for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,
d = a2 - a1
d = (6x - 2) - (8x + 4)
d = 6x - 8x - 2 - 4
d = -2x - 6 ...(1)
Also,
d = a3 - a2
d = (2x + 7) - (6x - 2)
d = 2x - 6x + 7 + 2
d = -4x + 9 ...(2)
Now, on equating (1) and (2), we get,
-2x - 6 = -4x + 9
4x - 2x = 9 + 6
2x = 15
x = 15/2
Therefore, for x = 15/2,these three terms will form an A.P.
Ques.2. If x + 1, 3x and 4x + 2 are in A.P., find the value of x.
Ans. Here, we are given three terms which are in A.P.,
First term (a1) = x + 1
Second term (a2) = 3x
Third term (a3) = 4x + 2
We need to find the value of x. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,
d = a2 - a1
d = (3x) - (x + 1)
d = 3x - x - 1
d = 2x - 1 ...(1)
Also,
d = a3 - a2
d = (4x + 2) - (3x)
d = 4x - 3x + 2
d = x + 2 ...(2)
Now, on equating (1) and (2), we get,
2x - 1 = x + 2
2x - x = 2 + 1
x = 3
Therefore, for x = 3, these three terms will form an A.P.
Ques.3. Show that (a − b)2, (a2 + b2) and (a + b)2 are in A.P.
Ans. Here, we are given three terms and we need to show that they are in A.P.,
First term (a1) = (a - b)2
Second term (a2) = (a2 + b2)
Third term (a3) = (a + b)2
So, in an A.P. the difference of two adjacent terms is always constant. So, to prove that these terms are in A.P. we find the common difference, we get,
d = a2 - a1
d = (a2 + b2) - (a - b)2
d = a2 + b2 - (a2 + b2 - 2ab)
d = a2 + b2 - a2 - b2 + 2ab
d = 2ab ...(1)
Also,
d = a3 - a2
d = (a + b)2 - (a2 + b2)
d = a2 + b2 + 2ab - a2 - b2
d = 2ab ...(2)
Now, since in equations (1) and (2) the values of d are equal, we can say that these terms are in A.P. with 2ab as the common difference.
Hence proved.
Ques.4. Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.
Ans. In the given problem, the sum of three terms of an A.P is 27 and the product of the three terms is 648. We need to find the three terms.
Here,
Let the three terms be (a - d),a,(a + d) where a is the first term and d is the common difference of the A.P
So,
(a - d) + a + (a + d) = 27
3a = 27
a = 9 ...(1)
Also,
(a - d)a(a + d) = a + 6
a(a2 - d2) = 648 [Using a2 - b2 = (a + b)(a - b)]
9(92 - d2) = 648
81 - d2 = 72
Further solving for d,
81 - d2 = 72
81 - 72 = d2
d = √9
d = 3 ...(2)
Now, substituting (1) and (2) in three terms
First term = a - d
So,
a - d = 9 - 3 = 6
Also,
Second term = a
So,
a = 9
Also,
Third term = a + d
So,
a + d = 9 + 3 = 12
Therefore, the three terms are 6, 9 and 12.
Ques.5. The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.
Ans. In the given problem, the sum of three terms of an A.P is 21 and the product of the first and the third term exceeds the second term by 6.
We need to find the three terms.
Here,
Let the three terms be (a - d),a,(a + d) where, a is the first term and d is the common difference of the A.P
So,
(a - d) + a + (a + d) = 21
3a = 21
a = 7 …(1)
Also,
(a - d)(a + d) = a + 6
a2 - d2 = a + 6 (Using a2 - b2 = (a + b)(a - b))
(7)2 - d2 = 7 + 6 (Using 1)
49 - 13 = d2
Further solving for d,
d2 = 36
d = √36
d = 6 ...(2)
Now, using the values of a and d in the expressions of the three terms, we get,
First term = a - d
So,
a - d = 7 - 6 = 1
Second term = a
So,
a = 7
Also,
Third term = a + d
So,
a + d = 7 + 6 = 13
Therefore, the three terms are 1, 7 and 13.
Ques.6. Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.
Ans. Here, we are given that four number are in A.P., such that there sum is 50 and the greatest number is 4 times the smallest.
So, let us take the four terms as a - d, a, a + d, a + 2d.
Now, we are given that sum of these numbers is 50, so we get,
(a - d) + (a) + (a + d) + (a + 2d) = 50
a - d + a + a + d + a + 2d = 50
4a + 2d = 50
2a + d = 25 ...(1)
Also, the greatest number is 4 times the smallest, so we get,
a + 2d = 4(a - d)
a + 2d = 4a - 4d
4d + 2d = 4a - a
6d = 3a
d = (3/6)a ...(2)
Now, using (2) in (1), we get,
2a + (3/6)a = 25
15a = 150
a = 150/15
a = 10
Now, using the value of a in (2), we get
d =(3/6)10
d = 10/2
d = 5
So, first term is given by,
a - d = 10 - 5 = 5
Second term is given by,
a = 10
Third term is given by,
a + d = 10 + 5 = 15
Fourth term is given by,
a + 2d = 10 + (2)(5)
= 10 + 10 = 20
Therefore, the four terms are 5, 10, 15, 20.
Ques.7. The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.
Ans. In the given problem, the sum of three terms of an A.P is 12 and the sum of their cubes is 288.
We need to find the three terms.
Here,
Let the three terms be (a - d),a,(a + d) where, a is the first term and d is the common difference of the A.P
So,
(a - d) + a + (a + d) = 12
3a = 12
a = 12/3
a = 4
Also, it is given that
(a - d)3 + a3 + (a + d)3 = 288
So, using the properties:
(a - b)3 = a3 - b3 + 3ab2 - 3a2b
(a + b)3 = a3 + b3 + 3ab2 + 3a2b
We get,
(a - d)3 + a3 + (a + d)3 = 288
a3 - d3 - 3a2d + 3d2a + a3 + d3 +3a2d + 3d2a = 288
3a3 + 6d2a = 288
a3 + 2d2a = 96
Further solving for d by substituting the value of a, we get,
(4)3 + 2d2(4) = 96
64 + 8d2 = 96
8d2 = 96 - 64
d2 = 32/8
On further simplification, we get,
d = 4
d = √4
d = ±2
Now, here d can have two values +2 and -2.
So, on substituting the values of a = 4 and d = 2 in three terms, we get,
First-term = a - d
So,
a - d = 4 - 2 = 2
Second term = a
So,
a = 4
Third term = a + d
So,
a + d = 4 + 2 = 6
Also, on substituting the values of a = 4 and d = -2 in three terms, we get,
First term = a - d
So,
a - d = 4 - (-2)
= 4 + 2
= 6
Second term = a
So,
a = 4
Third term = a + d
So,
a + d = 4 + (-2)
= 4 - 2 = 2
Therefore, the three terms are 2, 4 and 6 or 6, 4 and 2.
Ques.8. Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.
Ans. Let the four terms of the AP be a − 3d, a − d, a + d and a + 3d.
Given:
(a − 3d) + (a − d) + (a + d) + (a + 3d) = 56
⇒ 4a = 56
⇒ a = 14
Also,
⇒ 1176 − 54d2 = 980 − 5d2
⇒ 196 = 49d2
⇒ d2 = 4
⇒ d = ±2
When d = 2, the terms of the AP are 8, 12, 16, 20. When d = −2, the terms of the AP are 20, 18, 12, 8.
Ques.9. The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and third term is 5 times the common difference, find the three numbers.
Ans. Let the first three numbers in an arithmetic progression be a − d, a, a + d.
The sum of the first three numbers in an arithmetic progression is 18.
a − d + a + a + d = 18
⇒ 3a = 18
⇒ a = 6
The product of the first and third term is 5 times the common difference.
(a − d)(a + d) = 5d
⇒ a2 − d2 = 5d
⇒ (6)2 − d2 = 5d
⇒ 36 − d2 = 5d
⇒ d2 + 5d − 36 = 0
⇒ d2 + 9d − 4d − 36 = 0
⇒ d(d + 9) − 4(d + 9) = 0
⇒ (d + 9)(d − 4) = 0
⇒ d = 4, −9
For d = 4,
The numbers are 6 − 4, 6, 6 + 4
i.e. 2, 6, 10
For d = −9,
The numbers are 6 + 9, 6, 6 − 9
i.e. 15, 6, −3
Hence, the three numbers are 2, 6, 10 or 15, 6, −3.
Ques.10. The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the numbers.
Ans. Here, we are given that the angles of a quadrilateral are in A.P, such that the common difference is 10°.
So, let us take the angles as a - d, a, a + d, a + 2d
Now, we know that the sum of all angles of a quadrilateral is 360°. So, we get,
(a - d) + (a) + (a + d) + (a + 2d) = 360
a - d + a + a + d + a + 2d = 360
4a + 2(10) = 360
4a = 360 - 20
On further simplifying for a, we get,
a = 340/4
a = 85
So, the first angle is given by,
a - d = 85 - 10 = 75°
Second angle is given by,
a = 85°
Third angle is given by,
a + d = 85 + 10
= 95°
Fourth angle is given by,
a + 2d = 85 + 2(10)
= 85 + 20 = 105°
Therefore, the four angles of the quadrilateral are 75°, 85°, 95°, 105°.
Ans. Here, we are given three terms,
First-term (a1) = 8x + 4
Second term (a2) = 6x - 2
Third term (a3) = 2x + 7
We need to find the value of x for which these terms are in A.P. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,
d = a2 - a1
d = (6x - 2) - (8x + 4)
d = 6x - 8x - 2 - 4
d = -2x - 6 ...(1)
Also,
d = a3 - a2
d = (2x + 7) - (6x - 2)
d = 2x - 6x + 7 + 2
d = -4x + 9 ...(2)
Now, on equating (1) and (2), we get,
-2x - 6 = -4x + 9
4x - 2x = 9 + 6
2x = 15
x = 15/2
Therefore, for x = 15/2,these three terms will form an A.P.
Ques.2. If x + 1, 3x and 4x + 2 are in A.P., find the value of x.
Ans. Here, we are given three terms which are in A.P.,
First term (a1) = x + 1
Second term (a2) = 3x
Third term (a3) = 4x + 2
We need to find the value of x. So, in an A.P. the difference of two adjacent terms is always constant. So, we get,
d = a2 - a1
d = (3x) - (x + 1)
d = 3x - x - 1
d = 2x - 1 ...(1)
Also,
d = a3 - a2
d = (4x + 2) - (3x)
d = 4x - 3x + 2
d = x + 2 ...(2)
Now, on equating (1) and (2), we get,
2x - 1 = x + 2
2x - x = 2 + 1
x = 3
Therefore, for x = 3, these three terms will form an A.P.
Ques.3. Show that (a − b)2, (a2 + b2) and (a + b)2 are in A.P.
Ans. Here, we are given three terms and we need to show that they are in A.P.,
First term (a1) = (a - b)2
Second term (a2) = (a2 + b2)
Third term (a3) = (a + b)2
So, in an A.P. the difference of two adjacent terms is always constant. So, to prove that these terms are in A.P. we find the common difference, we get,
d = a2 - a1
d = (a2 + b2) - (a - b)2
d = a2 + b2 - (a2 + b2 - 2ab)
d = a2 + b2 - a2 - b2 + 2ab
d = 2ab ...(1)
Also,
d = a3 - a2
d = (a + b)2 - (a2 + b2)
d = a2 + b2 + 2ab - a2 - b2
d = 2ab ...(2)
Now, since in equations (1) and (2) the values of d are equal, we can say that these terms are in A.P. with 2ab as the common difference.
Hence proved.
Ques.4. Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers.
Ans. In the given problem, the sum of three terms of an A.P is 27 and the product of the three terms is 648. We need to find the three terms.
Here,
Let the three terms be (a - d),a,(a + d) where a is the first term and d is the common difference of the A.P
So,
(a - d) + a + (a + d) = 27
3a = 27
a = 9 ...(1)
Also,
(a - d)a(a + d) = a + 6
a(a2 - d2) = 648 [Using a2 - b2 = (a + b)(a - b)]
9(92 - d2) = 648
81 - d2 = 72
Further solving for d,
81 - d2 = 72
81 - 72 = d2
d = √9
d = 3 ...(2)
Now, substituting (1) and (2) in three terms
First term = a - d
So,
a - d = 9 - 3 = 6
Also,
Second term = a
So,
a = 9
Also,
Third term = a + d
So,
a + d = 9 + 3 = 12
Therefore, the three terms are 6, 9 and 12.
Ques.5. The sum of three terms of an A.P. is 21 and the product of the first and the third terms exceeds the second term by 6, find three terms.
Ans. In the given problem, the sum of three terms of an A.P is 21 and the product of the first and the third term exceeds the second term by 6.
We need to find the three terms.
Here,
Let the three terms be (a - d),a,(a + d) where, a is the first term and d is the common difference of the A.P
So,
(a - d) + a + (a + d) = 21
3a = 21
a = 7 …(1)
Also,
(a - d)(a + d) = a + 6
a2 - d2 = a + 6 (Using a2 - b2 = (a + b)(a - b))
(7)2 - d2 = 7 + 6 (Using 1)
49 - 13 = d2
Further solving for d,
d2 = 36
d = √36
d = 6 ...(2)
Now, using the values of a and d in the expressions of the three terms, we get,
First term = a - d
So,
a - d = 7 - 6 = 1
Second term = a
So,
a = 7
Also,
Third term = a + d
So,
a + d = 7 + 6 = 13
Therefore, the three terms are 1, 7 and 13.
Ques.6. Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.
Ans. Here, we are given that four number are in A.P., such that there sum is 50 and the greatest number is 4 times the smallest.
So, let us take the four terms as a - d, a, a + d, a + 2d.
Now, we are given that sum of these numbers is 50, so we get,
(a - d) + (a) + (a + d) + (a + 2d) = 50
a - d + a + a + d + a + 2d = 50
4a + 2d = 50
2a + d = 25 ...(1)
Also, the greatest number is 4 times the smallest, so we get,
a + 2d = 4(a - d)
a + 2d = 4a - 4d
4d + 2d = 4a - a
6d = 3a
d = (3/6)a ...(2)
Now, using (2) in (1), we get,
2a + (3/6)a = 25
15a = 150
a = 150/15
a = 10
Now, using the value of a in (2), we get
d =(3/6)10
d = 10/2
d = 5
So, first term is given by,
a - d = 10 - 5 = 5
Second term is given by,
a = 10
Third term is given by,
a + d = 10 + 5 = 15
Fourth term is given by,
a + 2d = 10 + (2)(5)
= 10 + 10 = 20
Therefore, the four terms are 5, 10, 15, 20.
Ques.7. The sum of three numbers in A.P. is 12 and the sum of their cubes is 288. Find the numbers.
Ans. In the given problem, the sum of three terms of an A.P is 12 and the sum of their cubes is 288.
We need to find the three terms.
Here,
Let the three terms be (a - d),a,(a + d) where, a is the first term and d is the common difference of the A.P
So,
(a - d) + a + (a + d) = 12
3a = 12
a = 12/3
a = 4
Also, it is given that
(a - d)3 + a3 + (a + d)3 = 288
So, using the properties:
(a - b)3 = a3 - b3 + 3ab2 - 3a2b
(a + b)3 = a3 + b3 + 3ab2 + 3a2b
We get,
(a - d)3 + a3 + (a + d)3 = 288
a3 - d3 - 3a2d + 3d2a + a3 + d3 +3a2d + 3d2a = 288
3a3 + 6d2a = 288
a3 + 2d2a = 96
Further solving for d by substituting the value of a, we get,
(4)3 + 2d2(4) = 96
64 + 8d2 = 96
8d2 = 96 - 64
d2 = 32/8
On further simplification, we get,
d = 4
d = √4
d = ±2
Now, here d can have two values +2 and -2.
So, on substituting the values of a = 4 and d = 2 in three terms, we get,
First-term = a - d
So,
a - d = 4 - 2 = 2
Second term = a
So,
a = 4
Third term = a + d
So,
a + d = 4 + 2 = 6
Also, on substituting the values of a = 4 and d = -2 in three terms, we get,
First term = a - d
So,
a - d = 4 - (-2)
= 4 + 2
= 6
Second term = a
So,
a = 4
Third term = a + d
So,
a + d = 4 + (-2)
= 4 - 2 = 2
Therefore, the three terms are 2, 4 and 6 or 6, 4 and 2.
Ques.8. Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.
Ans. Let the four terms of the AP be a − 3d, a − d, a + d and a + 3d.
Given:
(a − 3d) + (a − d) + (a + d) + (a + 3d) = 56
⇒ 4a = 56
⇒ a = 14
Also,
⇒ 1176 − 54d2 = 980 − 5d2
⇒ 196 = 49d2
⇒ d2 = 4
⇒ d = ±2
When d = 2, the terms of the AP are 8, 12, 16, 20. When d = −2, the terms of the AP are 20, 18, 12, 8.
Ques.9. The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and third term is 5 times the common difference, find the three numbers.
Ans. Let the first three numbers in an arithmetic progression be a − d, a, a + d.
The sum of the first three numbers in an arithmetic progression is 18.
a − d + a + a + d = 18
⇒ 3a = 18
⇒ a = 6
The product of the first and third term is 5 times the common difference.
(a − d)(a + d) = 5d
⇒ a2 − d2 = 5d
⇒ (6)2 − d2 = 5d
⇒ 36 − d2 = 5d
⇒ d2 + 5d − 36 = 0
⇒ d2 + 9d − 4d − 36 = 0
⇒ d(d + 9) − 4(d + 9) = 0
⇒ (d + 9)(d − 4) = 0
⇒ d = 4, −9
For d = 4,
The numbers are 6 − 4, 6, 6 + 4
i.e. 2, 6, 10
For d = −9,
The numbers are 6 + 9, 6, 6 − 9
i.e. 15, 6, −3
Hence, the three numbers are 2, 6, 10 or 15, 6, −3.
Ques.10. The angles of a quadrilateral are in A.P. whose common difference is 10°. Find the numbers.
Ans. Here, we are given that the angles of a quadrilateral are in A.P, such that the common difference is 10°.
So, let us take the angles as a - d, a, a + d, a + 2d
Now, we know that the sum of all angles of a quadrilateral is 360°. So, we get,
(a - d) + (a) + (a + d) + (a + 2d) = 360
a - d + a + a + d + a + 2d = 360
4a + 2(10) = 360
4a = 360 - 20
On further simplifying for a, we get,
a = 340/4
a = 85
So, the first angle is given by,
a - d = 85 - 10 = 75°
Second angle is given by,
a = 85°
Third angle is given by,
a + d = 85 + 10
= 95°
Fourth angle is given by,
a + 2d = 85 + 2(10)
= 85 + 20 = 105°
Therefore, the four angles of the quadrilateral are 75°, 85°, 95°, 105°.
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Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-8) Notes
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Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-8) Class 10 Questions
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The content of Chapter 5 - Quadratic Equations, RD Sharma Solutions - (Part-8) is prepared as per the latest Class 10 syllabus.
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