RS Aggarwal Solutions: Exercise 5B - Arithmetic Progressions Class 10 Notes | EduRev

Mathematics (Maths) Class 10

Class 10 : RS Aggarwal Solutions: Exercise 5B - Arithmetic Progressions Class 10 Notes | EduRev

 Page 1


1. Determine k so that (3k -2), (4k 6) and (k +2) are three consecutive terms of an AP.
3 2 , 4 6 k k 2 k
Sol:
It is given that and are three consecutive terms of an AP.
4 3 4 k 6
4 6 3 2 4 6
4 3 8
3 8 4
4 12
3
k k k
k k k
k
k k
k
k
6 2 2
2 k
k
Hence, the value of k is 3. 
2.
1 5 2 , 4 x x 2 x
Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP.
Sol:
It is given that and are in AP.
4 5 2 4 x 1
4 1 5 2 2 4 1
3 3
3 3 3
2 x 6
3
x x x
x x x x
x 3 x
x x
x
1 2
Hence, the value of x is 3. 
Page 2


1. Determine k so that (3k -2), (4k 6) and (k +2) are three consecutive terms of an AP.
3 2 , 4 6 k k 2 k
Sol:
It is given that and are three consecutive terms of an AP.
4 3 4 k 6
4 6 3 2 4 6
4 3 8
3 8 4
4 12
3
k k k
k k k
k
k k
k
k
6 2 2
2 k
k
Hence, the value of k is 3. 
2.
1 5 2 , 4 x x 2 x
Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP.
Sol:
It is given that and are in AP.
4 5 2 4 x 1
4 1 5 2 2 4 1
3 3
3 3 3
2 x 6
3
x x x
x x x x
x 3 x
x x
x
1 2
Hence, the value of x is 3. 
 
3. If (3y 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value 
of y.
Sol:
It is given that 3 1 , 3 5 y y and 5 1 y are three consecutive terms of an AP.
3 5 3 1 5 1 3 5
3 5 3 1 5 1 3 5
6 2 4
2 6 4 10
5
y y y y
y y y y
y
y
y
Hence, the value of y is 5.
4. Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP.
Sol:
Since 2 ,2 x x and 2 3 x are in AP, we have
2 2 2 3 2
2 3
5
5
x x x x
x
x
x
5. Show that 
2
2 2
, a b a b and 
2 2
a b are in AP. 
Sol:
The given numbers are 
2
2 2
, a b a b and 
2
. a b
Now,
2
2 2 2 2 2 2 2 2 2 2
2
2 2 2 2 2 2
2 2 2
2 2
a b a b a b a a b b a b a a b b a b
a b a b a a b b a b a b
So, 
2 2
2 2 2 2
2 a b a b a b a b a b (Constant)
Since each term differs from its preceding term by a constant, therefore, the given numbers 
are in AP.
6. Find the three numbers in AP whose sum is 15 and product is 80.
Sol:
Let the required numbers be , a d a and . a d
Then 15 a d a a d
Page 3


1. Determine k so that (3k -2), (4k 6) and (k +2) are three consecutive terms of an AP.
3 2 , 4 6 k k 2 k
Sol:
It is given that and are three consecutive terms of an AP.
4 3 4 k 6
4 6 3 2 4 6
4 3 8
3 8 4
4 12
3
k k k
k k k
k
k k
k
k
6 2 2
2 k
k
Hence, the value of k is 3. 
2.
1 5 2 , 4 x x 2 x
Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP.
Sol:
It is given that and are in AP.
4 5 2 4 x 1
4 1 5 2 2 4 1
3 3
3 3 3
2 x 6
3
x x x
x x x x
x 3 x
x x
x
1 2
Hence, the value of x is 3. 
 
3. If (3y 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value 
of y.
Sol:
It is given that 3 1 , 3 5 y y and 5 1 y are three consecutive terms of an AP.
3 5 3 1 5 1 3 5
3 5 3 1 5 1 3 5
6 2 4
2 6 4 10
5
y y y y
y y y y
y
y
y
Hence, the value of y is 5.
4. Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP.
Sol:
Since 2 ,2 x x and 2 3 x are in AP, we have
2 2 2 3 2
2 3
5
5
x x x x
x
x
x
5. Show that 
2
2 2
, a b a b and 
2 2
a b are in AP. 
Sol:
The given numbers are 
2
2 2
, a b a b and 
2
. a b
Now,
2
2 2 2 2 2 2 2 2 2 2
2
2 2 2 2 2 2
2 2 2
2 2
a b a b a b a a b b a b a a b b a b
a b a b a a b b a b a b
So, 
2 2
2 2 2 2
2 a b a b a b a b a b (Constant)
Since each term differs from its preceding term by a constant, therefore, the given numbers 
are in AP.
6. Find the three numbers in AP whose sum is 15 and product is 80.
Sol:
Let the required numbers be , a d a and . a d
Then 15 a d a a d
 
3 15
5
a
a
Also, . . 80 a d a a d
2 2
2
2
80
5 25 80
25 16 9
3
a a d
d
d
d
Thus, 5 a and 3 d
Hence, the required numbers are 2,5 8 8,5 2 . a n d o r a n d
7. The sum of three numbers in AP is 3 and their product is -35. Find the numbers.
Sol:
Let the required numbers be , a d a and . a d
Then 3 a d a a d
3 3
1
a
a
Also, . . 35 a d a a d
2 2
2
2
35
1. 1 35
36
6
a a d
d
d
d
Thus, 1 a and 6 d
Hence, the required numbers are 5,1 7 7,1 5 . a n d o r a n d
8. Divide 24 in three parts such that they are in AP and their product is 440.
Sol:
Let the required parts of 24 be , a d a and a d such that they are in AP.
Then 24 a d a a d
3 24
8
a
a
Also, . . 440 a d a a d
2 2
2
440
8 64 440
a a d
d
Page 4


1. Determine k so that (3k -2), (4k 6) and (k +2) are three consecutive terms of an AP.
3 2 , 4 6 k k 2 k
Sol:
It is given that and are three consecutive terms of an AP.
4 3 4 k 6
4 6 3 2 4 6
4 3 8
3 8 4
4 12
3
k k k
k k k
k
k k
k
k
6 2 2
2 k
k
Hence, the value of k is 3. 
2.
1 5 2 , 4 x x 2 x
Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP.
Sol:
It is given that and are in AP.
4 5 2 4 x 1
4 1 5 2 2 4 1
3 3
3 3 3
2 x 6
3
x x x
x x x x
x 3 x
x x
x
1 2
Hence, the value of x is 3. 
 
3. If (3y 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value 
of y.
Sol:
It is given that 3 1 , 3 5 y y and 5 1 y are three consecutive terms of an AP.
3 5 3 1 5 1 3 5
3 5 3 1 5 1 3 5
6 2 4
2 6 4 10
5
y y y y
y y y y
y
y
y
Hence, the value of y is 5.
4. Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP.
Sol:
Since 2 ,2 x x and 2 3 x are in AP, we have
2 2 2 3 2
2 3
5
5
x x x x
x
x
x
5. Show that 
2
2 2
, a b a b and 
2 2
a b are in AP. 
Sol:
The given numbers are 
2
2 2
, a b a b and 
2
. a b
Now,
2
2 2 2 2 2 2 2 2 2 2
2
2 2 2 2 2 2
2 2 2
2 2
a b a b a b a a b b a b a a b b a b
a b a b a a b b a b a b
So, 
2 2
2 2 2 2
2 a b a b a b a b a b (Constant)
Since each term differs from its preceding term by a constant, therefore, the given numbers 
are in AP.
6. Find the three numbers in AP whose sum is 15 and product is 80.
Sol:
Let the required numbers be , a d a and . a d
Then 15 a d a a d
 
3 15
5
a
a
Also, . . 80 a d a a d
2 2
2
2
80
5 25 80
25 16 9
3
a a d
d
d
d
Thus, 5 a and 3 d
Hence, the required numbers are 2,5 8 8,5 2 . a n d o r a n d
7. The sum of three numbers in AP is 3 and their product is -35. Find the numbers.
Sol:
Let the required numbers be , a d a and . a d
Then 3 a d a a d
3 3
1
a
a
Also, . . 35 a d a a d
2 2
2
2
35
1. 1 35
36
6
a a d
d
d
d
Thus, 1 a and 6 d
Hence, the required numbers are 5,1 7 7,1 5 . a n d o r a n d
8. Divide 24 in three parts such that they are in AP and their product is 440.
Sol:
Let the required parts of 24 be , a d a and a d such that they are in AP.
Then 24 a d a a d
3 24
8
a
a
Also, . . 440 a d a a d
2 2
2
440
8 64 440
a a d
d
 
2
64 55 9
3
d
d
Thus, 8 a and 3 d
Hence, the required parts of 24 are 5,8,11 11,8,5 . o r
9. The sum of three consecutive terms of an AP is 21 and the sum of the squares of these 
terms is 165. Find these terms
Sol:
Let the required terms be , a d a and . a d
Then 21 a d a a d
3 21
7
a
a
Also,
2 2
2
165 a d a a d
2 2
2
2
2
3 2 165
3 49 2 165
2 165 147 18
9
3
a d
d
d
d
d
Thus, 7 a and 3 d
Hence, the required terms are 4,7,10 10,7,4 . o r
10. The angles of quadrilateral are in whose AP common difference is 10 . Find the angles.
Sol:
Let the required angles be 15 , 5 , 5 15 , a a a a n d a as the common 
difference is 10 (given).
Then 15 5 5 15 360 a a a a
4 360
90
a
a
Hence, the required angles of a quadrilateral are
90 15 , 90 5 , 90 5 90 15 ; 75 ,85 ,95 105 . a n d o r a n d
11. Find four numbers in AP whose sum is 8 and the sum of whose squares is 216.
Sol:
(4, 6, 8, 10) or (10, 8, 6, 4)
Page 5


1. Determine k so that (3k -2), (4k 6) and (k +2) are three consecutive terms of an AP.
3 2 , 4 6 k k 2 k
Sol:
It is given that and are three consecutive terms of an AP.
4 3 4 k 6
4 6 3 2 4 6
4 3 8
3 8 4
4 12
3
k k k
k k k
k
k k
k
k
6 2 2
2 k
k
Hence, the value of k is 3. 
2.
1 5 2 , 4 x x 2 x
Find the value of x for which the numbers (5x + 2), (4x - 1) and (x + 2) are in AP.
Sol:
It is given that and are in AP.
4 5 2 4 x 1
4 1 5 2 2 4 1
3 3
3 3 3
2 x 6
3
x x x
x x x x
x 3 x
x x
x
1 2
Hence, the value of x is 3. 
 
3. If (3y 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value 
of y.
Sol:
It is given that 3 1 , 3 5 y y and 5 1 y are three consecutive terms of an AP.
3 5 3 1 5 1 3 5
3 5 3 1 5 1 3 5
6 2 4
2 6 4 10
5
y y y y
y y y y
y
y
y
Hence, the value of y is 5.
4. Find the value of x for which (x + 2), 2x, ()2x + 3) are three consecutive terms of an AP.
Sol:
Since 2 ,2 x x and 2 3 x are in AP, we have
2 2 2 3 2
2 3
5
5
x x x x
x
x
x
5. Show that 
2
2 2
, a b a b and 
2 2
a b are in AP. 
Sol:
The given numbers are 
2
2 2
, a b a b and 
2
. a b
Now,
2
2 2 2 2 2 2 2 2 2 2
2
2 2 2 2 2 2
2 2 2
2 2
a b a b a b a a b b a b a a b b a b
a b a b a a b b a b a b
So, 
2 2
2 2 2 2
2 a b a b a b a b a b (Constant)
Since each term differs from its preceding term by a constant, therefore, the given numbers 
are in AP.
6. Find the three numbers in AP whose sum is 15 and product is 80.
Sol:
Let the required numbers be , a d a and . a d
Then 15 a d a a d
 
3 15
5
a
a
Also, . . 80 a d a a d
2 2
2
2
80
5 25 80
25 16 9
3
a a d
d
d
d
Thus, 5 a and 3 d
Hence, the required numbers are 2,5 8 8,5 2 . a n d o r a n d
7. The sum of three numbers in AP is 3 and their product is -35. Find the numbers.
Sol:
Let the required numbers be , a d a and . a d
Then 3 a d a a d
3 3
1
a
a
Also, . . 35 a d a a d
2 2
2
2
35
1. 1 35
36
6
a a d
d
d
d
Thus, 1 a and 6 d
Hence, the required numbers are 5,1 7 7,1 5 . a n d o r a n d
8. Divide 24 in three parts such that they are in AP and their product is 440.
Sol:
Let the required parts of 24 be , a d a and a d such that they are in AP.
Then 24 a d a a d
3 24
8
a
a
Also, . . 440 a d a a d
2 2
2
440
8 64 440
a a d
d
 
2
64 55 9
3
d
d
Thus, 8 a and 3 d
Hence, the required parts of 24 are 5,8,11 11,8,5 . o r
9. The sum of three consecutive terms of an AP is 21 and the sum of the squares of these 
terms is 165. Find these terms
Sol:
Let the required terms be , a d a and . a d
Then 21 a d a a d
3 21
7
a
a
Also,
2 2
2
165 a d a a d
2 2
2
2
2
3 2 165
3 49 2 165
2 165 147 18
9
3
a d
d
d
d
d
Thus, 7 a and 3 d
Hence, the required terms are 4,7,10 10,7,4 . o r
10. The angles of quadrilateral are in whose AP common difference is 10 . Find the angles.
Sol:
Let the required angles be 15 , 5 , 5 15 , a a a a n d a as the common 
difference is 10 (given).
Then 15 5 5 15 360 a a a a
4 360
90
a
a
Hence, the required angles of a quadrilateral are
90 15 , 90 5 , 90 5 90 15 ; 75 ,85 ,95 105 . a n d o r a n d
11. Find four numbers in AP whose sum is 8 and the sum of whose squares is 216.
Sol:
(4, 6, 8, 10) or (10, 8, 6, 4)
 
12. Divide 32 into four parts which are the four terms of an AP such that the product of the first 
and fourth terms is to product of the second and the third terms as 7:15.
Sol:
Let the four parts in AP be 3 , , 3 . a d a d a d a n d a d Then,
3 3 32
4 32
8 ......... 1
a d a d a d a d
a
a
Also,
2
2
2 2
2 2
2
2
2
3 3 : 7 :15
8 3 8 3
7
(1)
8 8 15
64 9 7
64 15
15 64 9 7 64
960 135 448 7
135 7 960 448
128 512
4
2
a d a d a d a d
d d
F rom
d d
d
d
d d
d d
d d
d
d
d
When 8 a and 2, d
3 8 3 2 8 6 2
8 2 6
8 2 10
3 8 3 2 8 6 14
a d
a d
a d
a d
When 8 a and 2, d
3 8 3 2 8 6 14
8 2 8 2 10
8 2 6
3 8 3 2 8 6 2
a d
a d
a d
a d
Hence, the four parts are 2, 6, 10 and 14.
13. The sum of first three terms of an AP is 48. If the product of first and second terms exceeds 
4 times the third term by 12. Find the AP.
Sol:
Let the first three terms of the AP be , . a d a a n d a d Then,
Read More
Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

Related Searches

Important questions

,

Previous Year Questions with Solutions

,

mock tests for examination

,

Semester Notes

,

pdf

,

Free

,

study material

,

Sample Paper

,

shortcuts and tricks

,

past year papers

,

RS Aggarwal Solutions: Exercise 5B - Arithmetic Progressions Class 10 Notes | EduRev

,

RS Aggarwal Solutions: Exercise 5B - Arithmetic Progressions Class 10 Notes | EduRev

,

Exam

,

RS Aggarwal Solutions: Exercise 5B - Arithmetic Progressions Class 10 Notes | EduRev

,

Objective type Questions

,

Summary

,

ppt

,

Extra Questions

,

video lectures

,

Viva Questions

,

MCQs

,

practice quizzes

;