RS Aggarwal Solutions Chapter 11 Arithmetic Progression Exercise 11B Class 10 Maths
Chapter Name  RS Aggarwal Chapter 11 Arithmetic Progression 
Book Name  RS Aggarwal Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 11B Solutions
1. Determine k so that (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.
Solution
It is given that (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP.
∴ (4k – 6) – (3k – 2) = (k + 2) – (4k – 6)
⇒ 4k – 6 – 3k + 2 = k + 2 – 4k + 6
⇒ k – 4 = 3k + 8
⇒ k + 3k = 8 + 4
⇒ 4k = 12
⇒ k = 3
Hence, the value of k is 3.
2. Find the value of x for which the numbers (5x + 2), (4x – 1) and (x + 2) are in AP.
Solution
It is given that (5x + 2), (4x – 1) and (x + 2) are in AP.
∴ (4x – 1) – (5x + 2) = (x + 2) – (4x – 1)
⇒ 4x – 1 – 5x – 2 = x + 2 – 4x + 1
⇒  x – 3 =  3x + 3
⇒ 2x = 6
⇒ x = 3
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.
Solution
It is given that (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP.
∴ (3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)
⇒ 3y + 5 – 3y + 1 = 5y + 1 – 3y – 5
⇒ 6 = 2y – 4
⇒ 2y = 6 + 4 = 10
⇒ y = 5
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.
Solution
Since (x + 2), 2x and (2x + 3) are in AP, we have
2x – (x + 2) = (2x + 3) – 2x
⇒ x – 2 = 3
⇒ x = 5
∴ x = 5
5. Show that (a – b)^{2}, (a^{2 }+ b^{2}) and (a^{2} + b^{2}) are in AP.
Solution
Now,
(a^{2} + b^{2}) – (a – b)^{2} = a^{2} + b^{2} – (a^{2} – 2ab + b^{2}) = a^{2 }+ b^{2} – a^{2} + 2ab – b^{2} = 2ab
(a + b)^{2} – (a^{2} + b^{2}) = a^{2} + 2ab + b^{2 }– a^{2} – b^{2} = 2ab
So, (a^{2} + b^{2}) – (a – b)^{2} = (a + b)^{2} – (a^{2 }+ b^{2}) = 2ab (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.
Solution
Let the required numbers be (a – d), a and (a + d).
Then (a – d) + a + (a + d) = 15
⇒ 3a = 15
⇒ a = 5
Also, (a – d).a.(a + d) = 80
⇒ a(a^{2} – d^{2}) = 80
⇒ 5(25 – d^{2}) = 80
⇒ d^{2} = (25 – 16) = 9
⇒ d = ± 3
Thus, a = 5 and d = ±3
Hence, the required numbers are (2, 5 and 8) or (8, 5 and 2).
7. The sum of three numbers in AP is 3 and their product is 35. Find the numbers.
Solution
Let the required numbers be (a – d), a and (a + d).
Then (a – d) + a + (a + d) = 3
⇒ 3a = 3
⇒ a = 1
Also, (a – d), a.(a + d) =  35
⇒ a(a^{2} – d^{2}) =  35
⇒ 1.(1 – d^{2}) =  35
⇒ d^{2} = 36
⇒ d = ± 6
Thus, a = 1 and d = ± 6
Hence, the required numbers are (5, 1 and 7) or (7, 1 and 5).
8. Divide 24 in three parts such that they are in AP and their product is 440.
Solution
Let the required parts of 24 be (a – d), a and (a + d) such that they are in AP.
Then (a – d) + a + (a + d) = 24
⇒ 3a = 24
⇒ a = 8
Also, (a – d).a.(a + d) = 440
⇒ a(a^{2} – d^{2}) = 440
⇒ 8(64 – d^{2}) = 440
⇒ d^{2} = 64 – 55 = 9
⇒ d = ±3
Thus, a = 8 and d = ±3
Hence, the required parts of 24 are (5, 8, 11) or (11, 8, 5).
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.
Solution
Let the required terms be (a – d), a and (a + d).
Then (a – d) + a + (a + d) = 21
⇒ 3a = 21
⇒ a = 7
Also, (a – d)^{2} + a^{2} + (a + d)^{2} = 165
⇒ 3a^{2 }+ 2d^{2} = 165
⇒ (3 × 49 + 2d^{2}) = 165
⇒ 2d^{2} = 165 – 147 = 18
⇒ d^{2} = 9
⇒ d = ±3
Thus, a = 7 and d = ±3
Hence, the required terms are (4, 7, 10) or (10, 7, 4).
10. The angles of quadrilateral are in whose AP common difference is 10°. Find the angles.
Solution
Let the required angles be (a – 15)°, (a – 5)°, (a + 5)° and (a + 15)°, as the common difference is 10 (given).
Then (a – 15)° + (a – 5)° + (a + 5)° + (a + 15)° = 360°
⇒ 4a = 360
⇒ a = 90
Hence, the required angles of a quadrilateral are (90 – 15)°, (90 – 5)°, (90 + 5)° and (90 + 15)°; or 75°, 85°, 95°, and 105°.
11. Find four numbers in AP whose sum is 8 and the sum of whose squares is 216.
Solution
(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.
Solution
Let the four parts in AP be (a – 3d), (a – d), (a + d) and (a + 3d). Then,
(a – 3d) + (a – d) + (a + d) + (a + 3d) = 32
⇒ 4a = 32
⇒ a = 8 ...(1)
Also,
(a – 3d)(a + 3d): (a – d)(a + d) = 7 : 15
⇒ (8 – 3d)(8 + 3d)/(8 – d)(8 + d) = 7/15 [From (1)]
⇒ (64 – 9d^{2})/(64 – d^{2}) = 7/15
⇒ 15(64 – 9d^{2}) = 7(64 – d^{2})
⇒ 960 – 135d^{2} = 448 – 7d^{2}
⇒ 135d^{2} – 7d^{2} = 960 – 448
⇒ 128d^{2} = 512
⇒ d^{2} = 4
⇒ d = ±
When a = 8 and d = 2,
a – 3d = 8 – 3 × 2 = 8 – 6 = 2
a – d = 8 – 2 = 6
a + d = 8 + 2 = 10
a + 3d = 8 + 3 × 2 = 8 + 6 = 14
When a = 8 and d =  2,
a – 3d = 8 – 3 × (2) = 8 + 6 = 14
a – d = 8 – (2) = 8 + 2 = 10
a + d = 8 – 2 = 6
a + 3d = 8 + 3 × (2) = 8 – 6 = 2
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 and third term by 12. Find the AP.
Solution
Let the first three terms of the AP be (a – d), a and (a + d). Then,
(a – d) + a + (a + d) = 48
⇒ 3a = 48
⇒ a = 16
Now,
(a – d) × a = 4(a + d) + 12 (Given)
⇒ (16 – d) × 16 = 4(16 + d) + 12
⇒ 256 – 16d = 64 + 4d + 12
⇒ 16d + 4d = 256 – 76
⇒ 20d = 180
⇒ d = 9
When a = 16 and d = 9,
a – d = 16 – 9 = 7
a + d = 16 + 9 = 25
Hence, the first three terms of the AP are 7, 16 and 25.