Selina Concise Class 10 Math Chapter 10 Arithmetic Progression Solutions
Exercise 10D
(Q1) Find three numbers in Arithmetic progression whose sum is 24 and whose product is 440.
Solution:
Let us consider three numbers in arithmetic progression –
(a – b), a and (a + b)
Given that sum is 24 –
Sn = 24
(a – b) + a + (a + b) = 24
a – b + a + a + b = 24
a + a + a = 24
3a = 24
a = 24/3
a = 8
And also given that,
Product of terms = 440
(a – b) × a × (a + b) = 440
a × (a – b) × (a + b) = 440
a (a2 – b2) = 440 (∵(a – b) (a + b) = a2 – b2
8 (82 – b2) = 440
8 (64 – b2) = 440
64 – b2 = 440/8
64 – b2 = 55
– b2 = 55 – 64
– b2 = – 9
b2 = 9
b = ± 3
b = d
d – Common difference.
∴ a = 8 and d = 3
∴ t2 = a + d
= 8 + 3
t2 = 11
(a – d) = 8 – 3
= 5
∴ Arithmetic progression –
The three terms are – 5, 8, 11
And If a = 8 and d = – 3
Then the arithmetic progression is – 11, 8, 5.
(Q2) The sum of three consecutive terms of an arithmetic progression is 21 and the sum of their squares is 165. Find these terms.
Solution:
Let us consider the three consecutive terms are in arithmetic progression are –
(a – b), a and (a + b)
Given that, The sum of three consecutive terms of an arithmetic progression is 21.
(a – b) + a + (a + b) = 21
a – b + a + a + b = 21
3a = 21
a = 21/3
a = 7
And also given that, the sum of squares is 165
(a – b)2 + a2 + (a + b)2 = 165
a2 – 2ab + b2 + a2 + a2 + 2ab + b2 = 165
a2 + b2 + a2 + a2 + b2 = 165
3a2 + 2b2 = 165
3(7)2 + 2b2 = 165
3 (49) + 2b2 = 165
147 + 2b2 = 165
2b2 = 165 – 147
2b2 = 18
b2 = 18/2
b2 = 9
b = ±3
∴ a = 7 and b = d = ±3
If a = 7 and d = 3 then we have arithmetic progression is –
(a – d) = 7 – 3 = 4
(a + d) = 7 + 3 = 10
∴ The arithmetic progression is 4, 7, 10.
And
If a = 7 and d = -3 then we have arithmetic progression is –
(a – d) = 7 – (-3) = 7 + 3 = 10
(a + d) = 7 + (-3) = 7 – 3 = 4
∴ The arithmetic progression is 10, 7, 4
(Q3) The angles of a quadrilateral are in arithmetic progression with common difference 20°. Find its angles.
= Solution:
Given that the angles of a quadrilateral are in arithmetic progression with common difference 20°.
Let us consider the angles of a quadrilateral are –
x, x + 20°, x + 40°, x + 60° (∵ Difference is 20°
We know that, The sum of all interior angles of a quadrilateral is 360°.
∴ Sum of all angles of quadrilateral –
x + x + 20° + x + 40° + x + 60° = 360°
4x + 120° = 360°
4x + 120° = 360°
4x = 360° – 120°
4x = 240°
x = 240°/4
x = 60°
∴ x = 60°
x + 20° = 60° + 20° = 80°
x + 40° = 60° + 40° = 100°
x + 60° = 60° + 60° = 120°
∴ The angles of a quadrilateral are – 60°, 80°, 100° and 120°.