RD Chapter 19 Arithmetic Progressions Ex 19.3 Solutions
Question - 1 : - The Sum of the three terms of an A.P. is 21 and the product of the first, and the third terms exceed the second term by 6, find three terms.
Answer - 1 : -
Given:
The sum of first threeterms is 21
Let us assume thefirst three terms as a – d, a, a + d [where a is the first term and d is thecommon difference]
So, sum of first threeterms is
a – d + a + a + d = 21
3a = 21
a = 7
It is also given thatproduct of first and third term exceeds the second by 6
So, (a – d)(a + d) – a= 6
a2 – d2 –a = 6
Substituting the valueof a = 7, we get
72 – d2 –7 = 6
d2 =36
d = 6 or d = – 6
Hence, the terms of APare a – d, a, a + d which is 1, 7, 13.
Question - 2 : - Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers
Answer - 2 : -
Given:
Sum of first threeterms is 27
Let us assume thefirst three terms as a – d, a, a + d [where a is the first term and d is thecommon difference]
So, sum of first threeterms is
a – d + a + a + d = 27
3a = 27
a = 9
It is given that theproduct of three terms is 648
So, a3 –ad2 = 648
Substituting the valueof a = 9, we get
93 –9d2 = 648
729 – 9d2 =648
81 = 9d2
d = 3 or d = – 3
Hence, the given termsare a – d, a, a + d which is 6, 9, 12.
Question - 3 : - Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.
Answer - 3 : -
Given:
Sum of four terms is50.
Let us assume thesefour terms as a – 3d, a – d, a + d, a + 3d
It is given that, sumof these terms is 4a = 50
So, a = 50/4
= 25/2 … (i)
It is also given thatthe greatest number is 4 time the least
a + 3d = 4(a – 3d)
Substitute the valueof a = 25/2, we get
(25+6d)/2 = 50 – 12d
30d = 75
d = 75/30
= 25/10
= 5/2 … (ii)
Hence, the terms of APare a – 3d, a – d, a + d, a + 3d which is 5, 10, 15, 20
Question - 4 : - The sum of three numbers in A.P. is 12, andthe sum of their cubes is 288. Find the numbers.
Answer - 4 : -
Given:
The sum of threenumbers is 12
Let us assume thenumbers in AP are a – d, a, a + d
So,
3a = 12
a = 4
It is also given thatthe sum of their cube is 288
(a – d)3 +a3 + (a + d)3 = 288
a3 – d3 –3ad(a – d) + a3 + a3 + d3 +3ad(a + d) = 288
Substitute the valueof a = 4, we get
64 – d3 –12d(4 – d) + 64 + 64 + d3 + 12d(4 + d) = 288
192 + 24d2 =288
d = 2 or d = – 2
Hence, the numbers area – d, a, a + d which is 2, 4, 6 or 6, 4, 2
Question - 5 : - If the sum of three numbers in A.P. is 24 and their product is 440, findthe numbers.
Answer - 5 : -
Given:
Sum of first threeterms is 24
Let us assume thefirst three terms are a – d, a, a + d [where a is the first term and d is thecommon difference]
So, sum of first threeterms is a – d + a + a + d = 24
3a = 24
a = 8
It is given that theproduct of three terms is 440
So a3 –ad2 = 440
Substitute the valueof a = 8, we get
83 –8d2 = 440
512 – 8d2 =440
72 = 8d2
d = 3 or d = – 3
Hence, the given termsare a – d, a, a + d which is 5, 8, 11
Question - 6 : - The angles of a quadrilateral are in A.P. whose common difference is 10.Find the angles
Answer - 6 : -
Given: d = 10
We know that the sumof all angles in a quadrilateral is 360
Let us assume theangles are a – 3d, a – d, a + d, a + 3d
So, a – 2d + a – d + a+ d + a + 2d = 360
4a = 360
a = 90… (i)
And,
(a – d) – (a – 3d) =10
2d = 10
d = 10/2
= 5
Hence, the angles area – 3d, a – d, a + d, a + 3d which is 75o, 85o, 95o,105o