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 тАУ d, a, a + d [where a is the first term and d is thecommon difference]
So, sum of first threeterms is
a тАУ 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 тАУ d)(a + d) тАУ a= 6
a2┬атАУ d2┬атАУa = 6
Substituting the valueof a = 7, we get
72┬атАУ d2┬атАУ7 = 6
d2┬а=36
d = 6 or d = тАУ 6
Hence, the terms of APare a тАУ 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 тАУ d, a, a + d [where a is the first term and d is thecommon difference]
So, sum of first threeterms is
a тАУ d + a + a + d = 27
3a = 27
a = 9
It is given that theproduct of three terms is 648
So, a3┬атАУad2┬а= 648
Substituting the valueof a = 9, we get
93┬атАУ9d2┬а= 648
729 тАУ 9d2┬а=648
81 = 9d2
d = 3 or d = тАУ 3
Hence, the given termsare a тАУ 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 тАУ 3d, a тАУ d, a + d, a + 3d
It is given that, sumof these terms is 4a = 50
So, a = 50/4
= 25/2 тАж (i)
It is also given thatthe greatest number is 4 time the least
a + 3d = 4(a тАУ 3d)
Substitute the valueof a = 25/2, we get
(25+6d)/2 = 50 тАУ 12d
30d = 75
d = 75/30
= 25/10
= 5/2 тАж (ii)
Hence, the terms of APare a тАУ 3d, a тАУ 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 тАУ d, a, a + d
So,
3a = 12
a = 4
It is also given thatthe sum of their cube is 288
(a тАУ d)3┬а+a3┬а+ (a + d)3┬а= 288
a3┬атАУ d3┬атАУ3ad(a тАУ d) + a3┬а+ a3┬а+ d3┬а+3ad(a + d) = 288
Substitute the valueof a = 4, we get
64 тАУ d3┬атАУ12d(4 тАУ d) + 64 + 64 + d3┬а+ 12d(4 + d) = 288
192 + 24d2┬а=288
d = 2 or d = тАУ 2
Hence, the numbers area тАУ 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 тАУ d, a, a + d [where a is the first term and d is thecommon difference]
So, sum of first threeterms is a тАУ d + a + a + d = 24
3a = 24
a = 8
It is given that theproduct of three terms is 440
So a3┬атАУad2┬а= 440
Substitute the valueof a = 8, we get
83┬атАУ8d2┬а= 440
512 тАУ 8d2┬а=440
72 = 8d2
d = 3 or d = тАУ 3
Hence, the given termsare a тАУ 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:┬аd = 10
We know that the sumof all angles in a quadrilateral is 360
Let us assume theangles are a тАУ 3d, a тАУ d, a + d, a + 3d
So, a тАУ 2d + a тАУ d + a+ d + a + 2d = 360
4a = 360
a = 90тАж (i)
And,
(a тАУ d) тАУ (a тАУ 3d) =10
2d = 10
d = 10/2
= 5
Hence, the angles area тАУ 3d, a тАУ d, a + d, a + 3d which is 75o, 85o, 95o,105o