Sometimes we require a certain number of terms in A.P. there are ways of selecting terms, which are very convenient. Exercise 19.3, discusses problems based on the selection of terms in an A.P. Most students are scared of Maths subject as it consists of a lot of formulas and requires a logical approach in solving them. For students who find difficulty in solving Class 11 Maths problems, our subject experts have solved the problems using shortcut techniques to help students understand the concepts clearly. For a better performance in the board exam, students can make use of RD Sharma Class 11 Maths Solutions pdf, from the links given below.
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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.
Solution:
Given:
The sum of first three terms is 21
Let us assume the first three terms as a – d, a, a + d [where a is the first term and d is the common difference]
So, sum of first three terms is
a – d + a + a + d = 21
3a = 21
a = 7
It is also given that product of first and third term exceeds the second by 6
So, (a – d)(a + d) – a = 6
a2Â – d2Â – a = 6
Substituting the value of a = 7, we get
72Â – d2Â – 7 = 6
d2Â = 36
d = 6 or d = – 6
Hence, the terms of AP are a – d, a, a + d which is 1, 7, 13.
2. Three numbers are in A.P. If the sum of these numbers be 27 and the product 648, find the numbers
Solution:
Given:
Sum of first three terms is 27
Let us assume the first three terms as a – d, a, a + d [where a is the first term and d is the common difference]
So, sum of first three terms is
a – d + a + a + d = 27
3a = 27
a = 9
It is given that the product of three terms is 648
So, a3Â – ad2Â = 648
Substituting the value of a = 9, we get
93Â – 9d2Â = 648
729 – 9d2Â = 648
81 = 9d2
d = 3 or d = – 3
Hence, the given terms are a – d, a, a + d which is 6, 9, 12.
3. Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.
Solution:
Given:
Sum of four terms is 50.
Let us assume these four terms as a – 3d, a – d, a + d, a + 3d
It is given that, sum of these terms is 4a = 50
So, a = 50/4
= 25/2 … (i)
It is also given that the greatest number is 4 time the least
a + 3d = 4(a – 3d)
Substitute the value of a = 25/2, we get
(25+6d)/2 = 50 – 12d
30d = 75
d = 75/30
= 25/10
= 5/2 … (ii)
Hence, the terms of AP are a – 3d, a – d, a + d, a + 3d which is 5, 10, 15, 20
4. The sum of three numbers in A.P. is 12, and the sum of their cubes is 288. Find the numbers.
Solution:
Given:
The sum of three numbers is 12
Let us assume the numbers in AP are a – d, a, a + d
So,
3a = 12
a = 4
It is also given that the sum of their cube is 288
(a – d)3Â + a3Â + (a + d)3Â = 288
a3Â – d3Â – 3ad(a – d) + a3Â + a3Â + d3Â + 3ad(a + d) = 288
Substitute the value of a = 4, we get
64 – d3Â – 12d(4 – d) + 64 + 64 + d3Â + 12d(4 + d) = 288
192 + 24d2Â = 288
d = 2 or d = – 2
Hence, the numbers are a – d, a, a + d which is 2, 4, 6 or 6, 4, 2
5. If the sum of three numbers in A.P. is 24 and their product is 440, find the numbers.
Solution:
Given:
Sum of first three terms is 24
Let us assume the first three terms are a – d, a, a + d [where a is the first term and d is the common difference]
So, sum of first three terms is a – d + a + a + d = 24
3a = 24
a = 8
It is given that the product of three terms is 440
So a3Â – ad2Â = 440
Substitute the value of a = 8, we get
83Â – 8d2Â = 440
512 – 8d2Â = 440
72 = 8d2
d = 3 or d = – 3
Hence, the given terms are a – d, a, a + d which is 5, 8, 11
6. The angles of a quadrilateral are in A.P. whose common difference is 10. Find the angles
Solution:
Given:Â d = 10
We know that the sum of all angles in a quadrilateral is 360
Let us assume the angles 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 are a – 3d, a – d, a + d, a + 3d which is 75o, 85o, 95o, 105o