The second,third, and fourth terms of a ap are ( x-2), 5,( X-2) find the value of X and 9 term
The second, third, and fourth terms of an arithmetic progression can be written as:
Second term: a + d
Third term: a + 2d
Fourth term: a + 3d
Where a is the first term, d is the common difference.
Given that the second term is (x-2), the third term is 5, and the fourth term is (x-2), we can write the following equations:
x - 2 = a + d ... (1)
5 = a + 2d ... (2)
x - 2 = a + 3d ... (3)
From equation (2), we can solve for a in terms of d:
a = 5 - 2d
Substitute this value of a into equations (1) and (3):
x - 2 = (5 - 2d) + d
Simplifying this equation, we get:
x - 2 = 5 - d ... (4)
x - 2 = (5 - 2d) + 3d
Simplifying this equation, we get:
x - 2 = 5 + d ... (5)
From equations (4) and (5), we can conclude that:
5 - d = 5 + d
This equation implies that d = 0, and hence the common difference is zero. Therefore, the arithmetic progression is actually a constant sequence.
Substituting d = 0 into equations (2) and (5), we get:
a = 5 - 2(0) = 5
x - 2 = 5 + 0 => x = 7
Therefore, x = 7 and the 9th term of the sequence is (x-2) = (7-2) = 5.