(x-2), 6, (2x-3) are terms of an A.P, find x.
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calendarEducator since 2010
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We have (x-2), 6, (2x-3) as consecutive terms of an AP. The consecutive terms of an AP have a common difference.
So 2x - 3 - 6 = 6 - (x - 2)
=> 2x - 9 = 6 - x + 2
=> 2x - 9 = -x + 8
=> 3x = 17
=> x = 17/3
Therefore x is equal to 17/3.
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calendarEducator since 2008
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Given the terms:
(x-2) , 6, (2x-3) are terms of an arithmetical progression.
Then, we will assume that "r" is the common difference between terms.
==> 6 = (x-2) + r
==> x+r = 8.............(1)
==> (2x-3) = 6 + r
==> 2x -r = 9 ..............(2)
Now we will solve the system using the elimination method.
We will add (1) and (2).
==> 3x = 17
==> x = 17/3
==> r = 6- 11/3 = 7/3
==> (x-2) = 17/3 - 6/3 = 11/3
==> 6 = 6
==> 2x-3 = 34/3 - 9/3 = 25/3
==> 11/3, 6, 25/3 are terms of A.P and the common difference is r = 7/3
In an arithmetic progression( A.P), the successive terms are increasing or decreasing by the same amount called the common difference d.
=> Second term - first term = third term - second term.
The given terms of A.P is (x-2), 6 and (2x-3).
6 - (x-2) = 2x-3 - 6.
6-x+2 = 2x-9
6+2+9 = 2x+x
17 = 3x.
17/3 = x.
So x= 17/3
Therefore the first term of the A.P are x-2 = 17/3 -2 = 11/6.
The third term = 2x-3 = 2(17/3)-3 = 25/3.
Therefore the terms of the A.P are 11/3, 6 and 25/3.
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