Find the value of $x$ for which $(8x + 4), (6x – 2)$ and $(2x + 7)$ are in A.P.
Given:
$(8x + 4), (6x – 2)$ and $(2x + 7)$ are in A.P.
To do:
We have to find the value of $x$.
Solution:
If the given terms are in A.P., then their common difference is equal.
Therefore,
$(6x-2)-(8x+4)=(2x+7)-(6x-2)$
$6x-8x-2-4=2x-6x+7+2$
$-2x-6=-4x+9$
$4x-2x=9+6$
$2x=15$
$x=\frac{15}{2}$
The value of $x$ is $\frac{15}{2}$.
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