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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Updated on: 10-Oct-2022

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