Question

Find the value of x that makes (x+8), (4x +6) and 3x the first three terms of an arithmetic sequence

Answer 1

Answer:

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Answer 2

To find:

The value of $x$ that makes $(x+8)$, $(4x+6)$ and $3x$ the first three terms of an Arithmetic sequence

Step-by-step explanation:

Considering $(x+8)$, $(4x+6)$ and $3x$ the terms of an Arithmetic sequence, we can say that the respective differences are equal. Then,

$\quad (4x+6)-(x+8)=3x-(4x+6)$

$\Rightarrow 4x+6-x-8=3x-4x-6$

$\Rightarrow 3x-2=-x-6$

$\Rightarrow 3x+x=-6+2$

$\Rightarrow 4x=-4$

$\Rightarrow x=-1$

Final Answer:

The value of $x$ is $(-1)$ that makes $(x+8)$, $(4x+6)$ and $3x$ the first three terms of an Arithmetic sequence.

NOTE:

When $x=-1$, the given Arithmetic sequence becomes $7,2,-3$ where common difference is $(-5)$.