Find the value of x so that the given terms are consecutive terms of an arithmetic sequence. `2x` , `3x+1` , and `x^2+2`

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lemjay | High School Teacher | (Level 3) Senior Educator

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In arithmetic sequence, two consecutive terms has a common difference (d). 

`d= a_(n+1)-a_n`

If the arithmetic sequence is 2x, 3x+1, x^2+2 , the common difference is:

`d= a_2-a_1 = 3x + 1 - 2x`

`d=x + 1`

And,

`d = a_3-a_2=x^2+2-(3x+1)`

`d=x^2-3x +1`

To solve for x,  set the two d's equal to each other.

`x^2-3x+1=x+1`

Express the equation in quadratic form `ax^2+bx+c=0` .

`x^2-3x+1-(x+1)=x+1-(x+1)`

`x^2-4x=0`

Factor left side.

`x(x-4)=0`

Set each factor equal to zero. And isolate x.

`x-4=0 `                and               `x=0`

`x=4`

Hence, there are two values of x that would give us an arithemtic sequence for the terms `2x` , `3x+1`   and  `x^2+2` . These are x=0 and x=4.

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