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If the first three terms of an arithmetic progression are (x+3). (3x-10) and (2x+10), find x.

Guest Feb 9, 2018
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 #1
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(x+3)+ (3x-10) + (2x+10) =6x + 3

Sum them up as an arithmetic series:

6x + 3=3/2 *[2*[x+3] + (3-1)*(2x - 13)], solve for x

 

Solve for x:

6 x + 3 = (3 (2 (x + 3) + 2 (2 x - 13)))/2

 

6 x + 3 = (3 (2 x + 6 + 2 (2 x - 13)))/2

 

6 x + 3 = (3 (4 x - 26 + 2 x + 6))/2

 

Grouping like terms, 4 x + 2 x - 26 + 6 = (2 x + 4 x) + (6 - 26):

6 x + 3 = (3 ((2 x + 4 x) + (6 - 26)))/2

 

6 x + 3 = (3 (6 x + (6 - 26)))/2

 

6 x + 3 = (3 (6 x + -20))/2

 

Multiply both sides by 2:

2 (6 x + 3) = (2×3 (6 x - 20))/2

 

(2×3 (6 x - 20))/2 = 2/2×3 (6 x - 20) = 3 (6 x - 20):

2 (6 x + 3) = 3 (6 x - 20)

Expand out terms of the left hand side:

12 x + 6 = 3 (6 x - 20)

 

Expand out terms of the right hand side:

12 x + 6 = 18 x - 60

12x - 18x = -60 - 6

- 6x = - 66

x = -66 / -6

x= 11

 

 

 

 

 

 

 

 

 

 

 

Guest Feb 9, 2018
edited by Guest  Feb 10, 2018
 #2
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+1

We have that :

 

(x + 3)  + d  =  (3x - 10)       (1)

 

(3x - 10) + d  = (2x + 10)     (2)

 

Subtract  (2)  from  (1)   and we have

 

( x + 3)  - (3x - 10)   = (3x - 10) - (2x + 10)     simplify

 

-2x + 13   =  x - 20      add 2x, 20 to both sides

 

33   =  3x       divide both sides by 3

 

11  = x

 

 

cool cool cool

CPhill  Feb 10, 2018

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