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