Paralyzing Progressions!

$From\quad AP\quad we\quad can\quad write,\\ (2x+1)\quad -\quad (x+y)\quad =\quad (3x-1)-(2x+1)\\ =>\quad x-y+1\quad =\quad x-2\\ =>\quad y\quad =\quad 1+2\\ So,\quad y\quad =\quad 3\\ \\ From\quad GP\quad we\quad can\quad write,\\ \frac { xy }{ x } \quad =\quad \frac { 3x+10y }{ xy } \\ =>\quad y\quad =\quad \frac { 3x+10y }{ xy } \\ =>\quad 3\quad =\quad \frac { 3x\quad +\quad 30 }{ 3x } \quad [since,\quad y=3]\\ =>\quad 3x\quad =\quad x\quad +\quad 10\\ =>\quad 2x\quad =\quad 10\\ So,\quad x\quad =\quad 5\\ \\ Now\quad Sum\quad of\quad x\quad \& \quad y\quad is,\\ x\quad +\quad y\quad =\quad 3\quad +\quad 5\quad =\quad 8$ .

Hello all,

as for arithmetic progression(AP), use common difference,

(2x+1) - (x+y) = (3x-1) - (2x+1)

x + 1 -y = x -1 -1

1-y = -2

y =1 + 2 = 3

as for geometric progression(GP),use common ratio,

xy / x = (3x+10y) / xy (1)

substitute y=3 into this (1),

3x / x = (3x + 30) / 3x

(3x)(3x) = x(3x + 30)

9x^2 = 3x^2 + 30x

6x^2 - 30x = 0

6x(x - 5) = 0

x=0, or x=5,

Therefore sum of values of x and y are = 0 + 5 + 3 =8,

Thanks...