Mértani sorozat

Legyen a két sorozat `a_1, a_2, a_3...` és `b_1, b_2, b_3...`. Ekkor
`a_1 = b_1`
`b_2-a_2 = 5`
`b_3-a_3 = -5/4`
`b_4-a_4 = 35/16`

Viszont tudjuk azt is, hogy bármilyen mértani sorozatra
`a_n = a_1*q^(n-1)`

Vagyis, mivel `a_1 = b_1`, ezért
`a_1*q_b-a_1*q_a = 5`
`"I. "a_1(q_b-q_a) = 5`

`a_1*q_b^2-a_1*q_a^2 = -5/4`
`a_1(q_b^2-q_a^2) = -5/4`
`"II. "a_1(q_b-q_a)(q_b+q_a) = -5/4`

`a_1*q_b^3-a_1*q_a^3 = 35/16`
`a_1(q_b^3-q_a^3) = 35/16`
`"III. "a_1(q_b-q_a)(q_b^2+q_bq_a+q_a^2) = 35/16`

Ha a `"II."`-at leosztjuk az `"I."`-vel:
`q_b+q_a = -1/4 => q_b = -q_a-1/4`

És ha a `"III."`-at leosztjuk az `"I."`-vel:
`q_b^2+q_bq_a+q_a^2 = 7/16`
`(-q_a-1/4)^2+(-q_a-1/4)q_a+q_a^2 = 7/16`
`q_a^2+1/2 q_a+1/16 cancel(-q_a^2)-1/4 q_a cancel(+q_a^2) = 7/16 " /"-7/16`
`q_a^2+1/4 q_a-3/8 = 0`
`q_(a1) = 1/2 ", " q_(a2) = -3/4`

`q_(b1) = -1/2-1/4 = -3/4 ", " q_(b2) = -(-3/4)-1/4 = 1/2`

`a_(11)(-3/4-1/2) = 5`
`-5/4 a_(11) = 5 " /"*-4/5`
`a_(11) = -4`

`a_(12)(1/2-(-3/4)) = 5`
`5/4 a_(12) = 5 " /"*4/5`
`a_(12) = 4`

Vagyis 2 ilyen sorozatpár van:
`"I. " a_1 = b_1 = -4 ", " q_a = 1/2 ", " q_b = -3/4`
`a_1 = -4 ", " a_2 = -2 ", " a_3 = -1 ", " a_4 = -1/2 ", "...`
`b_1 = -4 ", " b_2 = 3 ", " b_3 = -9/4 ", " b_4 = 27/16 ", " ...`

`"II. " a_1 = b_1 = 4 ", " q_a = -3/4 ", " q_b = 1/2`
`a_1 = 4 ", " a_2 = -3 ", " a_3 = 9/4 ", " a_4 = -27/16 ", "...`
`b_1 = 4 ", " b_2 = 2 ", " b_3 = 1 ", " b_4 = 1/2 ", " ...`