2016/110) The sum of 3 numbers of GP is 42. if the 1st 2 number are increased by 2 and 3rd decreased by 4 the resultant of an AP. find the numbers

let the 1st number be a ratio be r
we have 3 numbers $a, ar, ar^2$
the sum = $a + ar + ar^2 = 42\cdots(1)$
from the given condition $(a+2), (ar+2)$ and $(ar^2-4)$ form ap or $2(ar+2) = a +2 + ar^2 - 4$
or $2ar + 4 = ar^2 + a - 2$
or $ar^2 -2ar + a - 6 = 0\cdots(2)$
from (2) we have $ar^2 = 2ar - a +  6$
putting in (1) we get
$a+ ar + 2ar - a +  6 = 42$
o $3ar = 36$
or $ar = 12$
putting above in (1) we get $2(1+r^2) = 5r$
solving this we get $r = 2$ or $r = \frac{1}{2}$
so a= 6(for r =2) or 24 for $r = \frac{1}{2}$
so the numbers are $(24,12,6)$ or $(6,12,24)$