If the sum of first four terms in a G.P. is 30 and that of the last four terms is 960, then the sum of the first 8 terms is 990
We also know that a = 2
sum of first 4 terms
= 2(r^4 - 1)(r-1) = 30
r^4 - 1 = 15r - 15
r^4 - 15r + 14 = 0
r = 1 works for the equation , but r = 1 would make all the terms of the
GP the same, so clearly not a solution
r^4 - 15r + 14 = 0 , a bit of trial and error ....
(r-1)(r - 2)(r^2 + 3r + 7) = 0
so r = 2 or r = not real
check: sum(4) = 2(2^4 - 1)/(2-1) = 30 , as needed
so r = 2 , BUT it does not work for sum(8)
sum(8) = 2(2^8 - 1)(2-1) = 510 , not 990 as necessary according to your second part of the question.
check your question.