The sum of first four terms is a G.P. is 30 and that of the last four terms is 960. Is its term is 2, find the common ratio.

Answer 1

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.

Answer 2

who says there are 8 terms? Now that we have
a=2
r=2
if there are n terms, then we know that
2(2^n-1)/(2-1) - 2(2^(n-4)-1)/(2-1) = 960
2^n - 1 - 2^(n-4) + 1 = 480
2^n - 2^n/16 = 480
15/16 * 2^n = 480
2^n = 512
n = 9
so the sequence is
2, 4, 8, 16, 32, 64, 128, 256, 512
and the sum of the last 4 terms is 960

Answer 3

misread the question.
Interpreted the question as if there were 8 terms, but it simply said that the last 4 terms had a sum of 960
e.g. There could have been 20 terms.

However .....
suppose there were n+4 terms
we found from the first part, that a = 2, and r = 2
then
Sum(last 4 terms) = sum(n+4) - sum(n) = 960
2(2^(n+4) - 1)/(2-1) - 2(2^n - 1)/(2-1) = 960
2^(n+4) - 1 - 2^n + 1 = 480
2^n(2^4 - 1) = 480
2^n = 32
n = 5

So the sequence consists of 9 terms, (not asked for just clearing up my problem) Everything checks out

Answer 4

is there any alternative solution?? except this