Question

The first three terms of a geometric progression are K-3,2K-4,4K-3 in that order find the value of K and the sum of the first 8 terms of the progression

Answer 1

If geometric, then
(2k-4)/(k-3) = (4k-3)/(2k-4)
(2k-4)^2 = (k-3)(4k-3)
4k^2 - 16k + 16 = 4k^2 -15k + 9
-k = -7
k = 7

so the first 3 terms are: 4, 10, 25

a = 4, r = 2.5
sum(8) = a(r^8 - 1)/(r-1)
= 4(2.5^8 - 1)/(2.5-1)
= 4(390625/256 - 1)/(3/2)
= 4(390369/256)(2/3)
=4(130123/128)
= 130123/32 or 4066.34375

the actual first 8 terms are:
4, 10, 25, 62.5, 156.25, 390.625, 976.5625, 2441.40625
sum = 4066.634375
Yeahh

Answer 2

k-3 = a
r a = 2k-4
r^2 a = 4k-3
-------------------
r= (4k-3)/(2k-4) = (2k-4)/(k-3)
so
(2k-4)^2= (k-3)(4k-3)

4 k^2 -16 k + 16=4 k^2 - 15k +9
-k = -7
k = 7
go back and find a and r