GP Caluculus
posted by Watermelon .
The third term of a geometric progression is 16. The sum of the third term and the fouth term is 8. Find
(a)the first term and the common ration of the progression
(b)the sum of infinity of the progression

Calculus, AP?
caluculus, fouth, ration, are not words normally used by Calculus students. Choosing words, and checking the spelling, reflects on you personally. If you desire success, work on these types of details. You can do better than this.
Given that a3 is 16, and a4 is 8, the ratio must be r=.5
a3=a0*r^(31)
16=a0*(1/2)^(31)or a0=16*4
check: 16*4, 16*2, 16*1, 16*1/2, ...
sum:
Sum= ao/(1r)=16*4/.5=16*8
check my thinking. 
this is not arithmetic progression = =

third term = ar^2 and ar^2 = 16
fourth term = ar^3
ar^2 + ar^3 = 8
16 + ar^3 = 8
ar^3 = 8
ar^3/(ar^2) = 8/16
r = 1/2
then a = 64
terms are: 64 32 16 8 4 ...
check: is term3 = 16 ? , YES
is 16  8 = 8 ? , yes
sum_{∞} = a/(1r) = 64/(3/2) = 128/3 
Thanks, R. I misread the statement, I some how read into the question the fourth term was 8. Duh on me.