SOLUTION: Given that x+6,x and x-3 are the first three terms of geometric progression.calculate the volume of x?.(i)find the value of x. (ii)the fifth term .(iii) the sum of infinity

Question 1125984: Given that x+6,x and x-3 are the first three terms of geometric progression.calculate the volume of x?.(i)find the value of x. (ii)the fifth term .(iii) the sum of infinity

Answer by htmentor(1343) (Show Source): You can put this solution on YOUR website!
The n-th term of a geometric progression is given by a_n = a*r^(n-1)
where a = the first term, r = the common ratio
The first 3 terms are x+6, x, and x+6
The first term a = x+6
The second term is a_2 = ar = x
The third term is a_3 = ar^2 = x-3
The common ratio, r = x/(x+6)
We solve for x by eliminating a and r:
(x+6)r = x -> r = x/(x+6)
(x+6)(x/(x+6))^2 = x-3
x^2/(x+6) = x-3
x^2 = x^2 + 3x - 18
This gives x = 6, and thus a = 12, and r = 1/2
Thus the 5th term is a_5 = 12(1/2)^4 = 3/4
The infinite sum is S = a/(1-r) = 12/(1/2) = 24
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