It is given that 3^x , 2^x+1, 4^x-1 are the first three terms of a geometric progression.
(i) Find, to 3 significant figures, the value of x .
(ii) Find, to the nearest integer, the sum to infinity of the progression.
Assuming the usual carelessness with parentheses,
4^(x-1)/2^(x+1) = 2^(2x-2)/2^(x+1) = 2^(x-3)
so now we have
2^(x+1)/3^x = 2^(x-3)
2^4 = 3^x
x = log16/log3 = 2.523719
so r = 2^(-0.476281) = 0.7188
a/(1-r) = 3^2.527319/0.2812 = 57.1245