Answer to Question #336157 in Algebra for habibi

Question

Given geometric Progression 1/64,1/32,1/16,1/8,1/4,… how many terms such that Sn=2^20-1/64 find the last term sn?

Answer 1

$$a=\dfrac{1}{64}, r=2$$

$$S_n=\dfrac{a(1-r^n)}{1-r}$$

$$\dfrac{\dfrac{1}{64}(1-2^n)}{1-2}=\dfrac{2^{20}-1}{64}$$

$$2^n=2^{20}$$

$$n=20$$

$$a_{20}=\dfrac{1}{64}(2)^{20-1}=2^{13}=8192$$