Question #336157

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?


1
Expert's answer
2022-05-04T17:34:58-0400
a=164,r=2a=\dfrac{1}{64}, r=2

Sn=a(1rn)1rS_n=\dfrac{a(1-r^n)}{1-r}

164(12n)12=220164\dfrac{\dfrac{1}{64}(1-2^n)}{1-2}=\dfrac{2^{20}-1}{64}

2n=2202^n=2^{20}

n=20n=20

a20=164(2)201=213=8192a_{20}=\dfrac{1}{64}(2)^{20-1}=2^{13}=8192


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