First G.P.: (1,2,4,8,\ldots) → (a_n=2^{n-1})
Second: (256,128,64,\ldots) → (a_n=256\cdot\left(\dfrac{1}{2}\right)^{n-1}=2^{8-(n-1)}=2^{9-n})
[ 2^{n-1}=2^{9-n}\implies n-1=9-n\implies 2n=10\implies n=5 ]
Answer: (n=5)
First G.P.: (1,2,4,8,\ldots) → (a_n=2^{n-1})
Second: (256,128,64,\ldots) → (a_n=256\cdot\left(\dfrac{1}{2}\right)^{n-1}=2^{8-(n-1)}=2^{9-n})
[ 2^{n-1}=2^{9-n}\implies n-1=9-n\implies 2n=10\implies n=5 ]
Answer: (n=5)