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)
11TH•Mathematics•Sequences And Series•Question 3
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The $n^{th}$ terms of the sequences $1,\ 2,\ 4,\ 8,\ \ldots$ and $256,\ 128,\ 64,\ \ldots$ are equal. Find the value of $n$.
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