In an arithmetic sequence $$\{a_n\}$$, $$a_1=3$$, $$a_4=12$$. In a sequence $$\{b_n\}$$, $$b_1=4$$, $$b_4=20$$, and $$\{b_n-a_n\}$$ is a geometric sequence.

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Uniteasy Admin · Accepted Answer

Let $$d$$ as the common difference of the arithmetic sequence $$\{a_n\}$$ 
Using the formula for general terms of an arithmetic sequence,
$$a_4=a_1+(4-1)d$$
$$d=(a_4-a_)/3=(12-3)/3=3$$
Therefore,
$$a_n=a_1+(n-1)d$$
$$a_n=3+3n-3=3n$$
Let q as the common ratio of the geometric sequence $$\{b_n-a_n\}$$
$$q^3=\dfrac{b_4-a_4}{b_1-a_1} =\dfrac{20-12}{4-3}=8 $$
Therefore,
$$q=2$$
$$b_n-a_n=(b_1-a_1)q^{n-1}$$
$$b_n = 3n+2^{n-1}$$

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