Question

# Exponentials of an Arithmetic Sequence, If a_{1}, a_{2}, a_{3}, \cdots is an arithmetic sequence with common difference d, show that the sequence 10^{a_{1}}, 10^{a_{2}}, 10^{a_{3}}, \cdots is a geometric sequence, and find the common ratio.

Polynomial arithmetic
Exponentials of an Arithmetic Sequence, If $$\displaystyle{a}_{{{1}}},{a}_{{{2}}},{a}_{{{3}}},\cdots$$ is an arithmetic sequence with common difference d, show that the sequence $$\displaystyle{10}^{{{a}_{{{1}}}}},{10}^{{{a}_{{{2}}}}},{10}^{{{a}_{{{3}}}}},\cdots$$ is a geometric sequence, and find the common ratio.

2021-08-06
Given,
$$\displaystyle{a}_{{{1}}},{a}_{{{2}}},{a}_{{{3}}},\cdots$$ is a n arithmetic sequence with common difference d.
Now, $$\displaystyle{a}_{{{2}}}={a}_{{{1}}}+{d},{a}_{{{3}}}={a}_{{{1}}}+{2}{d}$$, ----- $$\displaystyle{\left[\therefore{a}_{{{n}}}={a}_{{{1}}}+{\left({n}-{1}\right)}{d}\right]}$$.
Now consider the sequence, $$\displaystyle{10}^{{{a}_{{{1}}}}},{10}^{{{a}_{{{2}}}}},{10}^{{{a}_{{{3}}}}},\cdots$$
Here, $$\displaystyle{\frac{{{10}^{{{a}_{{{2}}}}}}}{{{10}^{{{a}_{{{1}}}}}}}}={\frac{{{10}^{{{{a}_{{{2}}}^{{+{d}}}}}}}}{{{10}^{{{a}_{{{1}}}}}}}}$$
$$\displaystyle={10}^{{{d}}}$$
$$\displaystyle{\frac{{{10}^{{{a}_{{{3}}}}}}}{{{10}^{{{a}_{{{2}}}}}}}}={\frac{{{10}^{{{{a}_{{{1}}}^{{+{2}{d}}}}}}}}{{{10}^{{{{a}_{{{1}}}^{{+{d}}}}}}}}}$$
$$\displaystyle={\frac{{{10}^{{{2}{d}}}}}{{{10}^{{{d}}}}}}$$
$$\displaystyle={10}^{{{d}}}$$
Clearly,
$$\displaystyle{\frac{{\text{second term}}}{{\text{first term}}}}={\frac{{\text{third term}}}{{\text{second term}}}}={10}^{{{d}}}$$
Therefore $$\displaystyle{10}^{{{a}_{{{1}}}}},{10}^{{{a}_{{{2}}}}},{10}^{{{a}_{{{3}}}}},\cdots$$ is a geometric sequence.
Step 2
The common ratio is $$\displaystyle{10}^{{{d}}}$$.