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
ANSWERED
asked 2021-08-05
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.

Expert Answers (1)

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 \(\displaystyle{10}^{{{d}}}\).
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