deiteresfp

2022-01-04

If ${a}_{1},{a}_{2},{a}_{3},\dots \phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{b}_{1},{b}_{2},{b}_{3},\dots$ are arithmetic sequences, show that ${a}_{1}+{b}_{1},{a}_{2}+{b}_{2},{a}_{3}+{b}_{3},\dots$ is also an arithmetic sequence.

esfloravaou

Given that the sequences ${a}_{1},{a}_{2},{a}_{3},\dots$ and ${b}_{1},{b}_{2},{b}_{3},\dots$ are arithmetic sequences.
Let the sequence ${a}_{1},{a}_{2},{a}_{3},\dots$ be an arithmetic sequence with common difference ${d}_{1}$ (the difference of two consecutive terms of thesequence).
The common difference ${d}_{1}$ of the sequence ${a}_{1},{a}_{2},{a}_{3},\dots$ can be calculated as ${d}_{1}={a}_{2}-{a}_{1}$ or ${d}_{1}={a}_{3}-{a}_{2}$
Let the sequence ${b}_{1},{b}_{2},{b}_{3},\dots$ be an arithmetic sequence with common difference ${d}_{2}$. It can be calculated as ${d}_{2}={b}_{2}-{b}_{1}$ or ${d}_{2}={b}_{3}-{b}_{2}$
Consider the sequence ${a}_{1}+{b}_{1},{a}_{2}+{b}_{2},{a}_{3}+{b}_{3},\dots$
Let $d$ be the common difference of this sequence. It can be calculated as $d=\left({a}_{2}+{b}_{2}\right)-\left({a}_{1}+{b}_{1}\right)=\left({a}_{2}-{a}_{2}\right)+\left({b}_{2}-{b}_{1}\right)={d}_{1}+{d}_{2}$.
Therefore the common difference of the sequence ${a}_{1}+{b}_{1},{a}_{2}+{b}_{2},{a}_{3}+{b}_{3},\dots$ can be written as $d={d}_{1}+{d}_{2}$
Since, the expression ${a}_{1}+{b}_{1},{a}_{2}+{b}_{2},{a}_{3}+{b}_{3},\dots$ has same common difference between each two consecutive terms. Hence, this expression is an arithmetic sequence if the sequence ${a}_{1},{a}_{2},{a}_{3},\dots$ and the sequence ${b}_{1},{b}_{2},{b}_{3},\dots$ are arithmetic sequences.

Do you have a similar question?