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2022-01-04

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

esfloravaou

Beginner2022-01-05Added 43 answers

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 $d=({a}_{2}+{b}_{2})-({a}_{1}+{b}_{1})=({a}_{2}-{a}_{2})+({b}_{2}-{b}_{1})={d}_{1}+{d}_{2}$ .

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.

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