Chapter 12, Problem 32RE

To determine

Show that $a_1+b_1,a_2+b_2,a_3+b_3,\mathrm{.....}$ is also in arithmetic sequence if $a_1,a_2,a_3,\mathrm{.....}$ and $b_1,b_2,b_3,\mathrm{.....}$ are in arithmetic sequence

Answer to Problem 32RE

  $a_1+b_1,a_2+b_2,a_3+b_3,\mathrm{.....}$ is also an arithmetic sequence

Explanation of Solution

Given information:

It is given in the question that, $a_1,a_2,a_3,\mathrm{.....}$ and $b_1,b_2,b_3,\mathrm{.....}$ are in arithmetic sequence.

Formula used:

Common difference, $d\Rightarrow a_2-a_1=a_3-a_2$

Where,

  $a_{1}isthe1stterm$

  $a_{2}isthe2ndterm$

  $a_{3}isthe3rdterm$

Proof:

If $a_1,a_2,a_3,\mathrm{.....}$ and $b_1,b_2,b_3,\mathrm{.....}$ are in arithmetic sequence means the common difference of two consecutive term are same for both given arithmetic sequence that is

  $d\Rightarrow a_2-a_1=a_3-a_2$

  $a_2-a_1=a_3-a_2\mathrm{.......}(a)$

And,

  $d\Rightarrow b_2-b_1=b_3-b_2$

  $b_2-b_1=b_3-b_2\mathrm{.......}(b)$

Form the another given equation $a_1+b_1,a_2+b_2,a_3+b_3,\mathrm{.....}$

  $d\Rightarrow \left(a_2+b_2\right)-\left(a_1+b_1\right)=\left(a_3+b_3\right)-\left(a_2+b_2\right)$

  $\begin{array}{l}\Rightarrow \left(a_2+b_2\right)-\left(a_1+b_1\right)=\left(a_3+b_3\right)-\left(a_2+b_2\right)\\ \Rightarrow a_2+b_2-a_1-b_1=a_3+b_3-a_2-b_2\\ \Rightarrow a_2-a_1+b_2-b_1=a_3-a_2+b_3-b_2\\ \Rightarrow \left(a_2-a_1\right)+\left(b_2-b_1\right)=\left(a_3-a_2\right)+\left(b_3-b_2\right)\end{array}$

By using the equation $a_2-a_1=a_3-a_2\mathrm{.......}(a)$ and $b_2-b_1=b_3-b_2\mathrm{.......}(b)$

  $\Rightarrow \left(a_2-a_1\right)+\left(b_2-b_1\right)=\left(a_3-a_2\right)+\left(b_3-b_2\right)$

  $\Rightarrow \left(a_3-a_2\right)+\left(b_3-b_2\right)=\left(a_3-a_2\right)+\left(b_3-b_2\right)$

Here, the above equation is true so $a_1+b_1,a_2+b_2,a_3+b_3,\mathrm{.....}$ is also an arithmetic sequence