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State True or False. If $a$,$b$,$c$ are in A.P., then $b + c$,$c + a$,$a + b$ are also in A.P.
$A$. $True$

Last updated date: 17th Apr 2024
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Hint: The above problem is related to arithmetic progression. First, we should be able to define an arithmetic sequence. After that its properties can be used to solve the above problem.

Arithmetic Progression is a sequence of numbers such that the difference of any two consecutive numbers is constant.
General ${n^{th}}$ term of an A.P. is given by: ${a_n} = a + \left( {n - 1} \right)d$
Where $a$ is the first term of the A.P. sequence and $d$ is known as the common difference.
Given in the problem, $a$,$b$,$c$ are in A.P.
$ \Rightarrow $First term $ = a$
$ \Rightarrow $Second term $ = a + d = b$
$ \Rightarrow $Third term $ = a + 2d = c$
$ \Rightarrow b - a = c - b = d$ ……………………………….. (1)
Here $d$ is the common difference.
Since we need to check whether $b + c$,$c + a$,$a + b$ are in A.P. or not.
We need to find the common difference of the consecutive terms.
$ \Rightarrow c + a - \left( {b + c} \right) = c + a - b - c = a - b$
Using equation (1) in above, we get
$c + a - \left( {b + c} \right) = a - b = - d$ ………………………….(2)
$a + b - \left( {c + a} \right) = a + b - c - a = b - c$
$ \Rightarrow a + b - \left( {c + a} \right) = b - c = - d$ ………………...(3)
From (2) and (3) , the common difference of the consecutive terms is equal.
$ \Rightarrow $$b + c$,$c + a$,$a + b$ are in A.P.
Hence option $(A)$ is the correct answer.

Note: While solving problems related to A. P like above, always try to first write the general term of the A.P. sequence. It is to be observed that if each term of an A.P. is added with a fixed number, then the new obtained sequence is also in A.P.
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