If a, b, c are in A.P., then show that: b+c-a, c+a-b, a+b-c are in A.P.

T.P b+c-a, c+a-b, a+b-c are in A.P. b+c-c, c+a-b, a+b-c are in A.P only if (c+a-b)-(b+c-a)=(a+b-c)-(c+a-b) LHS=(c+a-b)-(b+c-a) 2a-2b .....(1) RHS=(a+b-c)-(c+a-b) 2b-2c .....(2) since, a, b, c are in A.P -a=c-b or a-b=b-c .....(3) From (1), (2) and (3) LHS = RHS Thus, given numbers b+c-a, c+a-b, a+b-c are in A.P

If a, b, c are in A.P., then show that: b+c-a, c+a-b, a+b-c are i…

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Question

If a, b, c are in A.P., then show that:
$\text{b}+\text{c}-\text{a},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P.

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Answer

T.P $\text{b}+\text{c}-\text{a},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P.
$\text{b}+\text{c}-\text{c},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P only if $(\text{c}+\text{a}-\text{b})-(\text{b}+\text{c}-\text{a})=(\text{a}+\text{b}-\text{c})-(\text{c}+\text{a}-\text{b})$
$\text{LHS}=(\text{c}+\text{a}-\text{b})-(\text{b}+\text{c}-\text{a})$
$\Rightarrow2\text{a}-2\text{b}\ .....(1)$
$\text{RHS}=(\text{a}+\text{b}-\text{c})-(\text{c}+\text{a}-\text{b})$
$\Rightarrow2\text{b}-\text{2c}\ .....(2)$
since,
$\text{a},\ \text{b},\ \text{c}$ are in A.P
$\therefore\text{b}-\text{a}=\text{c}-\text{b}$
or $\text{a}-\text{b}=\text{b}-\text{c}\ .....(3)$
From (1), (2) and (3)
LHS = RHS
Thus, given numbers
$\text{b}+\text{c}-\text{a},\ \text{c}+\text{a}-\text{b},\ \text{a}+\text{b}-\text{c}$ are in A.P