If a, b, c are in AP, show that

Solution:

(i) $(b+c-a),(c+a-b),(a+b-c)$ are in AP.

To prove: $(b+c-a),(c+a-b),(a+b-c)$ are in AP.

Given: a, b, c are in A.P.

Proof: Let d be the common difference for the A.P. a,b,c

Since a, b, c are in A.P

$\Rightarrow b-a=c-b=$ common difference

$\Rightarrow a-b=b-c=d$

$\Rightarrow 2(a-b)=2(b-c)=2 d \ldots$ (i)

Considering series $(b+c-a),(c+a-b),(a+b-c)$

For numbers to be in A.P. there must be a common difference between them

Taking (b + c – a) and (c + a – b)

Common Difference $=(c+a-b)-(b+c-a)$

$=c+a-b-b-c+a$

$=2 a-2 b$

$=2(a-b)$

$=2 \mathrm{~d}[$ from eqn. (i) $]$

Taking (c + a – b) and (a + b – c)

Common Difference $=(a+b-c)-(c+a-b)$

$=a+b-c-c-a+b$

$=2 b-2 c$

$=2(b-c)$

$=2 d$ [from eqn. (i)]

Here we can see that we have obtained a common difference between numbers i.e. 2d

Hence, $(b+c-a),(c+a-b),(a+b-c)$ are in AP.

(ii) $\left(b c-a^{2}\right),\left(c a-b^{2}\right),\left(a b-c^{2}\right)$ are in AP.

To prove: $\left(b c-a^{2}\right),\left(c a-b^{2}\right),\left(a b-c^{2}\right)$ are in AP.

Given: $a, b, c$ are in A.P.

Proof: Let d be the common difference for the A.P. a,b,c

Since a, b, c are in A.P.

$\Rightarrow b-a=c-b=$ common difference

$\Rightarrow a-b=b-c=d \ldots$ (i)

Considering series $\left(b c-a^{2}\right),\left(c a-b^{2}\right),\left(a b-c^{2}\right)$

For numbers to be in A.P. there must be a common difference between them

Taking $\left(b c-a^{2}\right)$ and $\left(c a-b^{2}\right)$

Common Difference $=\left(c a-b^{2}\right)-\left(b c-a^{2}\right)$

$=\left[c a-b^{2}-b c+a^{2}\right]$

$=\left[c a-b c+a^{2}-b^{2}\right]$

$=[c(a-b)+(a+b)(a-b)]$

$=[(a-b)(a+b+c)]$

$a-b=d$, from eqn. (i)

$\Rightarrow[(d)(a+b+c)]$

Taking $\left(c a-b^{2}\right)$ and $\left(a b-c^{2}\right)$

Common Difference $=\left(a b-c^{2}\right)-\left(c a-b^{2}\right)$

$=\left[a b-c^{2}-c a+b^{2}\right]$

$=\left[a b-c a+b^{2}-c^{2}\right]$

$=[a(b-c)+(b-c)(b+c)]$

$=[(b-c)(a+b+c)]$

$b-c=d$, from eqn. (i)

$\Rightarrow[(d)(a+b+c)]$

Here we can see that we have obtained a common difference between numbers

i.e. [(d) $(a+b+c)$ ]

Hence, $\left(b c-a^{2}\right),\left(c a-b^{2}\right),\left(a b-c^{2}\right)$ are in AP.