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

Solution:

(i) $(a-c)^{2}=4(a-b)(b-c)$

To prove: $(a-c)^{2}=4(a-b)(b-c)$

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

Proof: Since $a, b, c$ are in A.P.

$\Rightarrow \mathrm{c}-\mathrm{b}=\mathrm{b}-\mathrm{a}=$ common difference

$\Rightarrow \mathrm{b}-\mathrm{c}=\mathrm{a}-\mathrm{b} \ldots$ (i)

And, $2 b=a+c(a, b, c$ are in A.P. $)$

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

Taking LHS $=(a-c)^{2}$

$=(2 b-c-c)^{2}[$ from eqn. (ii) $]$

$=(2 b-2 c)^{2}$

$=4(b-c)^{2}$

$=4(b-c)(b-c)$

$=4(a-b)(b-c)[b-c=a-b$ from eqn. (i)]

= RHS

Hence Proved

(ii) $a^{2}+c^{2}+4 a c=2(a b+b c+c a)$

To prove: $a^{2}+c^{2}+4 a c=2(a b+b c+c a)$

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

Proof: Since $a, b, c$ are in A.P.

$\Rightarrow 2 b=a+c$

$\Rightarrow b=\frac{a+c}{2}$ … (i)

Taking RHS $=2(a b+b c+c a)$

Substituting value of b from eqn. (i)

$=2\left[\left\{a\left(\frac{a+c}{2}\right)\right\}+\left\{\left(\frac{a+c}{2}\right) c\right\}+\{c a\}\right]$

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

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

$=2\left[\frac{a^{2}+c^{2}+4 a c}{2}\right]$

$=a^{2}+c^{2}+4 a c$

= LHS

Hence Proved

(iii) $a^{3}+c^{3}+6 a b c=8 b^{3}$

To prove: $a^{3}+c^{3}+6 a b c=8 b^{3}$

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

Formula used: $(a+b)^{3}=a^{3}+3 a b(a+b)+b^{3}$

Proof: Since $a, b, c$ are in A.P.

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

Cubing both side,

$\Rightarrow(2 b)^{3}=(a+c)^{3}$

$\Rightarrow 8 b^{3}=a^{3}+3 a c(a+c)+c^{3}$

$\Rightarrow 8 b^{3}=a^{3}+3 a c(2 b)+c^{3}[a+c=2 b$ from eqn. (i)]

$\Rightarrow 8 b^{3}=a^{3}+6 a b c+c^{3}$

On rearranging,

$a^{3}+c^{3}+6 a b c=8 b^{3}$

Hence Proved