If a, b, c are in AP, prove that
(i) $(a-c)^{2}=4(a-b)(b-c)$
(ii) $a^{2}+c^{2}+4 a c=2(a b+b c+c a)$
(iii) $a^{3}+c^{3}+6 a b c=8 b^{3}$
(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
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