Progressions Part 3

As $(a)$ , $(b)$ and $(c)$ are in an arithmetic progression, we can say that $a=a$ , $b=a+d$ and $c=a+2d$

Let's first solve $(a)$ ,

$LHS$ , $=(a^{2}+c^{2}+4ac)$

$={ a }^{ 2 }+\left( a+2d \right) ^{ 2 }+\left[ \left( a \right) \left( a+2d \right) \right]$

$={ a }^{ 2 }+a^{ 2 }+4{ d }^{ 2 }+4ad+4{ a }^{ 2 }+8ad$

$=6{ a }^{ 2 }+4{ d }^{ 2 }+12ad$

$RHS$ , $2(ab+bc+ca)$

$=2\left[ a\left( a+d \right) +\left( a+d \right) \left( a+2d \right) +\left( a+2d \right) a \right]$

$=2\left( 3{ a }^{ 2 }+6ad+2{ d }^{ 2 } \right)$

$=3{ a }^{ 2 }+6ad+2{ d }^{ 2 }$

Hence as $LHS=RHS$ $(a)$ is true.

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Now, let's take $(b)$ ,

$LHS$ , $(a-c)^{2}$

$={ \left[ a-\left( a+2d \right) \right] }^{ 2 }$

$={ { \left[ a-a-2d \right] } }^{ 2 }$

$=4{ d }^{ 2 }$

$RHS$ , $4(a-b)(b-c)$

$=4\left\{ \left[ a-\left( a-d \right) \right] \left[ \left( a+d \right) -\left( a+2d \right) \right] \right\}$

$=4\left[ \left( -d \right) \left( -d \right) \right]$

$=4{ d }^{ 2 }$

Hence as $LHS=RHS$ $(b)$ is also true.

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Therefore, both the conditions are true!