# Given the terms of a arithmetic sequence a,b,c, verify if 3(a^2+b^2+c^2)=6(a-b)^2+(a+b+c)^2?

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b= a+ h

c = a + 2h

Left Hand Side = 3(a^2 +(a+h)^2 +(a+ 2h)^2)

= 3(a^2 +(a^2 +2ah + h^2) +(a^2 + 4ah +4h^2))

= 3(3a^2 + 6ah + 5h^2)

Right Hand Side = 6(a-b)^2+(a+b+c)^2

= 6 h ^2 +( a + a+h +a +2h)^2

= 6 h^2 +( 3a+3h) ^2

= 6 h^2 +9(a^2+ 2ah +h^2)

=3( 2h^2 + 3a^2 +6ah + 3h^2)

=3(3a^2 + 6ah + 5h^2)

Therefore LHS = RHS verified.

We notice that if we'll raise to square the sum a+b+c, we'll get:

`(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + ac + bc)`

The identity to be verified is:

`3(a^2+b^2+c^2)=6(a-b)^2+(a+b+c)^2`

`3(a^2+b^2+c^2) = 6(a-b)^2+a^2 + b^2 + c^2` + 2(ab + ac + bc)

`3(a^2+b^2+c^2) - (a^2 + b^2 + c^2) = 6(a-b)^2` + 2(ab + ac + bc)

`2(a^2+b^2+c^2) = 6(a-b)^2` + 2(ab + ac + bc)

We'll divide by 2 both sides:

`(a^2+b^2+c^2) = 3(a-b)^2` + (ab + ac + bc)

We'll expand the binomial from the right side:

`(a^2+b^2+c^2) = 3a^2 - 6ab + 3b^2` + (ab + ac + bc)

`-2a^2 - 2b^2+ c^2` = (ab + ac + bc) - 6ab

But b = `(a+c)/2`

2b = a + c

If we'll raise to square both sides, we'll get:

`4b^2 = a^2 + 2ac + c^2 =gt 2ac = 4b^2 - a^2 - c^2`

`-2a^2 - 2b^2+ c^2 = b(a+c) + ac - 6ab`

`-2a^2 - 2b^2+ c^2 = 2b^2 + (4b^2 - a^2 - c^2)/2 - 6ab`

`-2a^2 - 2b^2+ c^2 = 4b^2 - a^2/2 - c^2/2 - 6ab`

`-6b^2 = 3a^2/2 + 3c^2/2 - 6ab`

-`12b^2 = 3(a^2 + c^2) - 12ab`

`4b^2 = 4ab - a^2 - c^2`

`b = a+d =gt b^2 = a^2 + 2ad + d^2`

`c = a+2d =gt c^2 = a^2 + 4ad + 4d^2`

`ab = a^2 + ad`

`4a^2 + 8ad + 4d^2 = 4a^2 + 4ad- a^2 - a^2- 4ad` -`4d^2`

8ad + `4d^2 ` = - `2a^2` - `4d^2 8d^2 = -2a^2 - 8ad`

**We notice that the given expression does not represent an identity since `8d^2` `!=` `-2a^2 - 8ad` , if a,b,c are the consecutive terms of a arithmetical progression.**