Question

ANSWER FAST ASAP!!!!

BY PROVING BOTH SIDES...

Answer 1

Given that a,b,c,d are in GP such that.

Let the four terms of the GP be a,ar,ar^2,ar^3.

Here,

a = a, b = ar, c = ar^2, d = ar^3.

LHS:

= > (a^2 + b^2 + c^2)(b^2 + c^2 + d^2)

= > (a^2 + a^2r^2 + a^2r^4)(a^2r^2 + a^2r^4 + a^2r^6)

= > (a^2(1 + r^2 + r^4))(a^2r^2(1 + r^2 + r^4))

= > (a^2)(a^2r^2)(1 + r^2r^4)(1 + r^2r^4)

= > (a^2)(a^2r^2)(1 + r^2r^4)^2

= > (a^4r^2)(1 + r^2r^4)^2

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RHS:

= > (ab + bc + cd)^2

= > (a^2r + a^2r^3 + a^2r^5)^2

= > (a^2r(1 + r^2 + r^4)^2)

= > a^4r^2(1 + r^2 + r^4)^2

LHS = RHS

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Hope this helps!

Answer 2

Let the common ration of terms be r ,

Then,

$a = a \\ \\ b = ar^{} \\ \\ c =ar^{2} \\ \\ d = ar^{3} \\ \\ L.H.S \to \\ \\ =(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) \\ \\ = (a^{2} + a^{2}r^{2} + a^{2}r^{4})(a^{2}r^{2} + a^{2}r^{6}) \\ \\ = a^{2}(1 + r^{2} + r^{4})a^{2}r{2}(1 + r^{2} + r^{4}) \\ \\ = a^{4}r^{2}(1 + r^{2} + r^{4})^{2}$

Now,

$R.H.S. \to (ab + bc + cd)^{2} \\ \\ = (a^{2}r + a^{2}r{3} + a^{2}r{5})^{2} \\ \\ = a^{4}r^{2}(1 + r^{2} + r^{4})^{2}$

Hence,

L.H.S = R.H.S