Q.297. Geometric Progression

Question 297.
If a, b, c, d are in GP, show that (a^2 + b^2 + c^2)(b^2 + c^2 + d^2) = (ab + bc + cd)^2.

Answer 297.
Let a, b, c, d be
m, mr, mr^2, mr^3

=> (a^2 + b^2 + c^2) ( b^2 + c^2 + d^2)
= (m^2 + m^2r^2 + m^2r^4) (m^2r^2 + m^2r^4 + m^2r^6)
= m^4r^2 (1 + r^2 + r^4) (1 + r^2 + r^4)
= m^4r^2(1 + r^2 + r^4)^2
= (m^2r + m^2r^3 + m^2r^5)^2
= [m*(mr) + (mr)*(mr^2) + (mr^2)(mr^3)]^2
= (ab + bc + ca)^2.

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