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GP a, b, c, d: (a^2+b^2+c^2)(b^2+c^2+d^2) Identity | JEE

JEE Maths question with a full step-by-step solution.

Question
If a,b,c,da, b, c, d are in G.P., then (a2+b2+c2)(b2+c2+d2)(a^2+b^2+c^2)(b^2+c^2+d^2) equals
Aab+bc+cdab+bc+cd
B(ab+bc+cd)2(ab+bc+cd)^2correct
C(ab+bc+cd)4(ab+bc+cd)^4
Dnone of these
Solution
Step 1: Take the terms as a,ar,ar2,ar3a, ar, ar^2, ar^3:
(a2+b2+c2)(b2+c2+d2)=a2(1+r2+r4)⋅a2r2(1+r2+r4)=a4r2(1+r2+r4)2.(a^2+b^2+c^2)(b^2+c^2+d^2)=a^2(1+r^2+r^4)\cdot a^2r^2(1+r^2+r^4)=a^4r^2(1+r^2+r^4)^2.
 Step 2: Also
ab+bc+cd=a2r+a2r3+a2r5=a2r(1+r2+r4) ⇒ (ab+bc+cd)2=a4r2(1+r2+r4)2.ab+bc+cd=a^2r+a^2r^3+a^2r^5=a^2r(1+r^2+r^4)\ \Rightarrow\ (ab+bc+cd)^2=a^4r^2(1+r^2+r^4)^2.
 Step 3: The two agree, so the product equals (ab+bc+cd)2.(ab+bc+cd)^2. Correct answer: (2)
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