If a, b, c are in G.P., prove that :

(a2 + b2 + c2), (ab + bc + cd), (b2 + c2 + d2) are in G.P.

As,


a, b, c, d are in G.P, let r be the common ratio.


Therefore,


b = ar … (1)


c = ar2 … (2)


d = ar3 … (3)


If we show that: (ab + bc + cd)2 = (a2 + b2 + c2) (b2 + c2 + d2)


we can say that:


(a2 + b2 + c2), (ab + bc + cd), (b2 + c2 + d2) are in G.P


As, (ab + bc + cd)2 = (a2r + a2r3 + a2r5)2


(ab + bc + cd)2 = a4r2(1 + r2 + r4)2 …(4)


As,


(a2 + b2 + c2)( b2 + c2 + d2) = (a2 + a2r2 + a2r4)(a2r2 + a2r4 + a2r6)


(a2 + b2 + c2)( b2 + c2 + d2) = a4r2(1 + r2 + r4)(1 + r2 + r4)


(a2 + b2 + c2)( b2 + c2 + d2) = a4r2(1 + r2 + r4)2 …(5)


From equation 4 and 5, we have:


(ab + bc + cd)2 = (a2 + b2 + c2)(b2 + c2 + d2)


Hence,


We can say that (a2 + b2 + c2), (ab + bc + cd), (b2 + c2 + d2) are in G.P.


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