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.