Q. 11 D5.0( 2 Votes )

# If a, b, c are in

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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