Q. 11 D5.0( 2 Votes )

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

(a^{2} + b^{2} + c^{2}), (ab + bc + cd), (b^{2} + c^{2} + d^{2}) are in G.P.

Answer :

As,

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

Therefore,

b = ar … (1)

c = ar^{2} … (2)

d = ar^{3} … (3)

If we show that: (ab + bc + cd)^{2} = (a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2})

we can say that:

(a^{2} + b^{2} + c^{2}), (ab + bc + cd), (b^{2} + c^{2} + d^{2}) are in G.P

As, (ab + bc + cd)^{2} = (a^{2}r + a^{2}r^{3} + a^{2}r^{5})^{2}

⇒ (ab + bc + cd)^{2} = a^{4}r^{2}(1 + r^{2} + r^{4})^{2} …(4)

As,

(a^{2} + b^{2} + c^{2})( b^{2} + c^{2} + d^{2}) = (a^{2} + a^{2}r^{2} + a^{2}r^{4})(a^{2}r^{2} + a^{2}r^{4} + a^{2}r^{6})

⇒ (a^{2} + b^{2} + c^{2})( b^{2} + c^{2} + d^{2}) = a^{4}r^{2}(1 + r^{2} + r^{4})(1 + r^{2} + r^{4})

⇒ (a^{2} + b^{2} + c^{2})( b^{2} + c^{2} + d^{2}) = a^{4}r^{2}(1 + r^{2} + r^{4})^{2} …(5)

From equation 4 and 5, we have:

(ab + bc + cd)^{2} = (a^{2} + b^{2} + c^{2})(b^{2} + c^{2} + d^{2})

Hence,

We can say that (a^{2} + b^{2} + c^{2}), (ab + bc + cd), (b^{2} + c^{2} + d^{2}) are in G.P.

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