Given a,b,c,d are in G.P, then $(a^{2} + a c + c^{2}) (b^{2} + b d + d^{2}) = (a b + b c + c d)^{2}$ If you think this is true write 1 otherwise 0 ?

Open in App

Solution

Since $a, b, c, d$ are in G.P.
We check
$(a^{2} + a c + c^{2}) (b^{2} + b d + d^{2}) = (a b + b c + c d)^{2}$
$b = a r, c = a r^{2}, d = a r^{3}$
L.H.S $= a^{2} (1 + r^{2} + r^{4}) a^{2} (r^{2} + r^{4} + r^{6}) = a^{4} r^{2} (1 + r^{2} + r^{4})^{2}$
R.H.S $= [a^{2} (r + r^{3} + r^{5})]^{2} = a^{4} r^{2} (1 + r^{2} + r^{4})^{2} = L . H . S$

Suggest Corrections

Similar questions

(a - d)^{2} = (b - c)^{2} + (c - a)^{2} + (d - b)^{2}

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

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

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

If a, b, c, d are in GP, prove that

$(a^{2} + b^{2} + c^{2}) (b^{2} + c^{2} + d^{2}) = (a b + b c + c d)^{2}$

Join BYJU'S Learning Program