#### Question

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

#### Solution

*a*, *b*, *c*, *d* are in G.P.

Therefore,

*bc* = *ad* … (1)

*b*^{2} = *ac *… (2)

*c*^{2} = *bd* … (3)

It has to be proved that,

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

R.H.S.

= (*ab* + *bc* + *cd*)^{2}

= (*ab* + *ad *+ *cd*)^{2} [Using (1)]

= [*ab* + *d* (*a* + *c*)]^{2}

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

= *a*^{2}*b*^{2} +2*a*^{2}*bd* + 2*acbd* + *d*^{2}(*a*^{2} + 2*ac* + *c*^{2})

= *a*^{2}*b*^{2} + 2*a*^{2}*c*^{2} + 2*b*^{2}*c*^{2} + *d*^{2}*a*^{2} + 2*d*^{2}*b*^{2} + *d*^{2}*c*^{2} [Using (1) and (2)]

= *a*^{2}*b*^{2} + *a*^{2}*c*^{2} + *a*^{2}*c*^{2} + *b*^{2}*c*^{2 }+ *b*^{2}*c*^{2} + *d*^{2}*a*^{2} + *d*^{2}*b*^{2} + *d*^{2}*b*^{2} + *d*^{2}*c*^{2}

= *a*^{2}*b*^{2} + *a*^{2}*c*^{2} + *a*^{2}*d*^{2 }+ *b*^{2 }× *b*^{2} + *b*^{2}*c*^{2} + *b*^{2}*d*^{2} + *c*^{2}*b*^{2} + *c*^{2 }× *c*^{2} + *c*^{2}*d*^{2}

[Using (2) and (3) and rearranging terms]

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

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

= L.H.S.

∴ L.H.S. = R.H.S.