Answer :

(i) (b + c)(b + d) = (c + a)(c + a)


To prove: (b + c)(b + d) = (c + a)(c + a)


Given: a, b, c, d are in GP


Proof: When a,b,c,d are in GP then



From the above, we can have the following conclusion


bc = ad … (i)


b2 = ac … (ii)


c2 = bd … (iii)


Taking LHS = (b + c)(b + d)


= b2 + bd + bc + cd


Using eqn. (i) , (ii) and (iii)


= ac + c2 + ad + cd


= c(a + c) + d(a + c)


= (a + c) (c + d)


Hence Proved


(ii)


To prove:


Given: a, b, c, d are in GP


Proof: When a,b,c,d are in GP then



From the above, we can have the following conclusion


bc = ad … (i)


b2 = ac … (ii)


c2 = bd


d = … (iii)


Taking LHS =


= [From eqn. (iii)]


=


=


=


= [From eqn. (ii)]


=


=


=


=


= RHS


Hence Proved


(iii) (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2


To prove: (a + b + c + d)2 = (a + b)2 + 2(b + c)2 + (c + d)2


Given: a, b, c, d are in GP


Proof: When a,b,c,d are in GP then



From the above, we can have the following conclusion


bc = ad … (i)


b2 = ac … (ii)


c2 = bd … (iii)


Taking LHS = (a + b + c + d)2


(a + b + c + d) (a + b + c + d)


a2 + ab + ac + ad + ba + b2 + bc + bd + ca + cb + c2 + cd + da + db + dc + d2


On rearranging


a2 + ab + ba +b2 + ac + ad + bc + bd + ca + cb + c2 + cd + da + db + dc + d2


On rearranging


(a + b)2 + ac + ad + bc + bd + ca + cb + da + db + c2 + cd + dc + d2


On rearranging


(a + b)2 + ac + ad + bc + bd + ca + cb + da + db + (c + d)2


On rearranging


(a + b)2 + ac + ca + ad + bc + cb + da + bd + db + (c + d)2


Using eqn. (i)


(a + b)2 + ac + ca + bc + bc + bc + bc + bd + db + (c + d)2


Using eqn. (ii)


(a + b)2 + b2 + b2 + bc + bc + bc + bc + bd + db + (c + d)2


Using eqn. (iii)


(a + b)2 + 2b2 + 4bc + c2 + c2 + (c + d)2


On rearranging


(a + b)2 + 2b2 + 4bc + 2c2 + (c + d)2


(a + b)2 + 2[b2 + 2bc + c2 ] + (c + d)2


(a + b)2 + 2(b + c)2 + (c + d)2


= RHS


Hence proved


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