(i) a(b2 + c2) = c(a2 + b2)
Given that a, b, c are in GP.
By using the property of geometric mean,
b2 = ac
Let us consider LHS: a(b2 + c2)
Now, substituting b2 = ac, we get
a(ac + c2)
a2c + ac2
c(a2 + ac)
Substitute ac = b2 we get,
c(a2 + b2) = RHS
∴ LHS = RHS
Hence proved.
(ii) a2b2c2 [1/a3 + 1/b3 + 1/c3] = a3 + b3 + c3
Given that a, b, c are in GP.
By using the property of geometric mean,
b2 = ac
Let us consider LHS: a2b2c2 [1/a3 + 1/b3 + 1/c3]
a2b2c2/a3 + a2b2c2/b3 + a2b2c2/c3
b2c2/a + a2c2/b + a2b2/c
(ac)c2/a + (b2)2/b + a2(ac)/c [by substituting the b2 = ac]
ac3/a + b4/b + a3c/c
c3 + b3 + a3 = RHS
∴ LHS = RHS
Hence proved.
(iii) (a + b + c)2 / (a2 + b2 + c2) = (a + b + c) / (a - b + c)
Given that a, b, c are in GP.
By using the property of geometric mean,
b2 = ac
Let us consider LHS: (a + b + c)2 / (a2 + b2 + c2)
(a + b + c)2 / (a2 + b2 + c2) = (a + b + c)2 / (a2 – b2 + c2 + 2b2)
= (a + b + c)2 / (a2 – b2 + c2 + 2ac) [Since, b2 = ac]
= (a + b + c)2 / (a + b + c)(a - b + c) [Since, (a + b + c)(a - b + c) = a2 – b2 + c2 + 2ac]
= (a + b + c) / (a - b + c)
= RHS
∴ LHS = RHS
Hence proved.
(iv) 1/(a2 – b2) + 1/b2 = 1/(b2 – c2)
Given that a, b, c are in GP.
By using the property of geometric mean,
b2 = ac
Let us consider LHS: 1/(a2 – b2) + 1/b2
Let us take LCM
1/(a2 – b2) + 1/b2 = (b2 + a2 – b2)/(a2 – b2)b2
= a2 / (a2b2 – b4)
= a2 / (a2b2 – (b2)2)
= a2 / (a2b2 – (ac)2) [Since, b2 = ac]
= a2 / (a2b2 – a2c2)
= a2 / a2(b2 – c2)
= 1/ (b2 – c2)
= RHS
∴ LHS = RHS
Hence proved.
(v) (a + 2b + 2c) (a – 2b + 2c) = a2 + 4c2
Given that a, b, c are in GP.
By using the property of geometric mean,
b2 = ac
Let us consider LHS: (a + 2b + 2c) (a – 2b + 2c)
Upon expansion we get,
(a + 2b + 2c) (a – 2b + 2c) = a2 – 2ab + 2ac + 2ab – 4b2 + 4bc + 2ac – 4bc + 4c2
= a2 + 4ac – 4b2 + 4c2
= a2 + 4ac – 4(ac) + 4c2 [Since, b2 = ac]
= a2 + 4c2
= RHS
∴ LHS = RHS
Hence proved.