Question -
Answer -
(i) a(b2 +c2) = c(a2 + b2)
Given that a, b, c arein GP.
By using the propertyof 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 weget,
c(a2 +b2) = RHS
∴ LHS = RHS
Hence proved.
(ii) a2b2c2 [1/a3 +1/b3 + 1/c3] = a3 + b3 +c3
Given that a, b, c arein GP.
By using the propertyof 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 arein GP.
By using the propertyof 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 arein GP.
By using the propertyof 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 arein GP.
By using the propertyof 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.