Question

Q 8

Answer 1

Let r be the common ratio of this G.P.
Given : a, b, c, d are in G.P.
⇒ 1^st = a,

2^nd term = b = ar,

3^rd term = c = ar²

4^th term = d = ar³

Now, `(1/("b"^2+"c"^2))^2=[1/(("ar")^2+("ar"^2)^2)]^2`

`=[1/("a"^2"r"^2+"a"^2"r"^4)]^2`

`=1/("a"^2"r"^4)[1/(1+"r"^2)]^2`

`=1/("a"^2"r"^4)xx1/(1+"r"^2)^2`

And, `(1/(a^2+b^2))xx(1/(c^2+d^2))=[1/(a^2+(ar)^2)]xx[1/((ar^2)^2+(ar^3)^2)]`

`=[1/(a^2+a^2r^2)]xx[1/(a^2r^4+a^2r^6)]`

`=1/a^2(1/(1+r^2))xx1/(a^2r^4)(1/(1+r^2))`

`=1/(a^4r^4)xx1/(1+r^2)^2`

`=>(1/(b^2+c^2))^2(1/(a^2+b^2))xx(1/(c^2+d^2))`

Hence, `1/(a^2+b^2),1/(b^2+c^2) "and" 1/(c^2+d^2)` are in G.P.