If a, b, c are in G.P., prove that : \(\frac{1}{a^2 + b^2}, \frac{1}{

1.	If a, b, c are in G.P., prove that : \(\frac{1}{a^2 + b^2}, \frac{1}{b^2 +c^2}, \frac{1}{c^2 + d^2}\) are in G.P
Answer» a, b, c, d are in G.P. Therefore, bc = ad … (1) b² = ac … (2) c²= bd … (3) To prove: \(\frac{1}{a^2 + b^2}\), \(\frac{1}{b^2 +c^2},\)\(\frac{1}{c^2 + d^2}\) are in G.P, we need to prove that: \(\frac{1}{(c^2 + d^2)(a^2 + b^2)}\) = \(\frac{1}{(b^2 + c^2)^2}\) {deduced using GM relation} Or, (b² + c²)² = (a² + b²)(c² + d²) Take LHS and proceed to prove LHS = (b² + c²)² = b⁴ + c⁴ + 2b²c² = a²c² + b²d² + a²d² + b²c² {using equation 2 and 3} = c²(a ² + b²) + d²(a² + b²) = (a² + b²) (c² + d²) = RHS \(\therefore\) \(\frac{1}{a^2 + b^2}\), \(\frac{1}{b^2 +c^2},\)\(\frac{1}{c^2 + d^2}\) are in GP Hence proved

If a, b, c are in G.P., prove that : \(\frac{1}{a^2 + b^2}, \frac{1}{b^2 +c^2}, \frac{1}{c^2 + d^2}\) are in G.P

Answer»

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

bc = ad … (1)

b² = ac … (2)

c²= bd … (3)

To prove: \(\frac{1}{a^2 + b^2}\), \(\frac{1}{b^2 +c^2},\)\(\frac{1}{c^2 + d^2}\) are in G.P, we need to prove that:

\(\frac{1}{(c^2 + d^2)(a^2 + b^2)}\) = \(\frac{1}{(b^2 + c^2)^2}\) {deduced using GM relation}

Or, (b² + c²)² = (a² + b²)(c² + d²)

Take LHS and proceed to prove

LHS = (b² + c²)²

= b⁴ + c⁴ + 2b²c²

= a²c² + b²d² + a²d² + b²c² {using equation 2 and 3}

= c²(a ² + b²) + d²(a² + b²)

= (a² + b²) (c² + d²) = RHS

\(\therefore\) \(\frac{1}{a^2 + b^2}\), \(\frac{1}{b^2 +c^2},\)\(\frac{1}{c^2 + d^2}\) are in GP

Hence proved