If a, b, c are in G.P., prove that : (a2 + b2 + c2), (ab + bc + cd),

1.	If a, b, c are in G.P., prove that : (a2 + b2 + c2), (ab + bc + cd), (b2 + c2 + d2) are in G.P.
Answer» As, a, b, c, d are in G.P, let r be the common ratio. Therefore, b = ar … (1) c = ar² … (2) d = ar³ … (3) If we show that: (ab + bc + cd)² = (a² + b² + c²) (b² + c² + d²) we can say that: (a² + b² + c²), (ab + bc + cd), (b² + c² + d²) are in G.P As, (ab + bc + cd)² = (a²r + a²r³ + a²r⁵)² ⇒ (ab + bc + cd)² = a⁴r²(1 + r² + r⁴)² …(4) As, (a² + b² + c²)( b²+ c² + d²) = (a² + a²r² + a²r⁴)(a²r² + a²r⁴ + a²r⁶) ⇒ (a² + b² + c²)( b² + c² + d ²) = a⁴r²(1 + r² + r⁴)(1 + r² + r⁴) ⇒ (a² + b² + c²)( b² + c ² + d²) = a⁴r²(1 + r² + r⁴)² …(5) From equation 4 and 5, we have: (ab + bc + cd)² = (a² + b² + c²)(b² + c² + d²) Hence, We can say that (a² + b² + c²), (ab + bc + cd), (b² + c² + d²) are in G.P.

If a, b, c are in G.P., prove that : (a2 + b2 + c2), (ab + bc + cd), (b2 + c2 + d2) are in G.P.

Answer»

As,

a, b, c, d are in G.P, let r be the common ratio.

Therefore,

b = ar … (1)

c = ar² … (2)

d = ar³ … (3)

If we show that: (ab + bc + cd)² = (a² + b² + c²) (b² + c² + d²)

we can say that:

(a² + b² + c²), (ab + bc + cd), (b² + c² + d²) are in G.P

As, (ab + bc + cd)² = (a²r + a²r³ + a²r⁵)²

⇒ (ab + bc + cd)² = a⁴r²(1 + r² + r⁴)² …(4)

As,

(a² + b² + c²)( b²+ c² + d²) = (a² + a²r² + a²r⁴)(a²r² + a²r⁴ + a²r⁶)

⇒ (a² + b² + c²)( b² + c² + d ²) = a⁴r²(1 + r² + r⁴)(1 + r² + r⁴)

⇒ (a² + b² + c²)( b² + c ² + d²) = a⁴r²(1 + r² + r⁴)² …(5)

From equation 4 and 5, we have:

(ab + bc + cd)² = (a² + b² + c²)(b² + c² + d²)

Hence,

We can say that (a² + b² + c²), (ab + bc + cd), (b² + c² + d²) are in G.P.