Let r be the common ratio of this G.P.Given : a, b, c, d are in G.P. ⇒ 1st = a,2nd term = b = ar, 3rd term = c = ar2 4th term = d = ar3 Now, (b2+c2) = [(ar)2 +(ar2)2]2= [a2r2 +a2r4]2= [a2r2 (1+r2)]2= a4r4 (1+r2)2And, (a2 + b2) x (c2 + d2) = [a2 + (ar)2] x [(ar2)2 + (ar3)2]= [a2 + a2r2] x [a2r4 + a2r6]= a2 (1 + r2) x a2r4(1 + r2)= a4r4(1+r2)2 ⇒ (b2 + c2)2 = (a2 + b2) x (c2 + d2) i.e. `("b"^2+"c"^2)/("a"^2+"b"^2)=("c"^2+"d"^2)/("b"^2+"c"^2)` Hence, (a2 + b2), (b2 + c2) and (c2 + d2) are in G.P.

English

Q 7 -

Question

Q 7

Sum

Solution

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, (b²+c²) = [(ar)² +(ar²)²]²
= [a²r² +a²r⁴]²
= [a²r² (1+r²)]²
= a⁴r⁴ (1+r²)²
And, (a² + b²) x (c² + d²) = [a² + (ar)²] x [(ar²)² + (ar³)²]
= [a² + a²r²] x [a²r⁴ + a²r⁶]
= a² (1 + r²) x a²r⁴(1 + r²)
= a⁴r⁴(1+r²)²