Progressions-8 (Reasoning)(IIT-JEE 2008)

Suppose four distinct positive numbers a₁, a₂, a₃,a₄ are in G.P. Let b₁= a₁, b₂= b₁+a₂, b₃= b₂+a₃ and b₄ = b₃+a_4.

Statement-1: The numbers b₁, b₂, b₃,b₄ are neither in A.P. nor in G.P.

Statement-2: The numbers b₁, b₂, b₃,b₄ are in H.P.

(a) Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for statement-1;

(b)  Statement-1 is True, Statement-2 is True; Statement-2 is not a correct explanation for statement-1;

(c) Statement-1 is True, Statement-2 is False;

(d) Statement-1 is False, Statement-2 is True;

Answer follows here:

Solution:

As a₁, a₂, a₃,a₄ are in G.P., let a₁ = a, a₂ = ar, a₃ = ar², a₄ = ar³  

b₁= a₁ = a, b₂= b₁+a₂ = a+ar, b₃= b₂+a₃ = a+ar+ar², b₄= b₃+a₄ = a+ar+ar²+ar³

Let us check for A.P.:

b₂-b₁ = a+ar-a = ar,   b₃-b₂ = a+ar+ar²-a-ar = ar² => b₂-b₁ ≠ b₃-b₂ => b₁, b₂, b₃,b₄ are not in A.P.

Let us check for G.P.:

b₂/b₁ = (a+ar)/a = 1+r,         b₃/b₂ = (a+ar+ar²)/(a+ar) = (1+r+r²)/(1+r) => b₂/b₁ ≠ b₃/b₂ => b₁, b₂, b₃,b₄ are not in G.P.

Let us check for H.P.:

(1/b₂)-(1/b₁) = (1/a+ar)-(1/a) = -r/(a+ar),  (1/b₃)-(1/b₂) = (1/a+ar+ar²)-(1/a+ar) = -r²/a(1+r)(1+r+r²) => 1/b₁, 1/b₂, 1/b₃,1/b₄ are not in A.P. => b₁, b₂, b₃,b₄ are not in H.P.

Answer (c)