## Tuesday, 8 November 2011

### 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+a4.
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;
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.