IIT-JEE 2008 · Paper 2 · Mathematics Q10

← IIT-JEE 2008 · Paper 2

MathematicsAssertion and reason+3 / -1Question 10 of 66

Suppose four distinct positive numbers $a_1,a_2,a_3,a_4$ are in G.P. Let $b_1=a_1$ , $b_2=b_1+a_2$ , $b_3=b_2+a_3$ and $b_4=b_3+a_4$ .

STATEMENT-1: The numbers $b_1,b_2,b_3,b_4$ are neither in A.P. nor in G.P.

and

STATEMENT-2: The numbers $b_1,b_2,b_3,b_4$ are in H.P.

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is True; STATEMENT-2 is NOT a correct explanation for STATEMENT-1

STATEMENT-1 is True, STATEMENT-2 is False

STATEMENT-1 is False, STATEMENT-2 is True

Answer and worked solution

Official answer: C

Let $a_k=ar^{k-1}$ with $r\ne1$ , $r>0$ . Then $b_1=a$ , $b_2=a(1+r)$ , $b_3=a(1+r+r^2)$ , $b_4=a(1+r+r^2+r^3)$ .

Statement 1. The differences $ar$ , $ar^2$ , $ar^3$ are unequal, so not an AP. Ratios $1+r$ and $\frac{1+r+r^2}{1+r}$ are unequal (their equality would need $r=0$ ), so not a GP. True.

Statement 2. For an HP, $\frac1{b_1}+\frac1{b_3}=\frac2{b_2}$ , i.e. $1+\frac1{1+r+r^2}=\frac2{1+r}$ , which reduces to $r^3=0$ . False.

(C)

Topics:Geometric progression

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IIT-JEE 2008 · Paper 2, Mathematics question 10