JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Sequences and Series

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.

(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✓ Correct
(D)STATEMENT-1 is False, STATEMENT-2 is True

Answer: (C)

Solution

Let

a_{1}, a_{2}, a_{3}, a_{4}

be in G.P. with common ratio

r

. Then

b_{1} = a_{1}

b_{2} = a_{1} + a_{1} r = a_{1} (1 + r)

b_{3} = a_{1} (1 + r + r^{2})

b_{4} = a_{1} (1 + r + r^{2} + r^{3})

. For A.P., we need

b_{2} - b_{1} = b_{3} - b_{2} = b_{4} - b_{3}

, which gives

a_{1} r = a_{1} r^{2} = a_{1} r^{3}

, impossible for distinct positive numbers. For G.P., we need

\frac{b _{2}}{b _{1}} = \frac{b _{3}}{b _{2}} = \frac{b _{4}}{b _{3}}

, which also fails. For H.P., we need

\frac{1}{b _{1}}, \frac{1}{b _{2}}, \frac{1}{b _{3}}, \frac{1}{b _{4}}

in A.P., which is not generally true. Hence STATEMENT-1 is true and STATEMENT-2 is false.

JEE Advanced · 2008 · 13 Apr 2008 · Paper 2 · Single correct option