JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Sequences and Series

Let

b_{i} > 1

for

i = 1, 2, \dots, 101

. Suppose

lo g_{e} b_{1}, lo g_{e} b_{2}, \dots, lo g_{e} b_{101}

are in Arithmetic Progression (A.P.) with the common difference

lo g_{e} 2

. Suppose

a_{1}, a_{2}, \dots, a_{101}

are in A.P. such that

a_{1} = b_{1}

and

a_{51} = b_{51}

. If

t = b_{1} + b_{2} + \dots + b_{51}

and

s = a_{1} + a_{2} + \dots + a_{51}

, then

Answer: (B)

Solution

Since

lo g_{e} b_{i}

are in A.P. with common difference

lo g_{e} 2

, the sequence

b_{i}

is a geometric progression with common ratio 2. The terms

a_{2}, a_{3}, \dots, a_{50}

are arithmetic means between

a_{1} = b_{1}

and

a_{51} = b_{51}

, while

b_{2}, b_{3}, \dots, b_{50}

are geometric means between the same endpoints. By the AM-GM inequality, arithmetic means exceed geometric means, so

a_{i} > b_{i}

for

i = 2, 3, \dots, 50

, giving

s > t

. For the 101st term:

a_{1}, a_{51}, a_{101}

form an A.P., while

b_{1}, b_{51}, b_{101}

form a G.P. Since

a_{1} = b_{1}

and

a_{51} = b_{51}

, and the G.P. grows faster than the A.P., we have

b_{101} > a_{101}

JEE Advanced · 2016 · 22 May 2016 · Paper 2 · Single correct option