JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Sequences and Series

Let

A_{1}, A_{2}, A_{3}, \dots

be squares such that for each

n \geq 1

, the length of the side of

A_{n}

equals the length of diagonal of

A_{n + 1}

. If the length of

A_{1}

is 12 cm then smallest value of

n

for which area of

A_{n}

is less than one, is ____

Answer: 9

Solution

a_{n}

is the side of

A_{n}

, then the side of

A_{n}

equals the diagonal of

A_{n + 1}

, so

a_{n} = a_{n + 1} 2

and the sides form a geometric progression with ratio

\frac{1}{2}

. With

a_{1} = 12

a_{n} = \frac{12}{( 2 ) ^{n - 1}}, a_{n}^{2} = \frac{144}{2 ^{n - 1}} .

This area falls below

1

when

2^{n - 1} > 144

, and since

2^{7} = 128

while

2^{8} = 256

, the least such value of

n - 1

8

. The smallest value of

n

9

JEE Main · 2021 · 25 Feb 2021 · Shift 1 · Integer answer