Let A1, A2, A3, ....... be squares such that for each n \ge 1, the length of the side of An equals the length of diagonal of An+1. If the length of A1 is 12 cm, then the smallest value of n for which area of An is less than one, is \underline\hspace2cm .

Sequences and Series PYQ — JEE Main 2021 · 25 Feb · Shift 1 · Q41 with Solution

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Correct answer: 9

Interpret the condition

Let the side length of square $A_n$ be $s_n$ .

Given:

$s_1 = 12$
For each $n \ge 1$ , the side of $A_n$ equals the diagonal of $A_{n+1}$ .

Now, diagonal of a square of side $s$ is $s\sqrt{2}$ . So,

s_n = s_{n+1}\sqrt{2}

Hence,

s_{n+1} = \frac{s_n}{\sqrt{2}}

Thus, side lengths form a geometric progression:

s_n = 12\left(\frac{1}{\sqrt{2}}\right)^{n-1}

Find the area of $A_n$

Area of square $A_n$ is

\text{Area}(A_n) = s_n^2

So,

\text{Area}(A_n) = \left[12\left(\frac{1}{\sqrt{2}}\right)^{n-1}\right]^2

= 144\left(\frac{1}{2}\right)^{n-1}

We need the smallest $n$ such that

144\left(\frac{1}{2}\right)^{n-1} < 1

Solve the inequality

\frac{144}{2^{n-1}} < 1

144 < 2^{n-1}

Now check powers of $2$ :

$2^7 = 128 < 144$
$2^8 = 256 > 144$

So the smallest value satisfying $2^{n-1} > 144$ is

n-1 = 8

Therefore,

n = 9

Verification

At $n=8$ :

\text{Area}(A_8)=144\left(\frac12\right)^7=\frac{144}{128}=1.125>1

At $n=9$ :

\text{Area}(A_9)=144\left(\frac12\right)^8=\frac{144}{256}=0.5625<1

Hence the smallest such $n$ is indeed $9$ .

Final Answer: $9$

More from Sequences and Series

View written solutionFree

Correct answer: 9

Interpret the condition

Let the side length of square $A_n$ be $s_n$ .

Given:

$s_1 = 12$
For each $n \ge 1$ , the side of $A_n$ equals the diagonal of $A_{n+1}$ .

Now, diagonal of a square of side $s$ is $s\sqrt{2}$ . So,

s_n = s_{n+1}\sqrt{2}

Hence,

s_{n+1} = \frac{s_n}{\sqrt{2}}

Thus, side lengths form a geometric progression:

s_n = 12\left(\frac{1}{\sqrt{2}}\right)^{n-1}

Find the area of $A_n$

Area of square $A_n$ is

\text{Area}(A_n) = s_n^2

So,

\text{Area}(A_n) = \left[12\left(\frac{1}{\sqrt{2}}\right)^{n-1}\right]^2

= 144\left(\frac{1}{2}\right)^{n-1}

We need the smallest $n$ such that

144\left(\frac{1}{2}\right)^{n-1} < 1

Solve the inequality

\frac{144}{2^{n-1}} < 1

144 < 2^{n-1}

Now check powers of $2$ :

$2^7 = 128 < 144$
$2^8 = 256 > 144$

So the smallest value satisfying $2^{n-1} > 144$ is

n-1 = 8

Therefore,

n = 9

Verification

At $n=8$ :

\text{Area}(A_8)=144\left(\frac12\right)^7=\frac{144}{128}=1.125>1

At $n=9$ :

\text{Area}(A_9)=144\left(\frac12\right)^8=\frac{144}{256}=0.5625<1

Hence the smallest such $n$ is indeed $9$ .

Final Answer: $9$

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