Let S1, S2,….be squares such that for each n≥1, the length of a side of Sn equals the length of adiagonal of Sn+1. If the length of a side of S1 is 10 cm, then the smallest value of n for which Area(Sn) <1 is

# Let be squares such that for each $n\ge 1,$ the length of a side of ${S}_{n}$ equals the length of adiagonal of ${S}_{n+1}.$ If the length of a side of ${S}_{1}$ is 10 cm, then the smallest value of $n$ for which Area(${S}_{n}$) $<1$ is

1. A

7

2. B

8

3. C

9

4. D

10

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### Solution:

Let ${a}_{n}$ denote the length of a side of ${S}_{n}.$ We are given

Length of a side of ${S}_{n}$ = Length of a diagonal of ${S}_{n+1}$

Thus, is a G.P. with first term 10 and common ratio $1/\sqrt{2}.$

Therefore,

${a}_{n}=10\left(1/\sqrt{2}{\right)}^{n-1}$

Also, Area $\left({S}_{n}\right)={a}_{n}^{2}<1$

$⇒{\left[10\left(1/\sqrt{2}{\right)}^{n-1}\right]}^{2}<1⇒100<{2}^{n-1}$

$⇒$smallest possible value of $n$ is 8.  