Correct Answer - B::C::D
We have, length of a side of `S_(n)` is equal to the length of a diagonal of `S_(n+1)`. Hence,
Length of a side of `S_(n)=sqrt2` (Length of a side of `S_(n+1)`)
or `("Length of a side of" S_(n+1))/("Length of side" S_(n))=1/sqrt2`, for all `nge1`
Hence, sides of `S_(1),S_(2),..,S_(n)` form a G.P. with common ratio `1//sqrt2` and first term 10.
`therefore` Side of `S_(n)=10(1/(sqrt2))^(n-1)=10/(2^((n-1)/2))`
`rArr` Area of `S_(n)=("side")^(2)=(10/(2^((n-1)/2)))^(2)=100/(2^(n-1))`
Now, area of `S_(n)lt1`
`rArr2^(n-1)gt100`
`rArrngt8`