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Let `ABCD` is a unit square and each side of the square is divided in the ratio `alpha : (1-alpha) (0 lt alpha lt 1)` . These points are connected to obtain another square. The sides of new square are divided in the ratio `alpha : (1-alpha)` and points are joined to obtain another square. The process is continued idefinitely. Let `a_(n)` denote the length of side and `A_(n)` the area of the `n^(th)` square
If`alpha=(1)/(3)`, then the least value of `n` for which `A_(n) gt (1)/(10)` is

Options

A. `4`
B. `5`
C. `6`
D. `7`

Answer

B

Explanation

`(b)`
Clearly from the diagram,
`a_(1)=1`
`(a_(n+1))^(2)=[(1-alpha)^(2)+alpha^(2)](a_(n))^(2)impliesA_(n+1)=[2alpha^(2)-2alpha+1]A_(n)`
For `alpha=(1)/(3)`, `(a_(n+1))^(2)=(5)/(9)(a_(n))^(2)`
`implies(A_(n+1))/(A_(n))=(5)/(9)`
`impliesA_(2)=(5)/(9)`, `A_(3)=((5)/(9))^(2)`,...
`impliesA_(n)=((5)/(9))^(n-1)`
`A_(n) lt (1)/(10) implies ((5)/(9))^(n-1) lt (1)/(10)`
`implies` Least value of `n` is `5`.
`sum_(n=1)^(oo)A_(n)=(8)/(3)`
`implies(1)/(1-(2alpha^(2)-2alpha+1))=(8)/(3)`
`implies16alpha^(2)-16alpha+3=0`
`implies16alpha^(2)-12alpha-4alpha+3=0`
`implies(4alpha-1)(4alpha-3)=0`
`impliesalpha=(1)/(4)`, `(3)/(4)`
The side of `n^(th)` square equal to the diagonal of `(n+1)^(th)` square. ,brgt `impliessqrt(2)a_(n)=a_(n+1)`
`implies((a_(n+1))^(2))/((a_(n))^(2))=(1)/(2)`
`implies4alpha^(2)-4alpha+1=0`
`impliesalpha=1//2`