Let A 1 , A 2 , A 3 , … . . be squares such that for each n ⩾ 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 the smallest value of n for which area of A n is less than one, is

Step-by-step solution

Let a n be the side length of A n . So, a n = 2 a n + 1 , a 1 = 12 ⇒   a n = 12 × 1 2 n - 1 Now, a n 2 < 1 ⇒ 144 2 n - 1 < 1 ⇒   2 n - 1 > 144 ⇒   n - 1 ≥ 8 ⇒   n ≥ 9