Jane started to knit a blanket. This blanket consists of knitted flowers. The diameter of one flower is 10 cm. The size of the blanket is 250cmX260cm. On the first day she knitted two flowers. How long does it take to knit the blanket (before combining the flowers) if she would always knit on the next day
a) two more flowers
This, also, is ambiguous. It could mean that she knitted two flowers a day- so she knitted "two more flowers" every day- or it could mean that she knitted "two more flowers"
than she had the previous day.
The blancket is 250 by 250 cm so 62500 square cm. The flowers are 10 cm in diameter, so 5 cm in radius so \(\displaystyle \pi(5)^2= 25\pi= 78.5 square centimeters. But, as others have pointed out, circles won't fill a square. There are two ways to handle that- (1) ignore the area about them- there are 62500/78.5= 795.8 or 796 roses or (2) assume the roses are actually knitted on 10 by 10 (100 square cm) squares so there are 62500/100= 625 squares. It is probably the latter that is intended.
If she knitted two roses a day, it would take 625/2= 312.5 days.
If she knitted two more roses
than the previous day the number or roses knitted is an "arithmetic series": 2+ 4+ 6+ 8+ ...= 2(1+ 2+ 3+ 4+...)
Notice 2 times the sum is the same as adding the sum to itself and that if we add (1+ 2+ 3+ 4+ ...+ (n-1)+ n) to the reversed sum (n+ (n-1)+ ... + 4+ 3+ 2+ 1) "term by term", each term adds to n+2: (n)+ (1), (n-1)+ 2, (n-2)+ 3, etc. are all equal to n+1. Since there are n terms, that is n(n+1) total roses knitted in n days.
So either n(n+1)= 625. That is the same as the quadratic equation \(\displaystyle n^2+ n- 625= 0\). That has two solutions but only one, 24.5 days, is positive.
b) two times more than in the previous day
This is a "geometric series": \(\displaystyle 2+ 4+ 6+ 8+ ...+ 2^n\) If we write \(\displaystyle S= 2+ 4+ 6+ 8+ ...+ 2^n\) then we can see that \(\displaystyle S- 2= 4+ 6+ 8+ ...+ 2^n= 2(2+ 3+ 4+ ...+ 2^{n-1})\). If we add \(\displaystyle 2^{n+1}= 2(2^n)\) to both sides, it becomes \(\displaystyle S- 2+ 2^{n+1}= 2(2+ 3+ 4+ ...+ 2^n)\) or \(\displaystyle S- 2+ 2^{n+1}= 2S. Subtracting S from both sides, \(\displaystyle S= 2^{n+1}- 2\).
That is, we must have \(\displaystyle 2^{n+1}- 2= 625\) or \(\displaystyle 2^{n+1}= 627\). Of course, \(\displaystyle 2^9= 512 and \(\displaystyle 2^10= 1024\) so this should require 10 days (9 days and a little work on the 10th day.)\)\)\)
