Answer :
This forms an arithmetic sequence, where the first term = 1, common difference = 1, and number of terms = n.
The formula for the sum of an arithmetic sequence is
[tex]S_{n} = \frac{1}{2}n(2a + (n - 1)d) [/tex]
Where a = the first term and d = the common difference.
We want the sum to be at least $50, so
[tex]50 \leq \frac{1}{2}n(2a + (n - 1)d) [/tex]
Substituting in a and d
[tex]50 \leq \frac{1}{2}n(2(1) + (n - 1)(1)) [/tex]
Rearranging
[tex]50 \leq \frac{1}{2}n(2 + (n - 1)) [/tex]
[tex]50 \leq \frac{1}{2}n(1 + n) [/tex]
[tex]100 \leq n(1 + n) [/tex]
[tex]100 \leq n^2 + n[/tex]
[tex]0 \leq n^2 + n - 100 [/tex]
[tex]\text{Let } n^2 + n -100 = 0[/tex]
[tex]n = \frac{-1 \pm \sqrt{1-4(1)(-100)}}{2} [/tex]
[tex]n = \frac{-1 \pm \sqrt{401}}{2} [/tex]
[tex]n \geq 0 \implies n = \frac{-1 + \sqrt{401}}{2} [/tex]
[tex]n \approx 9.51[/tex]
So it will take 9.51 weeks, or 10 weeks to the nearest week.
The formula for the sum of an arithmetic sequence is
[tex]S_{n} = \frac{1}{2}n(2a + (n - 1)d) [/tex]
Where a = the first term and d = the common difference.
We want the sum to be at least $50, so
[tex]50 \leq \frac{1}{2}n(2a + (n - 1)d) [/tex]
Substituting in a and d
[tex]50 \leq \frac{1}{2}n(2(1) + (n - 1)(1)) [/tex]
Rearranging
[tex]50 \leq \frac{1}{2}n(2 + (n - 1)) [/tex]
[tex]50 \leq \frac{1}{2}n(1 + n) [/tex]
[tex]100 \leq n(1 + n) [/tex]
[tex]100 \leq n^2 + n[/tex]
[tex]0 \leq n^2 + n - 100 [/tex]
[tex]\text{Let } n^2 + n -100 = 0[/tex]
[tex]n = \frac{-1 \pm \sqrt{1-4(1)(-100)}}{2} [/tex]
[tex]n = \frac{-1 \pm \sqrt{401}}{2} [/tex]
[tex]n \geq 0 \implies n = \frac{-1 + \sqrt{401}}{2} [/tex]
[tex]n \approx 9.51[/tex]
So it will take 9.51 weeks, or 10 weeks to the nearest week.
