Odd numbers take the form
, where
is an integer. When
, the last odd number would be 799. So we're adding
![S=1+3+5+\cdots+795+797+799](https://tex.z-dn.net/?f=S%3D1%2B3%2B5%2B%5Ccdots%2B795%2B797%2B799)
By reversing the order of terms, we have
![S=799+797+795+\cdots+5+3+1](https://tex.z-dn.net/?f=S%3D799%2B797%2B795%2B%5Ccdots%2B5%2B3%2B1)
and we can pair up terms in both sums at the same position to write
![2S=(1+799)+(3+797)+(5+795)+\cdots(795+5)+(797+3)+(799+1)](https://tex.z-dn.net/?f=2S%3D%281%2B799%29%2B%283%2B797%29%2B%285%2B795%29%2B%5Ccdots%28795%2B5%29%2B%28797%2B3%29%2B%28799%2B1%29)
so that we are basically adding 400 copies of 800, and from there we can find the value of the sum right away:
![2S=400\cdot800\implies S=160,000](https://tex.z-dn.net/?f=2S%3D400%5Ccdot800%5Cimplies%20S%3D160%2C000)
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We could also make use of the formulas,
![\displaystyle\sum_{i=1}^n1=n](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5En1%3Dn)
![\displaystyle\sum_{i=1}^ni=\dfrac{n(n+1)}2](https://tex.z-dn.net/?f=%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5Eni%3D%5Cdfrac%7Bn%28n%2B1%29%7D2)
We have
![S=\displaystyle\sum_{i=1}^{400}(2i-1)=2\sum_{i=1}^{400}i-\sum_{i=1}^{400}1=400(400+1)-400=400^2=160,000](https://tex.z-dn.net/?f=S%3D%5Cdisplaystyle%5Csum_%7Bi%3D1%7D%5E%7B400%7D%282i-1%29%3D2%5Csum_%7Bi%3D1%7D%5E%7B400%7Di-%5Csum_%7Bi%3D1%7D%5E%7B400%7D1%3D400%28400%2B1%29-400%3D400%5E2%3D160%2C000)
If PQ is 14, each segment, PR and RQ will be 7.
14 = PR + 7.
Approximately 21.22 yards.