Пусть наши член равны
![1.](https://tex.z-dn.net/?f=1.)
по первому условию , сумма равна
![\frac{a_{1}+a_{2}+...a_{13}}{a_{n-12}...+a_{n-1}+a_{n}}=0.5](https://tex.z-dn.net/?f=%5Cfrac%7Ba_%7B1%7D%2Ba_%7B2%7D%2B...a_%7B13%7D%7D%7Ba_%7Bn-12%7D...%2Ba_%7Bn-1%7D%2Ba_%7Bn%7D%7D%3D0.5)
это же условие можно переписать в виде
![S_{13}=(a_{1}+6d)*13 \\ ](https://tex.z-dn.net/?f=S_%7B13%7D%3D%28a_%7B1%7D%2B6d%29%2A13+%5C%5C%0A)
а последний 13 можно в виде
![S_{13}'=13(a_{1}+d(n-7))](https://tex.z-dn.net/?f=S_%7B13%7D%27%3D13%28a_%7B1%7D%2Bd%28n-7%29%29)
по условию следует что
![\frac{a_{1}+6d}{a_{1}+d(n-7)} = \frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7Ba_%7B1%7D%2B6d%7D%7Ba_%7B1%7D%2Bd%28n-7%29%7D+%3D+%5Cfrac%7B1%7D%7B2%7D)
![2.](https://tex.z-dn.net/?f=2.)
По второму условию задачи следует что
![S_{n}-(a_{1}+a_{2}+a_{3})](https://tex.z-dn.net/?f=S_%7Bn%7D-%28a_%7B1%7D%2Ba_%7B2%7D%2Ba_%7B3%7D%29)
ее можно переписать в виде
![\frac{2a_{1}+d(n-1)}{2}*n - (3a_{1}+3d)](https://tex.z-dn.net/?f=%5Cfrac%7B2a_%7B1%7D%2Bd%28n-1%29%7D%7B2%7D%2An+-+%283a_%7B1%7D%2B3d%29+++)
а последние без трех можно переписать в виде
![\frac{2a_{1}+d(n-1)}{2}*n-(3a_{1}+d(3n-6))](https://tex.z-dn.net/?f=+%5Cfrac%7B2a_%7B1%7D%2Bd%28n-1%29%7D%7B2%7D%2An-%283a_%7B1%7D%2Bd%283n-6%29%29)
заметим то что
![\frac{2a_{1}+d(n-1)}{2}*n - (3a_{1}+3d) = (\frac{n}{2}-\frac{3}{2})(dn+2d+2a_{1}) ](https://tex.z-dn.net/?f=%5Cfrac%7B2a_%7B1%7D%2Bd%28n-1%29%7D%7B2%7D%2An+-+%283a_%7B1%7D%2B3d%29+%3D+%28%5Cfrac%7Bn%7D%7B2%7D-%5Cfrac%7B3%7D%7B2%7D%29%28dn%2B2d%2B2a_%7B1%7D%29%0A)
![\frac{2a_{1}+d(n-1)}{2}*n-(3a_{1}+d(3n-6)) = (\frac{n}{2}-\frac{3}{2})(dn-4d+2a_{1})](https://tex.z-dn.net/?f=+%5Cfrac%7B2a_%7B1%7D%2Bd%28n-1%29%7D%7B2%7D%2An-%283a_%7B1%7D%2Bd%283n-6%29%29+%3D+%28%5Cfrac%7Bn%7D%7B2%7D-%5Cfrac%7B3%7D%7B2%7D%29%28dn-4d%2B2a_%7B1%7D%29)
по условию получаем
![\frac{dn+2d+2a_{1}}{dn-4d+2a_{1}}=\frac{4}{3}](https://tex.z-dn.net/?f=%5Cfrac%7Bdn%2B2d%2B2a_%7B1%7D%7D%7Bdn-4d%2B2a_%7B1%7D%7D%3D%5Cfrac%7B4%7D%7B3%7D+)
получаем систему уравнений
![\frac{a_{1}+6d}{a_{1}+d(n-7)} = \frac{1}{2}\\ \frac{dn+2d+2a_{1}}{dn-4d+2a_{1}}=\frac{4}{3}\\ \\ 2(a_{1}+6d)=a_{1}+dn-7d\\ 3(dn+2d+2a_{1})=4(dn-4d+2a_{1})\\ \\ a_{1}+19d=dn\\ 22d-2a_{1}=dn\\ \\ a_{1}+19d=22d-2a_{1}\\ 3a_{1}=3d\\ a_{1}=d\\ \\ \frac{7d}{d+dn-7d}=0.5\\ \frac{dn+4d}{dn-2d}=\frac{4}{3}\\\\ 7d=0.5d+0.5dn-3.5d\\ 3dn+12d=4dn-8d\\\\ n=20](https://tex.z-dn.net/?f=%5Cfrac%7Ba_%7B1%7D%2B6d%7D%7Ba_%7B1%7D%2Bd%28n-7%29%7D+%3D+%5Cfrac%7B1%7D%7B2%7D%5C%5C%0A%5Cfrac%7Bdn%2B2d%2B2a_%7B1%7D%7D%7Bdn-4d%2B2a_%7B1%7D%7D%3D%5Cfrac%7B4%7D%7B3%7D%5C%5C%0A%5C%5C%0A2%28a_%7B1%7D%2B6d%29%3Da_%7B1%7D%2Bdn-7d%5C%5C%0A3%28dn%2B2d%2B2a_%7B1%7D%29%3D4%28dn-4d%2B2a_%7B1%7D%29%5C%5C%0A%5C%5C%0Aa_%7B1%7D%2B19d%3Ddn%5C%5C%0A22d-2a_%7B1%7D%3Ddn%5C%5C%0A%5C%5C%0Aa_%7B1%7D%2B19d%3D22d-2a_%7B1%7D%5C%5C%0A3a_%7B1%7D%3D3d%5C%5C%0Aa_%7B1%7D%3Dd%5C%5C%0A%5C%5C%0A%5Cfrac%7B7d%7D%7Bd%2Bdn-7d%7D%3D0.5%5C%5C%0A%5Cfrac%7Bdn%2B4d%7D%7Bdn-2d%7D%3D%5Cfrac%7B4%7D%7B3%7D%5C%5C%5C%5C%0A7d%3D0.5d%2B0.5dn-3.5d%5C%5C%0A3dn%2B12d%3D4dn-8d%5C%5C%5C%5C%0An%3D20)
Ответ
![20](https://tex.z-dn.net/?f=20)