Answer:
Step-by-step explanation:
The sum of the 3rd and 8th term is:
[tex]a_3+a_8=1\\\\a_1+2d+a_1+7d=1\\\\\underline{2a_1+9d=1}[/tex]
The sum of the first seven terms is 35:
[tex]S_7=35\\\\\dfrac{a_1+a_7}2\cdot7=35\\{}\qquad\quad^{\div7\qquad\div7}\\ \dfrac{a_1+a_1+d}2=5\\{}\qquad\qquad^{\cdot2\qquad\cdot2}\\2a_1+6d=10\\{}\qquad\ ^{\div2\qquad\div2}\\a_1+3d=5\\{}\quad\ ^{-3d\quad\ -3d}\\a_1=5-3d[/tex]
[tex]2a_1+9d=1\\\\2(5-3d)+9d=1\\\\10-6d+9d=1\\{}\qquad\qquad^{-10\quad-10}\\{}\qquad3d=-9\\{}\qquad^{\div3\qquad\div3}\\{}\qquad d=-3\\\\\\a_1=5-3(-3)=5+9=14[/tex]
