Hello,
[tex]t_{1}\\
t_{2}=t_{1}+a=10\\
t_{3}=t_{1}+2a\\
t_{4}=t_{1}+3a\\
...\\
t_{10}=t_{1}+9a\\
s=120=t_{1}+(t_{1}+a)+(t_{1}+2a)+(t_{1}+3a)+...+(t_{1}+9a)\\
=10*t_{1}+a(0+1+2+3+...+9)\\
=10*(t_{2}-a)+a*\frac{10*9}{2}\\
=10(10-a)+45a\\
==\ \textgreater \ 10(10-a)+45a=120\\
==\ \textgreater \ 2(10-a)+9a=24\\
==\ \textgreater \ 7a=4\\
==\ \textgreater \ a=\frac{4}{7}
[/tex]
