Hi Farhaini,
Actually, they are writing the entire progression backward for the second S_n:
S_n= T_n+T_{n-1}+T_{n-2}+…T_1
So,
\space\space\space S_n=T_1 +T_2\space\space\space\space + T_3+.....T_n
+S_n= T_n+T_{n-1}+T_{n-2}+...T_1
This is because it’s easier to sum the pairs up together,
T_1 + T_n,
T_2+T_{n-1}
T_3+ T_{n-2} … etc
why do we do this?
because all of these pairs have the same value, which is 2a+(n-1)d as shown in the diagram.
T_1 + T_n
(T_1 +d)+(T_n -d) =T_1+T_n
(T_1 + 2d)+(T_n - 2d)=T_1+T_n … and so on.
The most common example is when you want to find 1+2+3…+ 10,
we add up the terms at both ends of the progression and then slowly work our way towards the middle.
1+10=11
2+9=11
3+8=11
4+7=11
5+6=11
Where 1 is T_1, 10 is T_n where n =10, 2 is T_2, 9 is T_{n-1} and so on.
So 1+2+3…+ 10=55
Hope this helps! 
Ohhh I understood Thank you very much !! 