Tn = a + (n-1)d
where a denotes first term and d denotes the common difference ( difference between any term and the previous term)

so T2 = a + d
T5 = a + 4d

but T5 = 2*T2

so a+4d = 2(a+d)

which simplifies to a = 2d

Sn the sum of n terms is given by

Sn = n/2[2a+ (n-1)d]

so S6 = 6/2[2a+5d] = 36

so 6a + 15d = 36

but a= 2d from above

so 12d + 15 d = 27d = 36

d = 4/3

a = 2d = 8/3