How to find first three terms in arithmetic series if given first and last terms and sum?

Start with Sn = n*(A1+An)/2, to solve for n.

Then use the definition of An = A1 + (n-1)d, to solve for d.

Then you already know A1, and A2=A1+d, and A3=A2+d

Sum = n (first + last) / 2, where n = number of terms

n (3 + 136) / 2 = 1390
n 139/2 = 1390
n = 1390 * 2 / 139
n = 20

1st term = 3
2nd term = 3 + d
3rd term = 3 + 2d
nth term = 3 + (n-1)d

20th term = 3 + 19d = 136
19d = 133
d = 7

First three terms: 3, 3+7, 3+2*7 -----> 3, 10, 17

a=3,  f(n)=an=136  Sn=1390,   An=a+(n-1)d,  Sn=n/2[2a+(n-1)d]

136=3+(n-1)d,      1390=n/2[6+(n-1)d]
133=(n-1)d,          1390=n/2[6+133]
                           1390=n/2*139
                           20=n                133=(20-1)d,   133==19d, d=7
   
a, a+d,a+2d   3,10,17

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First, we need to know the number of terms, n
using the formula for the sum of an arithmetic series;
S_n = ( n / 2 ) ( a_1 + a_n )

1390 = ( n / 2 ) ( 3 + 136 )
.........= ( n / 2 ) ( 139 )
.......................divide both sides of the equation by 139;
1390 / 139 =  ( n / 2 ) ( 139 ) / 139
10 = n / 2
.......................multip ly both sides of the equation by 2
10 ? 2 = ( n / 2 ) ? 2
20 = n
...or...
n = 20

then, we have to compute for the difference, d.
a_n = a + ( n - 1 ) d

d = ( a_n - a ) / ( n - 1 )
...= ( 136 - 3 ) / ( 20 - 1 )
...= 133 / 19
...= 7


the formula for the nth term of an arithmetic sequence is
a_n = a + ( n - 1 ) d

a_2 = 3 + ( 2 - 1 ) ( 7 )
.......= 3 + ( 1 ) ( 7 )
.......= 3 + 7
.......= 10
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a_3 = 3 + ( 3 - 1 ) ( 7 )
.......= 3 + ( 2 ) ( 7 )
.......= 3 + 14
.......= 17
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first three terms are 3, 10, 17
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