Question 1039350
You know that the sum of an arithmetic progression in terms of common difference and first term is, 
{{{S[n]=(n/2)(2t[1]+(n-1)*d)}}}
So,
{{{S[98]=(98/2)(2t[1]+(98-1)*1)}}}
{{{137=49(2t[1]+97)}}}
{{{2t[1]+97=137/49}}}
{{{2t[1]=137/49-97}}}
{{{t[1]=137/98-97/2}}}
Now that you have the first term you can subtract it from the sum to get the value that you're looking for.
{{{t[1]+X=S[98]}}}
{{{137/98-97/2+X=137}}}
{{{X=137-137/98+97/2}}}
or with a common denominator,
{{{X=9021/49}}}