Question 692570
1st term = a, common difference = d

So terms are, a, a+d, a+2d, ...

3rd term = 0 means that {{{ a+2d=0 }}}  ............ Eq (1)

so {{{ s[n]= (n/2)(2a + (n-1)d) }}}
and {{{ s[15] = (15/2)(2a+14d) = -300 }}}
{{{ (15/2)(2a+14d) = -300 }}}
{{{ 2a+14d) = -40 }}}  ............ Eq (2)

Eq (2) - 7 Eq (1) gives:
{{{ (2-7)a +(14-14)d=-40-0 }}}
{{{ -5a =-40 }}}
{{{ a=8 }}}, which is the first term


NB, If you need d then Eq (2) - 2 Eq (1) gives:
{{{ (2-2)a +(14-4)d=-40-0 }}}
{{{ 10d=-40 }}}
{{{ d=-4 }}}