Question 998907
{{{S[n] = (n/2)(2a+(n-1)d)}}}

n=15
a=?
d=-3
{{{S[15]=120}}}

{{{S[15]= (15/2)(2a+(15-1)(-3))}}}

=(15/2)(2a-42)

120*2 = 15(2a-42)

120*2/15 = 2a-42

8*2=2a-42

16+42=2a
58=2a

a=29 First term


{{{t[n]=a+(n-1)d}}}

{{{t[15]= 29+(14)(-3)}}}

=29-41

=-12

{{{t[15] = -12}}}