Question 1178674
{{{ n}}}th term of an arithmetic sequence is:

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


if the sixth and eighth terms of an arithmetic sequence are

{{{a[6]=-11}}} 
{{{a[8]=-19}}}

we have


{{{-11=a[1]+d(6-1)}}}

{{{-11=a[1]+5d}}}.........solve for {{{a[1]}}}

{{{a[1]=-11-5d}}}.......eq.1


{{{-19=a[1]+d(8-1)}}}

{{{-19=a[1]+7d}}}.........solve for {{{a[1]}}}

{{{a[1]=-19-7d}}}.......eq.2

from eq.1 and eq.2 we have

{{{-11-5d=-19-7d}}}......solve for {{{d}}}


{{{7d-5d=-19+11}}}

{{{2d=-8}}}

{{{d=-4}}}

go to

{{{a[1]=-11-5d}}}.......eq.1, substitute {{{d}}}

{{{a[1]=-11-5(-4)}}}

{{{a[1]=-11+20}}}

{{{a[1]=9}}}


your formula is:

{{{a[n]=9-4(n-1)}}}



then the {{{a[15]}}} term is:

{{{a[15]=9-4(15-1)}}}

{{{a[15]=9-4(14)}}}

{{{a[15]=9-56}}}

{{{a[15]=-47}}}