Question 323764
Please help me solve this: A certain arithmetic sequence has {{{a[51]=23}}}
and {{{a[101]=-127}}}. Find {{{a[1]}}}, {{{d}}}, {{{a[n]}}}, and {{{a[78]}}}
<pre><b>
Use  {{{a[n]}}}{{{""=""}}}{{{a[1]+(n-1)d)}}}

Substitute {{{n=51}}}, {{{a[51]=23}}}

{{{a[51]}}}{{{""=""}}}{{{a[1]+(51-1)d)}}}
{{{23}}}{{{""=""}}}{{{a[1]+50d)}}}
{{{a[1]+50d)}}}{{{""=""}}}{{{23}}}

Substitute {{{n=101}}}, {{{a[101]=-127}}}

{{{a[101]}}}{{{""=""}}}{{{a[1]+(101-1)d)}}}
{{{-127}}}{{{""=""}}}{{{a[1]+100d)}}}
{{{a[1]+100d)}}}{{{""=""}}}{{{-127}}}


So we have this system of equations:

{{{system(a[1]+50d=23, a[1]+100d=-127)}}}

Solve that system of equation by substitution or addition.
If you don't know how post again asking how.  The
solution is {{{a[1]}}}{{{""=""}}}{{{173}}}, {{{d}}}""=""}}}{{{-3}}}

Therefore substituting those in

{{{a[n]}}}{{{""=""}}}{{{a[1]+(n-1)d)}}}
{{{a[n]}}}{{{""=""}}}{{{173+(n-1)(-3)}}}
{{{a[n]}}}{{{""=""}}}{{{173+(-3)(n-1))}}}
{{{a[n]}}}{{{""=""}}}{{{173-3(n-1))}}}
{{{a[n]}}}{{{""=""}}}{{{173-3n+3))}}}
{{{a[n]}}}{{{""=""}}}{{{176-3n)}}}

Substitute n=78
{{{a[n]}}}{{{""=""}}}{{{176-3n)}}}
{{{a[78]}}}{{{""=""}}}{{{176-3(78))}}}
{{{a[78]}}}{{{""=""}}}{{{176-234)}}}
{{{a[78]}}}{{{""=""}}}{{{-58)}}}

Edwin</pre>