Question 1072952
<pre>
{{{a[1]=28}}}                <--first term
{{{a[n]=a[1]+(n-1)d=-77}}}   <--formula for nth term
{{{a[n]=28+(n-1)d=-77}}}
{{{28+(n-1)d=-77}}}
{{{(n-1)d = -105}}}

{{{S[n]=expr(n/2)(2a[1]+(n-1)d)= -196}}}  <-- formula for nth term

 Substitute for a<sub>1</sub> and (n-1)d=-105

{{{expr(n/2)(2(28)-105)= -196}}}

{{{expr(n/2)(56-105)= -196}}}

{{{expr(n/2)(-49)= -196}}}

Multiply through by 2

{{{n(-49)= -392}}}

{{{n=(-392)/(-49)}}}

{{{n=8}}}          <--number of terms whose sum is -77

We'll need d, so

{{{(n-1)d=-105}}}
{{{(8-1)d=-105}}}
{{{7d=-105}}}
{{{d=-15}}}        <-- common difference, d

To find 10th term:

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

Substitute n=10, a<sub>1</sub> = 28, n=10, d=-15

{{{a[10]=28+(10-1)(-15)}}}

{{{a[10]=28+9(-15)}}}

{{{a[10]=28-135}}}

{{{a[10]=-107}}}     <-- 10th term

{{{S[n]=expr(n/2)(2a[1]+(n-1)d)}}} <-- formula for sum of first n terms

{{{S[10]=expr(10/2)(2(28)+(10-1)(-15))}}}

{{{S[10]=(5)(56+(9)(-15))}}}

{{{S[10]=(5)(56-135)}}}

{{{S[10]=(5)(-79)}}}

{{{S[10]=-395}}}

Checking, the sequence is:

28,13,-2,-17,-32,-47,-62,-77,-92,-107

Add them

28+13-2-17-32-47-62-77-92-107 = -395

Add the first 8 terms

28+13-2-17-32-47-62-77 = -196

Everything checks.

Edwin</pre>