Question 1055154
<pre><b><font size = 4> 
The solution above contains an error. He took 

"The sum of the second and the third terms"

as if it were

"The sum of the FIRST and the third terms"

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The sum of the second and the third terms of a 
geometric progression is 6,

 {{{a[2]+a[3]}}}{{{""=""}}}{{{6}}}
{{{a[1]r+a[1]r^2}}}{{{""=""}}}{{{6}}}
{{{a[1](r+r^2)}}}{{{""=""}}}{{{6}}}
{{{a[1]}}}{{{""=""}}}{{{6/(r+r^2)}}}

the sum of the third and the fourth term is -12.

 {{{a[3]+a[4]}}}{{{""=""}}}{{{-12}}}
 {{{a[1]r^2+a[1]r^3}}}{{{""=""}}}{{{-12}}}

 {{{a[1](r^2+r^3)}}}{{{""=""}}}{{{-12}}}
 {{{a[1]}}}{{{""=""}}}{{{(-12)/(r^2+r^3)}}}

Set the two expressions for a<sub>1</sub> equal:

 {{{6/(r+r^2)}}}{{{""=""}}}{{{(-12)/(r^2+r^3)}}}

Divide both sides by 6

 {{{1/(r+r^2)}}}{{{""=""}}}{{{(-2)/(r^2+r^3)}}}

Cross-multiply

 {{{r^2+r^3)}}}{{{""=""}}}{{{-2(r+r^2)}}}

 {{{r^2+r^3)}}}{{{""=""}}}{{{-2r-2r^2}}}

 {{{r^3+3r^2+2r}}}{{{""=""}}}{{{0}}}

Factor out r:

 {{{r(r^2+3r+2)}}}{{{""=""}}}{{{0}}}

Factor the expression in parentheses:

 {{{r(r+1)(r+2)}}}{{{""=""}}}{{{0}}}

Use zero-factor property.  Set each
factor = 0:

 r = 0;  r+1 = 0;  r+2 = 0
          r = -1     r = -2

So we have three potential values for r:

If we use r = 0

Substitute in {{{a[1]}}}{{{""=""}}}{{{6/(r+r^2)}}}

 {{{a[1]}}}{{{""=""}}}{{{6/((0)+(0)^2)}}}

 {{{a[1]}}}{{{""=""}}}{{{6/0}}}

That's undefined so we must discard r = 0

----

If we use r = -1

Substitute in {{{a[1]}}}{{{""=""}}}{{{6/(r+r^2)}}}

 {{{a[1]}}}{{{""=""}}}{{{6/((-1)+(-1)^2)}}}

 {{{a[1]}}}{{{""=""}}}{{{6/((-1)+1)}}}

 {{{a[1]}}}{{{""=""}}}{{{6/0}}}

That's also undefined. So we must also discard r = -1

----

If we use r = -2

Substitute in {{{a[1]}}}{{{""=""}}}{{{6/(r+r^2)}}}

 {{{a[1]}}}{{{""=""}}}{{{6/((-2)+(-2)^2)}}}

 {{{a[1]}}}{{{""=""}}}{{{6/((-2)+4)}}}

 {{{a[1]}}}{{{""=""}}}{{{6/2}}}

 {{{a[1]}}}{{{""=""}}}{{{3}}}   

So the first term is 3,

The sequence = 3, -6, 12, -24, 48, -96, ...

The sum of the first 20 terms:

 {{{S[n]}}}{{{""=""}}}{{{(a[1](r^n-1))/(r-1)}}}

 {{{S[20]}}}{{{""=""}}}{{{3(2^20-1)/(2-1)}}}

{{{S[20]}}}{{{""=""}}}{{{3(1048576-1)/1}}}

{{{S[20]}}}{{{""=""}}}{{{3(1048575)}}}

{{{S[20]}}}{{{""=""}}}{{{3145725}}}


Edwin</pre></font></b>