Question 1120323
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From the condition, we have

{{{S[2]}}} = {{{a[1] + a[2]}}} = {{{a + ar}}} = a*(1+r),              (1)

{{{S[3]}}} = {{{a[1] + a[2] + a[3]}}} = {{{a+ar+ar^2}}} = {{{a*(1+r+r^2)}}}.    (2)


Divide (1) by (2). You will get

{{{(1+r)/(1+r+r^2)}}} = {{{S[2]/S[3]}}} = {{{20/65}}} = {{{4/13}}}.

or

13(1+r) = 4*(1+r+r^2),

13 + 13r = 4 + 4r + 4r^2

4r^2 -9r - 9 = 0

{{{r[1,2]}}} = {{{(9 +- sqrt(9^2 + 4*4*9))/(2*4)}}} = {{{(9 +- 15)/8}}};

{{{r[1]}}} = {{{(9 + 15)/8}}} = 3;   {{{r[2]}}} = {{{(9 - 15)/8}}} = {{{-6/8}}} = {{{-3/4}}}.



Subtract (1) from (2). You will get 

{{{a[3]}}} = {{{S[3]}}} - {{{S[2]}}} = 65 - 20 = 45 = {{{a*r^2}}}.     (*)



<U>Thus we should consider TWO opportunities</U>:



    1.  If r = 3,  then according to (*)  {{{a[3]}}} = 45 = {{{a*3^2}}} = 9a,

        which implies  a = 5.


        Then  {{{a[1]}}} = 5,  {{{a[2]}}} = 3*5 = 15  and  {{{a[3]}}} = 3*15 = 45.

        Then  {{{S[2]}}} = {{{a[1]+a[2]}}} = 5 + 15 = 20  and  {{{S[3]}}} = 5 + 15 + 45 = 65    ! Correct !.


    2.  If r= {{{-3/4}}}, then according (*)  a = {{{45/((9/16))}}} = 5*16 = 80.  

        Then  {{{a[2]}}} = {{{80*(-3/4)}}} = -60  and  {{{a[3]}}} = {{{-60*(-3/4)}}} = 45.

        In this way,  {{{S[2]}}} = {{{a[1]+a[2]}}} = 80 - 60 = 20  

                and   {{{S[3]}}} = {{{a[1] + a[2] + a[3]}}} = 80 - 60 + 45 = 65.

        So, this opportunity does work, too.


<U>Answer</U>.  There are  <U>TWO solutions</U>.

         One solution is  a= 5, r= 3 and the three terms are  5, 15, 45.

         Another solution is  a= 80,  r= {{{-3/4}}}  and the three terms are  80, -60 and 45.
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