Question 1135338

In General we write a Geometric Sequence like this:

{ {{{a}}}, {{{ar}}}, {{{ar^2}}}, {{{ar^3}}}, ... } where:


{{{a}}} is the first term, and
{{{r}}} is the factor between the terms (called the "common ratio")


if the sum of three numbers in G.P is {{{21}}}, we have


 {{{a+ar+ar^2=21}}}....solve for {{{a}}}

{{{a(1+r+r^2)=21}}}

{{{a=21/(1+r+r^2)}}}.....eq.1



 and, if the sum of their squares is {{{189}}}, we have


 {{{a^2+(ar)^2+(ar^2)^2=189}}}...solve for {{{a}}}

 {{{a^2+a^2 r^2+a^2r^4=189}}}

{{{a^2(1+r^2+r^4)=189}}}

{{{a^2=189/(1+r^2+r^4)}}}

{{{a=sqrt(189/(1+r^2+r^4))}}}........eq.2


from eq.1 and eq.2 we have


{{{21/(1+r+r^2)=sqrt(189/(1+r^2+r^4))}}}........square both sides

{{{441/(1+r+r^2)^2=189/(1+r^2+r^4)}}}

{{{441(1+r^2+r^4)=189(1+r+r^2)^2}}}

{{{441(1+r^2+r^4)=189(r^4 + 2 r^3 + 3 r^2 + 2 r + 1)}}}

{{{441+441r^2+441r^4=189 r^4 + 378 r^3 + 567 r^2 + 378 r + 189}}}

{{{441+441r^2+441r^4-189 r^4 - 378 r^3 - 567 r^2 - 378 r - 189=0}}}

{{{252 r^4 - 378 r^3 - 126 r^2 - 378 r + 252=0}}}.........factor

{{{126 (2 r - 1) (r - 2) (r^2 + r + 1)=0}}}

solutions:

if {{{2r-1=0}}}=>{{{r=1/2}}}

if {{{r-2=0}}}=>{{{r=2}}}

if {{{(r^2 + r + 1)=0}}}=> use quadratic formula and you will get complex solutions
{{{r = -(-1)^(1/3)}}}->disregard


now find {{{a}}}:

{{{a=21/(1+r+r^2)}}}.....eq.1, plug in {{{r=1/2}}}

{{{a=21/(1+1/2+(1/2)^2)}}}

{{{a=21/(1+1/2+1/4)}}}

{{{a=21/(4/4+2/4+1/4)}}}

{{{a=21/(7/4)}}}

{{{a=(21*4)/7}}}

{{{a=3*4}}}

{{{a=12}}}


or

{{{a=21/(1+r+r^2)}}}.....eq.1, plug in {{{r=2}}}

{{{a=21/(1+2+2^2)}}}

{{{a=21/7}}}

{{{a=3}}}


then {{{a}}}, {{{ar}}}, {{{ar^2}}} will be

if {{{a=12}}} and {{{r=1/2}}}

{{{a=12}}}

{{{ar=12*(1/2)=6}}}

 {{{ar^2=12*(1/4)=3}}}


or

if {{{a=3}}} and {{{r=2}}}

{{{a=3}}}

{{{ar=3*2=6}}}

 {{{ar^2=3*4=12}}}

you actually got same terms in reverse order


check their sum: 

{{{12+6+3=21}}} and 

{{{12^2+6^2+3^2=189}}} 

{{{144+36+9=189}}} 

{{{189=189}}}