a G.S :the sum of first three terms equals 7 and the sum of their cubes equals 73
find the sequence

Let the the numbers http://latex.codecogs.com/gif.latex?a,ar,ar^{2} so there sum will be


http://latex.codecogs.com/gif.latex?a+ar+ar^{2}=7........................... ...........(1)


and the sum of there cubes will be


http://latex.codecogs.com/gif.latex?a^{3}+a^{3}r^{3}+a^{3}r^{6}=73.......... ...............(2)


here we have two equations in two variables so we can solve it
subtract (2) from (1)


http://latex.codecogs.com/gif.latex?a^{3}+a^{3}r^{3}+a^{3}r^{6}- a-ar-ar^{2}=66 .........................(3)


divide (2) by (1)


http://latex.codecogs.com/gif.latex?\frac{a^{3}+a^{3}r^{3}+a^{3}r^{6}}{ a+ar+ar^{2}}=\frac{73}{7}

I won't simplify this now because I will need it


http://latex.codecogs.com/gif.latex?7a^{3}+7a^{3}r^{3}+7a^{3}r^{6}-73a-73ar-73ar^{2}=0

put 73 = 66 + 7 we will get


http://latex.codecogs.com/gif.latex?7a^{3}+7a^{3}r^{3}+7a^{3}r^{6}-7a-7ar-7ar^{2}-66(a+ar+ar^{2})=0
............................(4)


substitute from (2) and (3) in (4)


http://latex.codecogs.com/gif.latex?7 \times 66 - 7 \times 66 a =0


http://latex.codecogs.com/gif.latex?1 - a = 0


so a = 1


substitute in number (1)


http://latex.codecogs.com/gif.latex?1+r+r^{2}=7


http://latex.codecogs.com/gif.latex?r^{2}+r-6=0

by factorization


http://latex.codecogs.com/gif.latex?(r - 2)(r + 3) = 0


so r = 2 or r = -3 "refused"


so the sequence is 1 , 2 , 4 , ....................







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شكرا جدا علي الحل الرائع

انا اسف لكن انا لا افهم من اين جاءت ال)a( في المعادلة التي قبل ال1=a

we take (a) common factor from the last part of the previous equation

a + ar + ar^(2) = 0

in the previous step in the last part so the equation will be in the form "of the last part"

a( 1 + r + r^(2)) = 0