geometric sequence problem ارجو المساعدة من استاذ متخصص

karimadel7 · #1 14-03-2011, 08:16 PM

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

MathPrince · #2 16-03-2011, 10:28 PM

Let the the numbers $a,ar,ar^{2}$ so there sum will be

$a+ar+ar^{2}=7$ ......................................(1)

and the sum of there cubes will be

$a^{3}+a^{3}r^{3}+a^{3}r^{6}=73$ .........................(2)

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

$a^{3}+a^{3}r^{3}+a^{3}r^{6}- a-ar-ar^{2}=66$ .........................(3)

divide (2) by (1)

$\frac{a^{3}+a^{3}r^{3}+a^{3}r^{6}}{ a+ar+ar^{2}}=\frac{73}{7}$

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

$7a^{3}+7a^{3}r^{3}+7a^{3}r^{6}-73a-73ar-73ar^{2}=0$

put 73 = 66 + 7 we will get

$7a^{3}+7a^{3}r^{3}+7a^{3}r^{6}-7a-7ar-7ar^{2}-66(a+ar+ar^{2})=0$
............................(4)

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

$7 \times 66 - 7 \times 66 a =0$

$1 - a = 0$

so a = 1

substitute in number (1)

$1+r+r^{2}=7$

$r^{2}+r-6=0$

by factorization

$(r - 2)(r + 3) = 0$

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

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

اللهم أنت ربى لا إله إلا أنت خلقتنى وأنا عبدك وأنا على عهدك ووعدك ما استطعت أعوذ بك من شر ما صنعت أبوء لك بنعمتك على وأبوء بذنبى فاغفر لى فإنه لا يغفر الذنوب إلا أنت

karimadel7 · #3 17-03-2011, 04:07 PM

شكرا جدا علي الحل الرائع

karimadel7 · #4 17-03-2011, 04:32 PM

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

MathPrince · #5 17-03-2011, 07:28 PM

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