geometric sequence problem ارجو المساعدة من استاذ متخصص

MathPrince · #1 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 , ....................

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