What is the sum of the first 10 terms of this geometric sequence? 3, –6, 12, –24, …

Mathematics

2 answers:

sukhopar [10]3 years ago

8 0

Answer:

(-1023)

Step-by-step explanation:

3+(-6)+12+(-24)+48+(-96)+192+(-384)+768+(-1536)

Andrews [41]3 years ago

8 0

Answer:

Step-by-step explanation:

a1=3

r=-6/3=-2

$S_{10}=a1\frac{1-r^n}{1-r}\\ =3\frac{1-(-2)^{10}}{1-(-2)} \\=3\frac{1-1024}{3} \\=3\frac{-1023}{3} \\=-1023$

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Answer:

$\large\boxed{\bold{NO\ REAL\ SOLUTIONS}}\\\boxed{x=4-2i,\ x=4+2i}$

Step-by-step explanation:

$Domain:\\c-4\neq0\ \wedge\ c-5\neq0\Rightarrow c\neq4\ \wedge\ c\neq5\\\\\dfrac{c}{c-5}=\dfrac{4}{c-4}\qquad\text{cross multiply}\\\\c(c-4)=4(c-5)\qquad\text{use the distributive property}\\\\c^2-4c=4c-20\qquad\text{subtract}\ 4c\ \text{from both sides}\\\\c^2-8c=-20\qquad\text{add 20 to both sides}\\\\c^2-8c+20=0\qquad\text{use the quadratic formula}$

$\text{for}\ ax^2+bx+c=0\\\\\text{if}\ b^2-4ac0,\ \text{then the equation has two solutions}\ x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

$x^2-8x+20=0\\\\a=1,\ b=-8,\ c=20\\\\b^2-4ac=(-8)^2-4(1)(20)=64-80=-16$

$\text{In the set of complex numbers:}\\\\i=\sqrt{-1}\\\\\text{therefore}\ \sqrt{b^2-4ac}=\sqrt{-16}=\sqrt{(16)(-1)}=\sqrt{16}\cdot\sqrt{-1}=4i\\\\x=\dfrac{-(-8)\pm4i}{2(1)}=\dfrac{8\pm4i}{2}=\dfrac{8}{2}\pm\dfrac{4i}{2}=4\pm2i$