5.
a)
[tex]a_5=0\ , \ a_8=-12\\\\a_8=a_5+3r\\\\-12=0+3r\\\\r=-4\\\\a_5=a_1+4r\\\\0=a_1-16\\\\a_1=16\\\\a_n=16+(n-1)(-4)\\\\a_n=-4n+20\\[/tex]
b)
[tex]a_1+a_3=0\\\\a_1+a_1+2r=0\\\\2a_1+2r=0\\\\a_1+r=0 \Rightarrow a_2\\\\a_2+a_4=-8\\\\0+a_1+3r=-8\\\\+\left \{ {{a_1+r=0} \atop {a_1+3r=-8}|*(-1)} \right.\\\\-2r=8\rightarrow r=-4\\\\a_1-4=0\rightarrow a_1=4\\\\a_n=4+(n-1)(-4)\\\\a_n=-4n+8\\[/tex]
c)
[tex]+\left \{ {{2a_1+r=5} \atop {2a_1+2r=-4}*(-1)} \right.\\\\-r=9\rightarrow r=-9\\\\2a_1-9=5\\\\a_1=7\\\\a_n=7+(n-1)(-9)\\\\a_n=-9n+16\\[/tex]
2)
[tex]a_1=1\ , \ r=2\ , \ a_n=2n-1\\\\a_9=2*9-1=17\\\\a_{18}=2*18-1=35\\\\S_9=\frac{1+17}{2}*9\\\\S_9=81\\\\S_{18}=\frac{1+35}{2}*35\\\\S_{18}=324\\[/tex]
3)
[tex]a)\\a_{7}=25*\frac{1}{5}^6\\\\a_7=\frac{1}{625}\\\\b)\\a_{15}=-7*(-1)^{14}\\\\a_{15}=-7\\\\c)\\a_5=-2*(-3)^4\\\\a_5=-162\\\\d)\\a_4=8*(\frac{2}{3})^3\\\\a_4=\frac{64}{27}\\\\e)\\a_9=5*(\sqrt{2})^8\\\\a_9=80\\\\f)\\a_6=128*(\frac{1}{2}) ^5\\\\a_6=4\\[/tex]
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2024 KUDO.TIPS - All rights reserved.
Verified answer
5.
a)
[tex]a_5=0\ , \ a_8=-12\\\\a_8=a_5+3r\\\\-12=0+3r\\\\r=-4\\\\a_5=a_1+4r\\\\0=a_1-16\\\\a_1=16\\\\a_n=16+(n-1)(-4)\\\\a_n=-4n+20\\[/tex]
b)
[tex]a_1+a_3=0\\\\a_1+a_1+2r=0\\\\2a_1+2r=0\\\\a_1+r=0 \Rightarrow a_2\\\\a_2+a_4=-8\\\\0+a_1+3r=-8\\\\+\left \{ {{a_1+r=0} \atop {a_1+3r=-8}|*(-1)} \right.\\\\-2r=8\rightarrow r=-4\\\\a_1-4=0\rightarrow a_1=4\\\\a_n=4+(n-1)(-4)\\\\a_n=-4n+8\\[/tex]
c)
[tex]+\left \{ {{2a_1+r=5} \atop {2a_1+2r=-4}*(-1)} \right.\\\\-r=9\rightarrow r=-9\\\\2a_1-9=5\\\\a_1=7\\\\a_n=7+(n-1)(-9)\\\\a_n=-9n+16\\[/tex]
2)
[tex]a_1=1\ , \ r=2\ , \ a_n=2n-1\\\\a_9=2*9-1=17\\\\a_{18}=2*18-1=35\\\\S_9=\frac{1+17}{2}*9\\\\S_9=81\\\\S_{18}=\frac{1+35}{2}*35\\\\S_{18}=324\\[/tex]
3)
[tex]a)\\a_{7}=25*\frac{1}{5}^6\\\\a_7=\frac{1}{625}\\\\b)\\a_{15}=-7*(-1)^{14}\\\\a_{15}=-7\\\\c)\\a_5=-2*(-3)^4\\\\a_5=-162\\\\d)\\a_4=8*(\frac{2}{3})^3\\\\a_4=\frac{64}{27}\\\\e)\\a_9=5*(\sqrt{2})^8\\\\a_9=80\\\\f)\\a_6=128*(\frac{1}{2}) ^5\\\\a_6=4\\[/tex]