brittyarellano
  • brittyarellano
what is the 32nd term of the arithmetic sequence where a1=13 and a13=-59
Mathematics
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SOLVED
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katieb
  • katieb
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jdoe0001
  • jdoe0001
recall \(\Large a_{\color{brown}{ n}}=a_1+({\color{brown}{ n}}-1)d ?\)
brittyarellano
  • brittyarellano
yes but I'm having trouble being able to find d
jdoe0001
  • jdoe0001
well, we know the 13th term is 59 and the 1st one is 13 so, we could say that \(\bf a_{\color{brown}{ 13}}={\color{blue}{ a_1}}+({\color{brown}{ 13}}-1)d\implies 59={\color{blue}{ 13}}+({\color{brown}{ 13}}-1)d\) so, to find "d", just solve for "d" :)

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brittyarellano
  • brittyarellano
yes but I'm having trouble being able to find d
jdoe0001
  • jdoe0001
once you find "d", you can find \(\Large a_{32}\)
brittyarellano
  • brittyarellano
yes but I'm having trouble being able to find d
brittyarellano
  • brittyarellano
i tried doing that and what I got wasn't one of the answer choices. It gave me the options of -185,-179,-173- and -167
jdoe0001
  • jdoe0001
well.. what did you get for "d"?
brittyarellano
  • brittyarellano
I got -2.36
jdoe0001
  • jdoe0001
hmmm actually is -59 but anyhow \(\bf a_{\color{brown}{ 13}}={\color{blue}{ a_1}}+({\color{brown}{ 13}}-1)d\implies -59={\color{blue}{ 13}}+({\color{brown}{ 13}}-1)d\)
brittyarellano
  • brittyarellano
I plugged it in with -59 and I got -2.36
jdoe0001
  • jdoe0001
well... hold the mayo.. .i messed up the signs again, shoot
jdoe0001
  • jdoe0001
\(\bf a_{\color{brown}{ 13}}={\color{blue}{ a_1}}+({\color{brown}{ 13}}-1)d\implies -59={\color{blue}{ 13}}+({\color{brown}{ 13}}-1)d \\ \quad \\ -59-13=12d\implies -72=12d\implies \cfrac{\cancel{-72}}{\cancel{12}}=d \\ \quad \\ -6=d\)
brittyarellano
  • brittyarellano
oh I see where I messed. I got -173 for the answer! thanks so much!

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