Question

what is the 32nd term of the arithmetic sequence where a1=13 and a13=-59

Answer 1

recall \(\Large a_{\color{brown}{ n}}=a_1+({\color{brown}{ n}}-1)d ?\)

Answer 2

yes but I'm having trouble being able to find d

Answer 3

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" :)

Answer 4

yes but I'm having trouble being able to find d

Answer 5

once you find "d", you can find \(\Large a_{32}\)

Answer 6

yes but I'm having trouble being able to find d

Answer 7

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

Answer 8

well.. what did you get for "d"?

Answer 9

I got -2.36

Answer 10

hmmm actually is -59
but anyhow \(\bf a_{\color{brown}{ 13}}={\color{blue}{ a_1}}+({\color{brown}{ 13}}-1)d\implies -59={\color{blue}{ 13}}+({\color{brown}{ 13}}-1)d\)

Answer 11

I plugged it in with -59 and I got -2.36

Answer 12

well... hold the mayo.. .i messed up the signs again, shoot

Answer 13

\(\bf a_{\color{brown}{ 13}}={\color{blue}{ a_1}}+({\color{brown}{ 13}}-1)d\implies -59={\color{blue}{ 13}}+({\color{brown}{ 13}}-1)d
\\ \quad \\
-59-13=12d\implies -72=12d\implies \cfrac{\cancel{-72}}{\cancel{12}}=d
\\ \quad \\
-6=d\)

Answer 14

oh I see where I messed. I got -173 for the answer! thanks so much!