## brittyarellano one year ago what is the 32nd term of the arithmetic sequence where a1=13 and a13=-59

1. jdoe0001

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

2. brittyarellano

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

3. 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" :)

4. brittyarellano

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

5. jdoe0001

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

6. brittyarellano

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

7. 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

8. jdoe0001

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

9. brittyarellano

I got -2.36

10. jdoe0001

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

11. brittyarellano

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

12. jdoe0001

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

13. jdoe0001

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

14. brittyarellano

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