Question

What is the 32nd term of the arithmetic sequence where a1 = 12 and a13 = -60?

Answer 1

We have :
\[a_{13}=12d+a_1\]
So :
\[-60=12d+12\]
And then :
\[12d=-60-12=-72\]
So :
\[d=-\frac{72}{12}=-6\]
Now we have :
\[a_{32}=31d+a_1\]

Answer 2

Wow! Ok, so I'm pretty confused! Hahaha so the answer is a22 = 31d + a1

Answer 3

It is
\[\Large a_{32}=31d+a_1\]
And you have all what you want

Answer 4

Our formula \[\large a_{n}= a_{1} +(n-1)d\]for Arithmetic Progression: were given our first term, a1 = 12,and our 13th term, a13 = -60 we can use this to find our difference between all the terms. \[\large -60= 12+(13-1)d\]\[\large -60 = 12 + 12d\]\[\large -60 - 12 = 12d\]\[\large -72 =12d\]\[\large d= 6\] Now we've found the difference between all our numbers to 32 by using the information given to us by cleverly manipulating the formula for Arithmetic progression. now to find the 32nd term of the sequence we have our

d = 6, a1 = 12 n=32

\[\large a_{32} = 12- (32-1)6\]\[\large a_{32}= 12 - 31(6)\] Ill leave the rest to you :)

Answer 5

Sorry d = -6, D:: i forgot my negative in there.

Answer 6

Haha, habit

Answer 7

It should be \[\large a_{32}= 12 +(32-1)(-6)\]\[\large a_{32}= 12 + 31(-6)\] *******************************************

Answer 8

Does this make sense? What @Noura11 and I did?

Answer 9

-174! Thank you for all of the time that you spent on this problem! :) wow! I appreciate it! Yes, it makes much more sense now.

Answer 10

Awesome :)