Nobody Special: In an arithmetic sequence a_8 = 9 and a_12 = 37. What is a_13?
Is there an equation or something that I'm missing here? Something to simplify it?
If anyone has any ideas, that would be awesome! ^_^
Hi Special,
I am going to look after you myself.
a_8 should be written as a
8 I used the 'sub' button above the smilies to get this.
Just press a 'sub' put the 8 where the curser is, then move curser to the end and keep going. The 'sub' stands for subtext.
Most likely it is written as a_8 because this is the code for subtext in LaTex which is a maths program - It is probably the code in other computer languages as well.
now we have
It is an AP and you need to find a
13 a
8 = 9 and a
12=37
the formula for the nth term of an AP is
T
n=a+(n-1)d SCRATCH THAT
sorry I just realized that you are using different letters from what I am used to, I had better change the formula accordingly
an = a1 + (n-1)d [where d is the common difference][maybe this is the equation that you are missing? - It is not normally expressed like this, not in my experience anyway]
and you have
a
8=a
1+(8-1)*d ==> a
1+7d = 9
a
12=a
1+(12-1)*d ==> a
1+11d = 37
that is
a
1+7d = 9 (1)
a
1+11d = 37 (2)
You just have to solve these simultaneously
(2)-(1) gives you
4d = 28
d=7
Now substitute that back into 1 of the equations to get the value of a
1 Then just substitute the values of a
1 and d into the AP equation.
Then sub in n=13 and find the answer that you actually want.
Have a go if you don't understand I will help some more.