Can you explain how to find a sub 1 in the following problem?
1. Given 2 terms in an arithmetic sequence, find the recursive formula.
a sub 18 = 3362 , a sub 38 = 7362
Steps:
a. Find difference:
(7632 - 3362)/(38 - 18) = 4000/20 = 200
b. plug into recursive formula:
a sub n = a sub(n -1) + 200
What is a sub 1?
Also explain how to find a sub 1 given a term and common difference below:
a sub 21 = -1.4, d = 0.6
Steps:
a. plug into recursive formula:
a sub n = a sub(n -1) + 0.6
How do I find a sub 1?
a<18> = a<1> + 17*200 = 3362
So, a<1> = -38
similarly,
a<21> = a<1> + 20*0.6 = -1.4
now again, just solve for a<1>
a<n> = a + (n-1)d
or
term(18) = 3362 ----> a+17d = 3362
term(38) = 7362 ----> a + 37d = 7362
subtract them:
20d = 4000
d = 200
sub into a+17d = 3362
a = -38
so recursive:
term(n) = term(n-1) + 200, term(1) = -38
general:
term(n) = a+(n-1)d
= -38 + 200(n-1)
= 200n - 238
THANKS SO MUCH! I get it now. Thanks for showing how to write notations correctly too. In the second problem, I got -13.4 for a<1> where a<21> = -1.4 and d=0.6
You're welcome.
As for the notation, I'm not sure it's "correct" or standard. I just thought of using it since making subscripts is a pain, and the parentheses sometimes get confused with function notation.
You will also see
An, An+1 (ambiguous: cf. (An)+1)
A_n, A_n+1 (same)
A[n], A[n+1]
and other even more obscure attempts
To find a sub 1 in the given problem, you will need to use the recursive formula and the given information. The recursive formula for an arithmetic sequence is:
a sub n = a sub(n-1) + d
where "a sub n" represents the nth term, "a sub(n-1)" represents the previous term, and "d" represents the common difference.
In the first problem, you are given two terms in the sequence: a sub 18 = 3362 and a sub 38 = 7362. Using the formula, you can find the common difference, which is the difference between any two terms in the sequence.
Step 1: Find the difference.
To find the common difference, subtract a sub 18 from a sub 38, and divide by the difference in positions:
(7362 - 3362) / (38 - 18) = 4000 / 20 = 200
Step 2: Plug into the recursive formula.
Now that you know the common difference is 200, you can plug it into the recursive formula:
a sub n = a sub(n-1) + 200
Step 3: Find a sub 1.
To find a sub 1, you need to trace the sequence back to the first term. Since the information provided is for a sub 18 and a sub 38, you can start from a sub 38 and work your way back using the recursive formula.
a sub 38 = a sub 37 + 200
a sub 37 = a sub 36 + 200
...
a sub 2 = a sub 1 + 200
Since a sub 1 is the first term, it does not have any term before it. Therefore, you will not add 200 to it.
By tracing back the sequence, you can find a sub 1.
In the second problem, you are given a sub 21 = -1.4 and the common difference d = 0.6. Using the same steps, you can find a sub 1.
Step 1: Plug into the recursive formula.
a sub n = a sub(n-1) + 0.6
Step 2: Find a sub 1.
Start from a sub 21 and work your way back:
a sub 21 = a sub 20 + 0.6
a sub 20 = a sub 19 + 0.6
...
a sub 2 = a sub 1 + 0.6
Again, since a sub 1 is the first term, you do not need to add 0.6 to it. By tracing back the sequence, you can find a sub 1.