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
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?
4 answers
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
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
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
1 answer