If you have an explicit formula, coming up with a particular term is easy. If not, you have to use the recursion several times.
#1. try (-2)^(n-1)
#2. You can use the explicit formula, but showing how to handle it as a general arithmetic sequence is also nice.
Naturally, any arithmetic/geometric sequence has an explicit formula.
#3 same as #2
#4 ratio is 2/3. 0.67 is only an approximation. Unless decimal values are given, I'd avoid them.
Can you check my homework please?
Given the following representations, tell whether it's arithmetic/geometric, find the difference/ratio, find the explicit formula, find the recursive formula, and the given term. (This whole question was in the form of a table)
1. -6, 12, 24....
Type:geometric
Rati0: -2
explicit form: a<n> = -6(-2)^n-1.
(not sure about notation. trying to write -6 times -2 raised to n-1)
recursive form: a<n> = -2(a<n-1>), and a<1> = -6
(trying to write -2 times a sub n-1)
I'm not sure which formula to use to find given term of a<10>.
- using explicit: (here's my try)
a<10>= -6(-2)^10-1
a<10>= -6(-2)^9
a<10>= -6(-512)
a<10>= 3072
- using recursive: I'm not sure how to do it. but here's my reasoning. need to find previous value, a<9>. so if a<10> = 3062 and r = -2; then a<9> = 3072/-2 = -1536. Use a<9> as previous value in recursive formula.
a<10> = r(a<n-1>)
a<10> = -2(-1536)
a<10> = 3072
Is there an easier way to figure it out?
2. 10, 20, 30, 40...
type: arithmetic
difference: 10
explicit form: a<n> = 10n
recursive form: a<n> = a<n-1> +10, and a<1> = 10
given term of a<32>:
- using explicit:
a<32>= 10 + (n-1)10
a<32>= 10 + (32-1)10
a<32>= 10 + 31(10)
a<32>= 10 + 310 = 320
or do I use formula i just solved for? a<n> = 10n
a<32> = 10(32)= 320
- using recursive: my reasoning is that I need to find a<31> which is previous value. so if a<32> = 320 and difference = 10; then a<31> = 320 - 10 = 310. Use a<31> in recursive form.
a<32> = a<n-1> + d
a<32> = 310 + 10
a<32> = 320
3. -10, -8, -6, -4...
Type: arithmetic
difference: 2
Explicit form: a<n> = 2n - 12
Recursive form: a<n> = a<n-1> + 2, and a<1> = -10
given term of a<56>:
- using explicit:
a<56>= -10 + (n-1)2
a<56>= -10 + (56-1)2
a<56>= -10 + (55)2
a<56>= -10 + 110 = 100
or a<56> = 2n - 12
a<56> = 2(56) - 12
a<56> = 112 - 12 = 100 ??
- using recursive: need to find a<55> which is previous value. so if a<56> = 100 and difference = 2; then a<55> = 100 - 2 = 98. Use a<55> in recursive form.
a<56> = a<n-1> + d
a<56> = 98 + 2 = 100 ??
4. 72, 48, 32...
Type: geometric
ratio: .67
explicit form: a<n>= 72(.67)^n-1
recursive form: a<n>= .67(a<n-1>) and a<1> = 72
given term of a<5>:
- using explicit:
a<5>= 72(.67)^5-1
a<5>= 72(.67)^4
a<5>= 72(.20)
a<5>= 14.5
-using recursive:
need to find a<4> which is previous value. so if a<5> = 14.5 and ratio = .67; then a<4> = 14.5/.67 = 21.64. Use a<4> in recursive form.
a<5>= r(a<n-1>)
a<5>= .67(21.64) = 14.51
Questions:
1. When the problem asks for finding a certain term, do I use the formula I just came up with or do I use the standard form depending on the type of sequence?
2. Is there an easier way to find a term using the recursive formula?
1 answer
1 answer