What is a30 for the arithmetic sequence presented in the table below? n 6 11 an 50 35 Hint: an = a1 + d(n − 1), where a1 is the first term and d is the common difference. a30 = −42 a30 = −38 a30 = −28 a30 = −22 Post New Answer
can you help me please? @mathmale
Hint: an = a1 + d(n − 1), where a1 is the first term and d is the common difference. Neither a1 nor d is given; we have to find both. Fortunately, we are given two examples: when n=6, an=50, and when n=11, an=35.
yeah thats what i did but i got these huge numbers
an = a1 + d(n − 1) Let's substitute 50 for an and 6 for n: an = 50 + d(6 − 1)
so it would be an = 50 + 5d right?
an = a1 + d(n − 1) now subst. 35 for an and 11 for n. Type out your result, please.
and an = 35 + 10d?
so it would be an = 50 + 5d right? try it! let n=6. But there's no "n" in your equation.
You need an equation for an that depends upon a1 and n. To obtain such an equation you must have a1 and d. What have you done so far to find a1 and d?
no there isn't an n
im confused on how to do the equation
Then your formula won't function as it should. If you had the correct formula, I could give y ou "n=6" and your formula would return 50. Know why that is?
no i don't, why is that?
You are told that if n=6, an=a6=50, right?
yess
and you are told that if n=11, an=a11=35. Right?
so what do i do
I was going thru the procedure of helping you set u p 2 equations from which you could obtain a1 and d. But you typed, "so it would be an = 50 + 5d right?" You might need to develop some patience here.
Starting over with the general formula: an=a1+d(n-1). Case 1: when n=6 an =50. The general formula becomes 50=a1+d(6-1), or 50=a1+d(5). Case 2: when n=11, an=35. the general formula then becomes what?
Making any progress?
35= a1+d(11-1)
or d(10)
Better yet, a1+10d=35. What was the other equation?
a1+5d=50
but how do i get the answer from these 2 equations
Yes. Now please solve these 2 equations simultaneously. You could find either d or a1. What methods have you used in the past to solve simultaneous linear equations?
by solving them do i divide?
like 35 divided by 10 and 50 divided by 5?
No. You'll need to eliminate one variable and then find the other. Which var. do you want to find first, d or a1?
a1
You have 2 equations: a1+5d=50 and a1+10d=35 How would you eliminate the variable d, so that you could find the value of a1?
multiplying ?
i really wanna figure this out fast because I'm timed in my assignment please and i just need this one question i don't understand @mathmale
Commonly used methods include 1) addition/subtraction; 2) substitution; 3) graphing.
You have 2 equations: a1+5d=50 and a1+10d=35 Multiply the 1st one by -2 and write out your result here.
multply -2 to what
-2(a1+5d=50) = ??
Multiply the first equation by -2.
Result?
where did you get the -2 from?
I'll explain that later if you like. You sound like you're in a hurry. Let's solve the problem and then come back to discuss the details.
By the way, the method we're using here is the "addition / subtraction method."
-2(a1+5d=50) = -2a1 -10d = -100
Can you see how I obtained that? If so, combine your two equations:
2a1 -10d = -100 a1 + 10d = 35 ---------------- ???
a(30)=65-3(30-1) so we solve for a(30)
so its -22
Excuse me, that should be -2a1 -10d = -100 a1 + 10d = 35 ----------------
Add each colum separately. Combine: -2a1 +a1 ----- ?
Hint: -2 + 1 = -1 Hint #2: -2k + 1k = -1k
I'm sorry this is going slowly. My goal is to help you understand this material, as well as to finish the problem in time to get credit for it. OpenStudy says you're "just looking around." I need your full attention.
a(30)=65-3(30-1) so we solve for a(30) is correct. Why didn't you say you had tentatively arrived at a conclusion? What's your final answer? a(30)= ?
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