Find an equation for the nth term of the arithmetic sequence. a15 = -53, a16 = -5. I have the answer I just want to make sure I'm right.
If the first term is a, and if the jump from one term to the next is d, n is the number of terms, l is the last term, then l = a + (x-1)d. Please apply this formula to your situation, showing what you've already done. Then I'd be happy to give you feedback on it.
an = -725 + 48(n - 1)
How did you find that -725? That 48?
Well because from -53 to -5 is 48. And -725 has to be apart of the equation
a part *
I agree: 'd', the jump from one term to the next, is +48. I can certainly verify for you your value of a (a1), but would like to know which method you used to determine that a = a1 = -725.
Well its the only option for a1, an = -725 - 48(n - 1) an = -725 + 48(n + 1) an = -725 + 48(n - 1) an = -725 - 48(n + 1)
I see...you're not actually doing math to determine the first term (-725), but are choosing your answer from four given choices. Highly unusual situation here, where the four choices are identical. Are you at all interested in knowing how to determine the first term if you're not given four possible answers from which to choose?
Sure!
This question was more about know the formula
I start with the most general formula applicable here: l = a + (n-1)d. I'd guess you've seen this formula or a different version of it. You've correctly identified the "jump," 'd', as being 48. Thus, we have l = a + (n-1)48.
Case 1: a(15)=-53. This tells us that when n=15, l = -53. Case 2: a(16) = -5. We've already used this info and thus don't need it again.
So, back to Case 1: l = a + (n-1) *48 Substitute l=-53 and n=15. This results in one equation in one unknown, and is therefore sufficient so that you can calculate a.
If l=a+(n-1)48, and we want a, then re-write this equation as a= l-48(n-1); substitute n=15 and l=-53.
This is all the info you need to calculate a for yourself. I've verified this; I get a = -725, as expected. Nice working with you. 'Til later! Thanks for the medal!
Join our real-time social learning platform and learn together with your friends!