Asked by Aly
Find the sum of the first 33 terms of the arithmetic sequence whose first term is 3 and whose 33rd term is -253.
Answers
Answered by
herp_derp
If you are working on arithmetic sequence, I assume you know this equation.
x = a + d(n-1)
Okay, so the first thing to do is to find the common difference.
-253 = 3 + d(33-1)
Solve for "d" and you get
d = -8
Next, I also assume you know the equation for the sum of an arithmetic sequence.
(n/2)(2a + (n-1)d)
So, plug the numbers in the equation and...
sum = (33/2)(2 x 3 - 8(33-1))
sum = -4125
x = a + d(n-1)
Okay, so the first thing to do is to find the common difference.
-253 = 3 + d(33-1)
Solve for "d" and you get
d = -8
Next, I also assume you know the equation for the sum of an arithmetic sequence.
(n/2)(2a + (n-1)d)
So, plug the numbers in the equation and...
sum = (33/2)(2 x 3 - 8(33-1))
sum = -4125
Answered by
Reiny
given:
a = 3
term(33) = a+32d = -253
32d = -256
d = -8
sum(33) = (33/2)(first + last)
= (33/2)(3 -253) = -4125
a = 3
term(33) = a+32d = -253
32d = -256
d = -8
sum(33) = (33/2)(first + last)
= (33/2)(3 -253) = -4125
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.