Question

(11940)/(4) = (4n)/(4) 2985 = n The 11th term of an arithmetic sequence is 63 and the 15th term is 87. Find the sequence and the 800th term. 87 - 63 = 24 The sequence is 63, 87, 111, 135, 159. 11, 12, 13, 14, 15

          (11940)/(4) = (4n)/(4)
2985 = n
The 11th term of an arithmetic sequence is 63 and the 15th term is 87. Find the sequence and the 800th term.
87 - 63 = 24
The sequence is 63, 87, 111, 135, 159.
11, 12, 13, 14, 15
        

Added by Purificaci-N S.

Calculus: Early Transcendentals
Calculus: Early Transcendentals
James Stewart 8th Edition
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(11940)/(4) = (4n)/(4) 2985 = n The 11th term of an arithmetic sequence is 63 and the 15th term is 87. Find the sequence and the 800th term. 87 - 63 = 24 The sequence is 63, 87, 111, 135, 159. 11, 12, 13, 14, 15
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Transcript

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00:01 Okay, in this problem, we're asked to find the eighth term of an arithmetic sequence.
00:05 So we're looking for a8, and we're told that the fourth term is 19, and the 15th term is 52.
00:22 Okay.
00:23 So what are the two things? what are some of the things we need for an arithmetic sequence? we need the common difference, and we need the first term.
00:31 So we don't have any of those.
00:32 If we can get those, we're kind of basically done.
00:36 Now, finding the common difference can be tricky when you have non -consecutive terms, but we've got a little formula for this.
00:43 Okay, so we can find the common difference here by doing a -15 minus a -4, subtracting our non -consecutive terms, and then dividing this by how far apart they are.
00:56 So essentially dividing this by 15 minus 4.
01:00 Okay? so in our problem, we'll have 52 minus 19, divided by 15 minus 4.
01:08 52 minus 19 should be 33.
01:16 15 minus 4 should give us 11...
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