Math Problem Statement

using the following sequence of numbers, extrapolate on the same pattern another 10 places 0,4000,9000,14980,21920,29800,38600,48310,58900,70360,82680,95820,109790,124550,140100,156420

Solution

To extrapolate the given sequence, we first analyze the pattern between the terms:

Differences between consecutive terms:

  1. 40000=40004000 - 0 = 4000
  2. 90004000=50009000 - 4000 = 5000
  3. 149809000=598014980 - 9000 = 5980
  4. 2192014980=694021920 - 14980 = 6940
  5. 2980021920=788029800 - 21920 = 7880
  6. 3860029800=880038600 - 29800 = 8800
  7. 4831038600=971048310 - 38600 = 9710
  8. 5890048310=1059058900 - 48310 = 10590
  9. 7036058900=1146070360 - 58900 = 11460
  10. 8268070360=1232082680 - 70360 = 12320
  11. 9582082680=1314095820 - 82680 = 13140
  12. 10979095820=13970109790 - 95820 = 13970
  13. 124550109790=14760124550 - 109790 = 14760
  14. 140100124550=15550140100 - 124550 = 15550
  15. 156420140100=16320156420 - 140100 = 16320

Observations:

The differences themselves form an increasing arithmetic sequence, with a common difference of 980980 (e.g., 50004000=1000,59805000=980,5000 - 4000 = 1000, 5980 - 5000 = 980, etc.).

Formula for the differences:

Let the difference at the nn-th position be dnd_n. The formula for dnd_n is: dn=4000+(n1)980d_n = 4000 + (n - 1) \cdot 980

Generating the next 10 terms:

Using the last term in the sequence (156420156420) as the starting point, we calculate the next differences and add them cumulatively.

  1. d16=4000+15980=16480d_{16} = 4000 + 15 \cdot 980 = 16480
  2. d17=4000+16980=17460d_{17} = 4000 + 16 \cdot 980 = 17460
  3. d18=4000+17980=18440d_{18} = 4000 + 17 \cdot 980 = 18440
  4. d19=4000+18980=19420d_{19} = 4000 + 18 \cdot 980 = 19420
  5. d20=4000+19980=20400d_{20} = 4000 + 19 \cdot 980 = 20400
  6. d21=4000+20980=21380d_{21} = 4000 + 20 \cdot 980 = 21380
  7. d22=4000+21980=22360d_{22} = 4000 + 21 \cdot 980 = 22360
  8. d23=4000+22980=23340d_{23} = 4000 + 22 \cdot 980 = 23340
  9. d24=4000+23980=24320d_{24} = 4000 + 23 \cdot 980 = 24320
  10. d25=4000+24980=25300d_{25} = 4000 + 24 \cdot 980 = 25300

Adding these to the previous term:

  1. 156420+16480=172900156420 + 16480 = 172900
  2. 172900+17460=190360172900 + 17460 = 190360
  3. 190360+18440=208800190360 + 18440 = 208800
  4. 208800+19420=228220208800 + 19420 = 228220
  5. 228220+20400=248620228220 + 20400 = 248620
  6. 248620+21380=270000248620 + 21380 = 270000
  7. 270000+22360=292360270000 + 22360 = 292360
  8. 292360+23340=315700292360 + 23340 = 315700
  9. 315700+24320=340020315700 + 24320 = 340020
  10. 340020+25300=365320340020 + 25300 = 365320

Final Sequence:

0,4000,9000,14980,21920,29800,38600,48310,58900,70360,82680,95820,109790,124550,140100,156420,172900,190360,208800,228220,248620,270000,292360,315700,340020,3653200, 4000, 9000, 14980, 21920, 29800, 38600, 48310, 58900, 70360, 82680, 95820, 109790, 124550, 140100, 156420, 172900, 190360, 208800, 228220, 248620, 270000, 292360, 315700, 340020, 365320

Would you like a detailed explanation of the process or a graph of this sequence?


5 Follow-Up Questions:

  1. Would you like to derive a formula for the nn-th term of the sequence?
  2. Can we analyze how quickly the sequence grows as nn increases?
  3. Do you want to calculate specific properties like the sum of these terms?
  4. Should we graph the sequence to visualize its growth?
  5. Can this pattern be applied to a real-world problem you're solving?

Tip:

When working with sequences, always check for patterns in differences or ratios to identify arithmetic or geometric progressions.

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Math Problem Analysis

Mathematical Concepts

Arithmetic Sequences
Difference Patterns
Extrapolation

Formulas

d_n = 4000 + (n - 1) * 980
Term_n = Previous_Term + d_n

Theorems

Arithmetic Sequence Formula

Suitable Grade Level

Grades 9-12