Arithmetic Progression - $@mEe
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An arithmetic progression is a sequence of numbers where the difference between consecutive terms is constant. The document provides examples of arithmetic progressions and discusses formulas for calculating the nth term in a sequence, the sum of terms in a finite arithmetic progression, and solving problems involving finding the number of terms, first term, or common difference given values in the sequence.
![The sum ofn terms, we find as,
Sum = n [(first term + last term) / 2]
=Now last term will be = a + (n-1) d
Therefore,
Sum(Sn) =n [{a + a + (n-1) d } /2 ]
= n/2 [ 2a + (n+1)d]](https://image.slidesharecdn.com/sameesproject-arithmeticprogression-141215101721-conversion-gate01/85/Arithmetic-Progression-mEe-13-320.jpg)
Arithmetic Progression - $@mEe
- 3. In mathematics, an arithmetic progression (AP) or arithmetic sequence is a sequence of numbers such that the difference between the consecutive terms is constant. For instance, the sequence 5, 7, 9, 11, 13, 15 … is an arithmetic progression with common difference of 2.
- 4. A finite portion of an arithmetic progression is called a finite arithmetic progression and sometimes just called an arithmetic progression. The sum of a finite arithmetic progression is called an arithmetic series.
- 5. 3 , 7 , 11 , 15 , 19 , 23 …… : Commom Difference (d) = 4 (11 , 15) : Starting Number (a) = 3
- 6. If the initial term of an arithmetic progression is and the common difference of successive members isd, then thenth term of the sequence:
- 7. 3 , 7 , 11 , 15 , 19 , 23 …… Each time you want another term in the sequence you’d add d. This would mean the second term was the first term plus d. The third term is the first term plus d plus d (added twice). The fourth term is the first term plus d plus d plus d (added three times). So you can see to get the nth term we’d take the first term and add d (n - 1) times. a a n d n 1 Try this to get the 5th term. 3 5 14 3 16 19 5 a
- 8. 1 2 3
- 9. a 1 2 3 4
- 11. a = 1 d = 3
- 12. The fourth term is 3 and the 20th term is 35. Find the first term and both a term generating formula and a recursive formula for this sequence. How many differences would you add to get from the 4th term to the 20th term? 3 35 4 20 a a a a 16d 20 4 Solve this for d d = 2 The fourth term is the first term plus 3 common differences. a a 3d 4 1 3 (2) 35 3 3 1 a We have all the info we need to express these sequences.
- 13. The sum ofn terms, we find as, Sum = n [(first term + last term) / 2] =Now last term will be = a + (n-1) d Therefore, Sum(Sn) =n [{a + a + (n-1) d } /2 ] = n/2 [ 2a + (n+1)d]
- 14. •Solution. 1) First term is a = 100 , an = 500 2) Common difference is d = 105 -100 = 5 3) nth term is an = a + (n-1)d 4) 500 = 100 + (n-1)5 5) 500 - 100 = 5(n – 1) 6) 400 = 5(n – 1) 7) 5(n – 1) = 400 8) 5(n – 1) = 400 9) n – 1 = 400/5 10) n - 1 = 80 11) n = 80 + 1 12) n = 81 Hence the no. of terms are 81.