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# Sequence Examples

### Sequence application 1

### Sequence application 2

### Sequence application 3

### Sequence application 4

- Question:
- Find the eleventh term of the arthmetic sequences whose first fifth terms are -2, 5, 12, 19 and 26

- Solution
- A sequence a
_{1}, a_{2}, a_{3},...,a_{n},... is called an arithmetic sequence if each term differs from the preceding term by a fixed amount. - The common difference of the given sequence is d = 7. The nth term a
_{n}of such a sequence is given by the equation: a_{n}= a_{1}+ (n - 1)d - Thus, a
_{11}= a_{1}+ (n - 1)d = -2 + (11 - 1)7 = -2 + 10 × 7 = -2 + 70 = 68 - Therefore, the eleventh term of this sequence is 68.

- Question:
- If the second term of an arithmetic sequence is -2 and the fifth term is -17, find the seventh term.

- Solution
- The nth term a
_{n}of an arithmetic sequence is given by the equation: - a
_{n}= a_{1}+ (n - 1)d, in which d = a_{n}- a_{n-1} - a
_{2}= a_{1}+ (2 - 1)d = a_{1}+ d = -2 ... (1) - a
_{5}= a_{1}+ (5 - 1)d = a_{1}+ 4d = -17 ... (2) - (2) - (1) obtain 3d = -17 - (-2) = -17 + 2 = -15
- d = -15/3 = -5 ... (3)
- from (3) and (1) obtain a
_{1}= -d - 2 = -(-5) - 2 = 5 - 2 = 3 - Thus, a
_{7}= a_{1}+ (7 - 1)d = 3 + 6 × (-5) = 3 + (-30) = -27 - Therefore, the seventh term is -27.

- Question:
- If the two terms of a geometric sequence are 3 and 9, respectively, which term of the sequence is equal to 729?

- Solution
- A sequence a
_{1}, a_{2}, a_{3},..., a_{n},... is called a geometric sequence if each term is obtained by multiplying the preceding term by a fixed amount. - In general, a sequence of the form a
_{1}, a_{1}r, a_{1}r^{2},...,a_{1}r^{n-1},...for every positive integer n is a geometric sequence. The number r is called the common ratio, and the nth term a_{n}is given by a_{n}= a_{1}r^{n-1} - Since a
_{1}= 3 and a_{2}= 9 , so the commom ratio r = a_{2}/a_{1}= 3 - a
_{n}= a_{1}r^{n-1}= 3 × 3^{n-1}= 3^{1+n-1}= 3^{n}and 729 = 3^{6} - Thus 3
^{n}= 3^{6}, so n = 6 - Therefore, the 6th term of the sequence is 729.

- Question:
- Find the sum of the first 10 terms of arithmetic sequence whose first term is -3 and whose common difference is 5.

- Solution
- The formula for the sum of the first n terms of an arithmetic sequence is
- S
_{n}= (a_{1}+ a_{n})n/2 - For the arithmetic sequence whose the nth term is
- a
_{n}= a_{1}+ (n - 1)d - a
_{10}= a_{1}+ (10 - 1)5 = -3 + 9 × 5 = -3 + 45 = 42 - S
_{n}= (a_{1}+ a_{n})n/2 = ( -3 + 42 ) × 10/2 = 39 × 10/2 = 195 - Therefore, the sum of the first 10 terms is 195.

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