EXERCISE 6 Determine the first three terms of each of the following arithmetic sequences of which: (a) the 3 rd term of the sequence is 23 and the 26 th term is 230 . (b) the 5 th term of the sequence is 19 and the 15 th term is 59 . The 15 th and 3 rd terms of an arithmetic sequence are 100 and 28 respectively. Determine the 100 th term.
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To find the first three terms of the sequences, we can use the formula for the \( n \)-th term of an arithmetic sequence: \[ a_n = a_1 + (n-1)d \] where \( a_1 \) is the first term, \( d \) is the common difference, and \( n \) is the term number. (a) The 3rd term \( a_3 = 23 \) gives us the equation: \[ a_1 + 2d = 23 \tag{1} \] The 26th term \( a_{26} = 230 \) gives us: \[ a_1 + 25d = 230 \tag{2} \] Subtracting (1) from (2) eliminates \( a_1 \): \[ (25d - 2d) = 230 - 23 \implies 23d = 207 \implies d = 9 \] Now substitute \( d \) back into (1): \[ a_1 + 2(9) = 23 \implies a_1 + 18 = 23 \implies a_1 = 5 \] So the first three terms are \( 5, 14, 23 \). (b) The 5th term \( a_5 = 19 \): \[ a_1 + 4d = 19 \tag{3} \] The 15th term \( a_{15} = 59 \): \[ a_1 + 14d = 59 \tag{4} \] Subtracting (3) from (4): \[ (14d - 4d) = 59 - 19 \implies 10d = 40 \implies d = 4 \] Substituting \( d \) back into (3): \[ a_1 + 4(4) = 19 \implies a_1 + 16 = 19 \implies a_1 = 3 \] Thus, the first three terms are \( 3, 7, 11 \). Now, for the third part with \( a_{15} = 100 \) and \( a_3 = 28 \): From \( a_3 \): \[ a_1 + 2d = 28 \tag{5} \] From \( a_{15} \): \[ a_1 + 14d = 100 \tag{6} \] Subtracting (5) from (6): \[ (14d - 2d) = 100 - 28 \implies 12d = 72 \implies d = 6 \] Substituting \( d \) into (5): \[ a_1 + 2(6) = 28 \implies a_1 + 12 = 28 \implies a_1 = 16 \] Finally, for the 100th term: \[ a_{100} = a_1 + 99d = 16 + 99(6) = 16 + 594 = 610 \] Thus, the 100th term is \( 610 \).
