Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 years ago

Arithmetic problem ! explain please?

the sum of the first 12 terms of an arithmetic series is 186 , and the 20th term is 83 .What is the sum of the first 40 terms?

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  • 8 years ago
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    1) Let the first term of AP be 'a' and its common difference be 'd'.

    2) Sum to n terms is: S_n = (n/2){2a + (n-1)d} and

    nth term is: a_n = a + (n-1)d

    Applying these,

    S_12: 6(2a + 11d) = 186; ==> 2a + 11d = 31 .......... (i)

    a_20: a + 19d = 83 ........... (ii)

    Solving the above two equations, (i) & (ii):

    a = -12 and d = 5.

    3) Sum of first 40 terms:

    S_40: (40/2){2*(-12) + (40-1)(5)} = 20(-24 + 195)

    ==> S_40 = 3420

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  • 8 years ago

    Below is the work that I have been able to do for this problem. I am hitting a brick wall as it seems there might be some additional data that you didn't list. As it is, using 2 textbook resources, I am unable to solve this problem without more information. Is it possible that there was more information given that you did not list?

    If we can determine the function t, we can solve for any term in the sequence by substituting the term for t.

    We know that

    t(1) + t(2) + t(3) + t(4) + t(5) + t(6) + t(7) + t(8) + t(9) + t(10) + t(11) + t(12) = 186

    We also know that t(20) = 83

    The sum of the first n terms in an arithmetic sequence is defined:

    Sn = [n(t(1) + t(n))]/2

    That is Sn is the sum of the first n terms, t(1) is the first term, t(n) is the nth term, and n is the number of terms in the sequence from t(1) to t(n).

    In our case,

    Sn = 186

    t(1) = ??

    t(n) = ??

    n = 12

    186 = [12(t(1) + t(12))]/2

    186 = 6(t(1) + t(12))

    31 = (t(1) + t(12))

    Source(s): Larson, R., Hostetler, R. P., & Edwards, B. H. (2006). Calculus with analytic geometry. (8 ed.). Boston, MA: Houghton Mifflin Harcourt (HMH). and Angel, A. R., Abbott, C. D., & Runde, D. C. (2007). A survey of mathematics with applications. (8 ed., p. 291). Boston, MA: Addison-Wesley.
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