Algebra 2 Arithmetic Series Help?

"Find the sum of the first 15 terms of the series if the 5th term is 29 and the 8th term is 56."

Comments

  • a5= 29, a8= 56.

    the nth term of an Arithmetic Progression is given by,

    an= a + (n-1)d

    so, you have to calculate the value of a(first term) and d (common difference) here.

    a5= a + 4d = 29 => a= 29 - 4d

    a8= a + 7d = 56 => a= 56 - 7d

    => 29 - 4d = 56 - 7d

    => 3d = 27

    => d= 9.

    a + 4d = 29

    => a + (4*9) = 29

    => a = -7.

    sum of first 15 terms of the A.P:

    S = 2a + (n-1)d

    => S = -14 + (15-1)*9

    => S = -14 + 126

    => S = 112.

  • An arithmetic series is one where the next number in the series is equal to the previous one plus a constant

    e.g. 1, 3, 5, 7, 9 adding 2 each time

    or 4,9,14,19,24 adding 5 each time

    if we assume the series starts with x, and adds y each time then the first five terms are:

    x, x+y, x+2y, x+3y, x+4y

    so the fifth term is x+4y = 29

    the eight term is x+?y = 56

    from these two you can work out what x and y are, and then you can write down the first 15 terms and add them together

  • 14 ? (-fifty 4 + 9i) i =5 nth term = 9i - fifty 4 nth term = 9(14)-fifty 4 an = seventy two an = 9i- fifty 4 a = 40 5 - fifty 4 a = - 9 -- first term worry-loose distinction= 9 an = a+ (n-a million)*d seventy two = -9 +9n - 9 seventy two = 9n - 18 9n = seventy two + 18 9n = ninety n = ninety/9 n = 10 Sum = n*(a+an)/2 Sum = 10*(-9+seventy two)/2 Sum = 5*(sixty 3) Sum = 315. . . .ans.

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