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chillhill

  • one year ago

Find the first two terms of an arithmetic sequence if the sixth term is 21 and the sum of the first 17th terms is 0.

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  1. mathstudent55
    • one year ago
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    Arithmetic sequence: a1 = a1 a2 = a1 + d a3 = a1 + 2d a4 = a1 + 3d a5 = a1 + 4d a6 = a1 + 5d The 6th term, a6, equal a1 + 5d. We know the 6th term is 21, so we get this equation: 21 = a1 + 5d a1 + 5d = 21 (Here is Eq. 1) Sum: Sn = [n(a1 + an)]/2 S17 = [17(a1 + a17)]/2 a17 = a1 + 16d S17 = [17(a1 + a1 + 16d)]/2 We are told that the sum of the first 17 term is 0, so we get the second equation. 0 = [17(a1 + a1 + 16d)]/2 0 = 17(2a1 + 16d)/2 0 = 17a1 + 136d 17a1 + 136d = 0 (Here is the 2nd equation) Now we solve these equations simultaneously: a1 + 5d = 21 (Eq. 1) 17a1 + 136d = 0 (Eq. 2) Multiply Eq. 1 by -17 and add to Eq. 2: 51d = -357 d = -7 a1 + 5(-7) = 21 a1 - 35 = 21 a1 = 56

  2. mathstudent55
    • one year ago
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    a1 = 56 a2 = a1 + d a2 = 56 + (-7) = 49 a1 = 56 & a2 = 49

  3. Jhannybean
    • one year ago
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    \[a_n = a_1+(n-1)d\]\[S_n = \frac{n(a_1+a_n)}{2}\] \[a_6 =a_1+(6-1)d =21 \implies \color{red}{21 = a_1+5d}\]\[S_{17} =\frac{17(a_1+\color{blue}{a_{17}})}{2} = 0\ \\ \qquad a_{17}=a_1 +(17-1)d \implies \color{blue}{a_{17} = a_1 +16d}\]\[S_{17} =\frac{17(a_1+a_1+16d)}{2}=0 \\ S_{17} =\frac{17(2a_1+16d)}{2}=0 \\ \qquad \implies \frac{17 \cdot 2(a_1+8d)}{2}=0 \\ \qquad \implies 17(a_1+8d)=0 \\ \qquad \implies a_1+8d = 0 \\ \qquad \implies \color{orange}{a_1 =-8d}\]We can use this to find d, then plug it into the red equation. \[\color{orange}{a_1=-8d} \implies \color{green}{d=-\frac{a_1}{8}}\]\[21=a_1+5\left(\color{green}{-\frac{a_1}{8}}\right)\]\[21=a_1-\frac{5a_1}{8}\qquad \implies 21=a_1\left(1-\frac{5}{8}\right) \qquad \implies 21 = \frac{3}{8}a_1\]\[\color{purple}{a_1=56}\]Now that we've found our first term, we can use that to find the difference, d, and then our second term using the formula for arithmetic sequence.\[\color{pink}{d=-\frac{a_1}{8} = -\frac{56}{8} = -7}\]\[a_2 =\color{purple}{ a_1}+(n-1)\color{pink}{d} \\ a_2=56+(2-1)(-7) \\ a_2=56-7 \\ \color{gold}{a_2=49}\]

  4. mathstudent55
    • one year ago
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    @Jhannybean That is beautiful. I'm almost in tears just looking at it!!! Just look at the use of colors!

  5. imqwerty
    • one year ago
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    looks like u got a color wand :D

  6. Jhannybean
    • one year ago
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    I had to.... I confused myself otherwise

  7. imqwerty
    • one year ago
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    :)

  8. imqwerty
    • one year ago
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    i make circles on useful results nd info to note them ( ͡° ͜ʖ ͡°)

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