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Math Help - Arithmetic Sequences - T5 = 3T2 and S6= 36, find the first three terms.

  1. #1
    Dev
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    Arithmetic Sequences - T5 = 3T2 and S6= 36, find the first three terms.

    Yo, some questions.

    As in how do u do these. Thanks.

    T5 = 3T2 and S6= 36, find the first three terms.

    also

    Sn = 5n2(sqaured that is) + 3n - Find the first three terms and thus find a formula for Tn

    Sweet as Thanks.
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  2. #2
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    a_5=a_2+(5-2)d and we know that a_5=3a_2 so 2a_2=3d. And we were also given: 36=S_6=\frac {6(a_1+a_6)}2=3((a_2-d)+(a_2+4d))=3(2a_2+3d)=18d\rightarrow d=2. Using this: a_2=6. So the first three terms: 4, 6, 8.

    For the second problem: S_n=n(3+5n)=\frac{n(6+10n)}2=\frac{n(6+10(n-1)+10)}2=\frac{n(2\cdot 8+(n-1)\cdot 10)}2. You can see that a_1=8 and d=10 therefore a_n=8+(n-1)\cdot 10 and the first three terms: 8, 18, 28.
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  3. #3
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    Hello, Dev!

    We are expected to know these formulas:

    . . n^{th}\text{ term: }\;T_n \:=\:a + (n-1)d

    . . \text{Sum of first }n\text{ terms: }\;S_n \;=\;\frac{n}{2}[2a + (n-1)d]

    . . . . where: . \begin{Bmatrix}a &=& \text{first term} \\ d &=& \text{common diff.} \\ n &=& \text{no. of terms} \end{Bmatrix}


    T_5 \:=\: 3\!\cdot\!T_2\:\text{ and }\:S_6\:=\: 36
    Find the first three terms.
    2nd term: . T_2 \:=\:a+d
    5th term: . T_5 \:=\:a+4d

    T_5 = 3\!\cdot\!T_2 \quad\Rightarrow\quad a + 4d \:=\:3(a+d) \quad\Rightarrow\quad d \:=\:2a .[1]

    S_6 = 36\!:\;\;36 \:=\:\frac{6}{2}[2a + 5d] \:=\:36 \quad\Rightarrow\quad 2x + 5d \:=\:12 .[2]

    Substitute [1] into [2]: . 2a + 5(2a) \:=\:12 \quad\Rightarrow\quad 12a \:=\:12 \quad\Rightarrow\quad\boxed{ a \:=\:1}

    Substitute into [1]: . d = 2(1) \quad\Rightarrow\quad\boxed{d \:= \:2}

    The first term is 1, the common difference is 2.


    Therefore, the first three terms are: .1, 3, 5.

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  4. #4
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    Hello again, Dev!

    S_n \:=\: 5n^2  + 3n
    Find the first three terms and thus find a formula for T_n
    The sum of the first n terms is: . S_n \:=\:\frac{n}{2}[2a + (n-1)d]

    The sum of the first one term is: . a
    The formula says: . S_1 \:=\:5(1^2) + 3(1) \:=\:8
    . . Hence: . \boxed{ a\:=\:8} .[1]

    The sum of the first two terms is: . a + (a+d) \:=\:2a+d
    The formula says: . S_2 \:=\:5(2^2) + 3(2) \:=\:26
    . . Hence: . 2a+d \:=\:26 .[2]

    Substitute [1] into [2]: . 2(8) + d \:=\:26 \quad\Rightarrow\quad\boxed{ d \:=\:10}

    The first term is: a = 8 . . . the common difference is: d = 10


    \text{The first three terms are: }\;{\bf{\color{red}8,\:18,\:28}}

    \text{The }n^{th}\text{ term is: }\;T_n \:=\:8 +10(n-1) \quad\Rightarrow\quad {\bf{\color{red} T_n \:=\:10n -2}}

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