Shakila B

30 Oktober 2023 16:04

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Shakila B

30 Oktober 2023 16:04

Pertanyaan

Kakk tolong bantu jawabb yaaa pada suatu deret aritmatika diketahui suku ke-3 adalah 22 dan jumlah deret suku ke-6 hingga Suku ke-9 adalah 268 - a Tentukan suku pertama deret aritmatika tersebut - b Tentukan beda dari deret tersebut - C tentukan jumlah 6 suku pertama dari deret tersebut

 

Kakk tolong bantu jawabb yaaa

 

pada suatu deret aritmatika diketahui suku ke-3 adalah 22 dan jumlah deret suku ke-6 hingga Suku ke-9 adalah 268 

- a Tentukan suku pertama deret aritmatika tersebut

- b Tentukan beda dari deret tersebut

- C tentukan jumlah 6 suku pertama dari deret tersebut 

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Jawaban terverifikasi

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Reyhan Y

30 Oktober 2023 22:33

Jawaban terverifikasi

<p>Dik U<sub>3</sub> = 22</p><p>&nbsp; &nbsp; &nbsp;U<sub>6</sub> +U<sub>7</sub> + U<sub>8</sub> + U<sub>9 </sub>= 268</p><p>Dit = a,b,S<sub>6</sub>?</p><p>Jawab&nbsp;</p><p>&nbsp; &nbsp; &nbsp; &nbsp;U<sub>3 </sub>= a+(3-1)b = 22</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;= a+2b = 22..............(1)</p><p>&nbsp; &nbsp; &nbsp;&nbsp;</p><p>&nbsp; &nbsp; &nbsp; &nbsp;U6+ U7 + U8 + U9 = 268</p><p>&nbsp; &nbsp; &nbsp; &nbsp;a+(6-1)b+ a+(7-1)b + a+(8-1)b + a+(9-1)b = 268</p><p>&nbsp; &nbsp; &nbsp; &nbsp;a+5b+a+6b+a+7b+a+8b = 268</p><p>&nbsp; &nbsp; &nbsp; &nbsp; 4a+26b = 268.............(2)</p><p>&nbsp;</p><p>&nbsp;</p><p>Gunaka metode SPLDV untuk persamaan (1)&amp;(2)</p><p>a+2b = 22. &nbsp; &nbsp; &nbsp; &nbsp; |X3| &nbsp; &nbsp;4a+8b = 88</p><p>4a+26b = 268. &nbsp;|X1|. &nbsp; 4a+26b = 268 &nbsp; _</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; -----------------------</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;-18b = -180</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; b= 10</p><p>Maka beda dari deret tersebut adalah 10</p><p>&nbsp;</p><p>U3 = a+2b = 22</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; a+2(10) = 22</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; a+20= 22</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;a=22-20 = 2</p><p>&nbsp;</p><p>Maka suku pertamanya adalah 2</p><p>&nbsp;</p><p>Sn = ½n(2a+(n-1)b)</p><p>S6 = ½.6(2.2+(6-1)10)</p><p>&nbsp; &nbsp; &nbsp; = 3(4 + (5)10)</p><p>&nbsp; &nbsp; &nbsp; &nbsp;= 3(4+50)</p><p>&nbsp; &nbsp; &nbsp; &nbsp;=3(54) = 162</p><p>&nbsp;</p><p>Maka jumlah 6 suku pertamanya adalah 162</p>

Dik U3 = 22

     U6 +U7 + U8 + U9 = 268

Dit = a,b,S6?

Jawab 

       U3 = a+(3-1)b = 22

             = a+2b = 22..............(1)

      

       U6+ U7 + U8 + U9 = 268

       a+(6-1)b+ a+(7-1)b + a+(8-1)b + a+(9-1)b = 268

       a+5b+a+6b+a+7b+a+8b = 268

        4a+26b = 268.............(2)

 

 

Gunaka metode SPLDV untuk persamaan (1)&(2)

a+2b = 22.         |X3|    4a+8b = 88

4a+26b = 268.  |X1|.   4a+26b = 268   _

                                      -----------------------

                                             -18b = -180

                                                  b= 10

Maka beda dari deret tersebut adalah 10

 

U3 = a+2b = 22

          a+2(10) = 22

          a+20= 22

           a=22-20 = 2

 

Maka suku pertamanya adalah 2

 

Sn = ½n(2a+(n-1)b)

S6 = ½.6(2.2+(6-1)10)

      = 3(4 + (5)10)

       = 3(4+50)

       =3(54) = 162

 

Maka jumlah 6 suku pertamanya adalah 162


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Sumber W

Community

31 Oktober 2023 00:39

Jawaban terverifikasi

<p>Jawaban yang tepat :</p><p>a. Suku pertama (a) = 2</p><p>b. Beda (b) = 10</p><p>c. Jumlah 6 suku pertama (S<sub>6</sub>) = 162</p><p>&nbsp;</p><p><strong>Pembahasan</strong> :&nbsp;</p><p>U<sub>3</sub> = 22</p><p>Jumlah deret suku ke-6 hingga Suku ke-9 adalah 268</p><p>&nbsp;</p><p>a. <strong>Menentukan suku pertama (a) dan beda (b)</strong></p><p>&nbsp; &nbsp; S<sub>n</sub> = n/2(2a + (n - 1)b))</p><p>&nbsp; &nbsp; S<sub>5</sub> = 5/2(2a + (5 - 1)b)</p><p>&nbsp; &nbsp; S<sub>5</sub> = 5/2(2a + 4b)</p><p>&nbsp; &nbsp; S<sub>5</sub> = 5a + 10b &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; ...(1)</p><p>&nbsp;</p><p>&nbsp; &nbsp; S<sub>9</sub> = 9/2(2a + (9 - 1)b)</p><p>&nbsp; &nbsp; S<sub>9</sub> = 9/2(2a + 8b)</p><p>&nbsp; &nbsp; S<sub>9</sub> = 9a + 36b &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; ...(2)</p><p>&nbsp;</p><p>&nbsp; &nbsp; S<sub>9</sub> - S<sub>5</sub> = 268</p><p>&nbsp; &nbsp;(9a + 36b) - (5a + 10b) = 268</p><p>&nbsp; &nbsp;9a + 36b - 5a - 10b = 268</p><p>&nbsp; &nbsp;4a + 26b = 268 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;...(3)</p><p>&nbsp;</p><p>&nbsp; &nbsp; U<sub>n</sub> = a + (n - 1)b</p><p>&nbsp; &nbsp; U<sub>3</sub> = 22&nbsp;</p><p>&nbsp; &nbsp; a + (3 - 1)b = 22</p><p>&nbsp; &nbsp; a + 2b = 22 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; ...(4)</p><p>&nbsp;</p><p>&nbsp; &nbsp;<strong> Eliminasi a dari persamaan (3) dan (4)</strong></p><p><strong>&nbsp; &nbsp;</strong> 4a + 26b = 268 &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; --&gt; 4a + 26b = 268</p><p>&nbsp; &nbsp; &nbsp;a + 2b = 22 &nbsp; &nbsp; &nbsp; (dikali 4) --&gt; 4<u>a + 8b = 88 &nbsp; &nbsp;&nbsp;</u> _</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;18b = 180</p><p>&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; b = <strong>10</strong></p><p><strong>Substitusikan b = 10 ke persamaan (4) untuk mendapatkan nilai a</strong></p><p>a + 2b = 22</p><p>a + 2(10) = 22&nbsp;</p><p>a + 20 = 22</p><p>a = 22 - 20</p><p>a = <strong>2</strong></p><p><strong>Menentukan jumlah 6 suku pertama (S<sub>6</sub>)</strong></p><p>&nbsp;S<sub>n</sub> = n/2(2a + (n - 1)b))</p><p>&nbsp;S<sub>6</sub> = 6/2(2(2) + (6 - 1)10)</p><p>&nbsp; &nbsp; &nbsp; &nbsp;= 3(4 + &nbsp;50)</p><p>&nbsp; &nbsp; &nbsp; &nbsp;= 3 x 54</p><p>&nbsp; &nbsp; &nbsp; &nbsp;= <strong>162</strong></p>

Jawaban yang tepat :

a. Suku pertama (a) = 2

b. Beda (b) = 10

c. Jumlah 6 suku pertama (S6) = 162

 

Pembahasan

U3 = 22

Jumlah deret suku ke-6 hingga Suku ke-9 adalah 268

 

a. Menentukan suku pertama (a) dan beda (b)

    Sn = n/2(2a + (n - 1)b))

    S5 = 5/2(2a + (5 - 1)b)

    S5 = 5/2(2a + 4b)

    S5 = 5a + 10b                 ...(1)

 

    S9 = 9/2(2a + (9 - 1)b)

    S9 = 9/2(2a + 8b)

    S9 = 9a + 36b                 ...(2)

 

    S9 - S5 = 268

   (9a + 36b) - (5a + 10b) = 268

   9a + 36b - 5a - 10b = 268

   4a + 26b = 268                        ...(3)

 

    Un = a + (n - 1)b

    U3 = 22 

    a + (3 - 1)b = 22

    a + 2b = 22                               ...(4)

 

    Eliminasi a dari persamaan (3) dan (4)

    4a + 26b = 268                   --> 4a + 26b = 268

     a + 2b = 22       (dikali 4) --> 4a + 8b = 88      _

                                                                   18b = 180

                                                                    b = 10

Substitusikan b = 10 ke persamaan (4) untuk mendapatkan nilai a

a + 2b = 22

a + 2(10) = 22 

a + 20 = 22

a = 22 - 20

a = 2

Menentukan jumlah 6 suku pertama (S6)

 Sn = n/2(2a + (n - 1)b))

 S6 = 6/2(2(2) + (6 - 1)10)

       = 3(4 +  50)

       = 3 x 54

       = 162


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