SBMPTN Zone : Barisan dan Deret Geometri

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Tipe:

Standar SBMPTN HOTS

No.

Misalkan U_n menyatakan suku ke-n suatu barisan geometri dengan rasio positif. Jika diketahui U_5=16 dan {\log U_4+\log U_5-\log U_6=\log 4}, maka nilai U_4 adalah....

1. 2
2. 4
3. 6

4. 8
5. 16

Ganesha Operation

$$\begin{array}{rl}\log U_4+\log U_5-\log U_6& =\log 4\\ \log U_4+2\log U_5-\log U_6& =\log 4+\log U_5\\ \log U_4+\log {U_5}^2-\log U_6& =\log 4+\log 16\\ \log U_4+\log (U_4\cdot U_6)-\log U_6& =\log 64\\ \log U_4+\log U_4+\log U_6-\log U_6& =\log 64\\ 2\log U_4& =\log 64\\ \log {U_4}^2& =\log 64\\ {U_4}^2& =64\\ U_4& =8\end{array}$$

Jadi, nilai U_4 adalah 16.

No.

Suku ke-n suatu deret geometri adalah U_n. Jika U_4 = 2p-18; U_8 = p+40 dan \dfrac{U_5-U_3}{U_3}= 3 maka jumlah 4 suku pertama deret tesebut adalah

1. 14
2. 21{,}5
3. 23{,}75

4. 26{,}25
5. 29{,}50

$$\begin{array}{rl}{\displaystyle \frac{U_5-U_3}{U_3}}& =3\\ {\displaystyle \frac{a{r}^4-a{r}^2}{a{r}^2}}& =3\\ r^2-1& =3\\ r^2& =4\\ r& =2\end{array}$$

$$\begin{array}{rl}{\displaystyle \frac{U_8}{U_4}}& ={\displaystyle \frac{a{r}^7}{a{r}^3}}\\ {\displaystyle \frac{p+40}{2p-18}}& =r^4\\ {\displaystyle \frac{p+40}{2p-18}}& =16\\ p+40& =32p-288\\ p& ={\displaystyle \frac{328}{31}}\end{array}$$

$$\begin{array}{rl}{U}_4& =2p-18\\ a{r}^3& =2\left({\displaystyle \frac{328}{31}}\right)-18\\ a(8)& ={\displaystyle \frac{656}{3}}1-18\\ 8a& ={\displaystyle \frac{98}{31}}\\ a& ={\displaystyle \frac{49}{124}}\end{array}$$

$$\begin{array}{rl}{S}_n& ={\displaystyle \frac{a(r^n-1)}{r-1}}\\ s_4& ={\displaystyle \frac{\left({\displaystyle \frac{49}{124}}\right)(2^4-1)}{2-1}}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{49}{124}}\right)(16-1)}{1}}\\ & =\left({\displaystyle \frac{49}{124}}\right)\left(15\right)\\ & ={\displaystyle \frac{735}{124}}\\ & =5,93\end{array}$$

Jadi, jumlah 4 suku pertama deret tesebut adalah \dfrac{735}{124}. TIDAK ADA JAWABAN

No.

Suku pertama dan suku kedua suatu deret geometri berturut-turut adalah p^2 dan p^x. Jika suku ke lima deret tersebut adalah p^{18} maka x=...

1. 1
2. 2
   
3. 4
   

4. 6
5. 8
   

U_1=a=p^2,
U_2=p^x,
U_5=p^{18}

$$\begin{array}{rl}r& ={\displaystyle \frac{U_2}{U_1}}\\ & ={\displaystyle \frac{p^x}{p^2}}\\ & =p^{x-2}\end{array}$$

$$\begin{array}{rl}{U}_5& =p^{18}\\ a{r}^4& =p^{18}\\ \left(p^2\right){\left(p^{x-2}\right)}^4& =p^{18}\\ \left(p^2\right)\left(p^{4x-8}\right)& =p^{18}\\ p^{2+4x-8}& =p^{18}\\ p^{4x-6}& =p^{18}\\ 4x-6& =18\\ 4x& =24\\ x& =\boxed{{\displaystyle \boxed{{\displaystyle 6}}}}\end{array}$$

Jadi, x=6. JAWAB: D

No.

Jika persamaan 2x^2+x+k mempunyai akar-akar x_1 dan x_2. Jika x_1, x_2 dan \dfrac12 merupakan suku pertama, kedua dan ketiga suatu deret geometri, maka tentukanlah suku ke empat deret tersebut.

Grup Telegram

$$\begin{array}{rl}{x}_1+x_2& =-{\displaystyle \frac{b}{a}}\\ & =-{\displaystyle \frac{1}{2}}\\ x_1& =-x_2-{\displaystyle \frac{1}{2}}\end{array}$$

$$\begin{array}{rl}{x_2}^2& =x_1\cdot {\displaystyle \frac{1}{2}}\\ {x_2}^2& =(-x_2-{\displaystyle \frac{1}{2}})\cdot {\displaystyle \frac{1}{2}}\\ {x_2}^2& =-{\displaystyle \frac{1}{2}}{x}_2-{\displaystyle \frac{1}{4}}\\ 4{x_2}^2& =-2x_2-1\\ 4{x_2}^2+2x_2+1& =0\\ x_2& ={\displaystyle \frac{-2\pm \sqrt{2^2-4(4)(1)}}{2(4)}}\\ & ={\displaystyle \frac{-2\pm \sqrt{4-16}}{8}}\\ & ={\displaystyle \frac{-2\pm \sqrt{-12}}{8}}\\ & ={\displaystyle \frac{-2\pm 2i\sqrt{3}}{8}}\\ & ={\displaystyle \frac{-1\pm i\sqrt{3}}{4}}\end{array}$$

$$\begin{array}{rl}{x}_1& =-x_2-{\displaystyle \frac{1}{2}}\\ & =-\left({\displaystyle \frac{-1\pm i\sqrt{3}}{4}}\right)-{\displaystyle \frac{1}{2}}\\ & ={\displaystyle \frac{1\mp i\sqrt{3}}{4}}-{\displaystyle \frac{2}{4}}\\ & ={\displaystyle \frac{-1\mp i\sqrt{3}}{4}}\end{array}$$

$$\begin{array}{rl}r& ={\displaystyle \frac{x_2}{x_1}}\\ & ={\displaystyle \frac{{\displaystyle \frac{-1\pm i\sqrt{3}}{4}}}{{\displaystyle \frac{-1\mp i\sqrt{3}}{4}}}}\\ & ={\displaystyle \frac{-1\pm i\sqrt{3}}{-1\mp i\sqrt{3}}}\cdot {\displaystyle \frac{-1\pm i\sqrt{3}}{-1\pm i\sqrt{3}}}\\ & ={\displaystyle \frac{1\mp 2i\sqrt{3}\mp 3}{1+3}}\\ & ={\displaystyle \frac{1\mp 2i\sqrt{3}\mp 3}{4}}\end{array}$$

$$\begin{array}{rl}{U}_4& ={\displaystyle \frac{1}{2}}\cdot r\\ & ={\displaystyle \frac{1}{2}}\cdot {\displaystyle \frac{1\mp 2i\sqrt{3}\mp 3}{4}}\\ & ={\displaystyle \frac{1\mp 2i\sqrt{3}\mp 3}{8}}\end{array}$$

• U_4=\dfrac{1-2i\sqrt3-3}8
  $$\begin{array}{rl}{U}_4& ={\displaystyle \frac{-2-2i\sqrt{3}}{8}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{-1-i\sqrt{3}}{4}}}}}}\end{array}$$


• U_4=\dfrac{1+2i\sqrt3+3}8
  $$\begin{array}{rl}{U}_4& ={\displaystyle \frac{4+2i\sqrt{3}}{8}}\\ & =\boxed{{\displaystyle \boxed{{\displaystyle {\displaystyle \frac{2+i\sqrt{3}}{4}}}}}}\end{array}$$

Jadi, suku ke empat deret tersebut adalah \dfrac{-1-i\sqrt3}4 atau \dfrac{2+i\sqrt3}4.