SNBT Zone : Barisan dan Deret Geometri
Table of Contents
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Tipe:
No.
Misalkan Un menyatakan suku ke-n suatu barisan geometri dengan rasio positif. Jika diketahui U5 = 16 dan log U4 + log U5 − log U6 = log 4, maka nilai U4 adalah....- 2
- 4
- 6
- 8
- 16
ALTERNATIF PENYELESAIAN
\begin{aligned}
\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+\log16\\
\log U_4+\log \left(U_4\cdot U_6\right)-\log U_6&=\log64\\
\log U_4+\log U_4+\log U_6-\log U_6&=\log64\\
2\log U_4&=\log64\\
\log{U_4}^2&=\log64\\
{U_4}^2&=64\\
U_4&=8
\end{aligned}
Jadi, nilai U4 adalah 16.
No.
Suku ke-n suatu deret geometri adalah Un. Jika U4 = 2p − 18; U8 = p + 40 dan- 14
- 21,5
- 23,75
- 26,25
- 29,50
ALTERNATIF PENYELESAIAN
\begin{aligned}
\dfrac{U_5-U_3}{U_3}&= 3\\
\dfrac{ar^4-ar^2}{ar^2}&= 3\\
r^2-1&=3\\
r^2&=4\\
r&=2
\end{aligned}
\begin{aligned}
\dfrac{U_8}{U_4}&=\dfrac{ar^7}{ar^3}\\
\dfrac{p+40}{2p-18}&=r^4\\
\dfrac{p+40}{2p-18}&=16\\
p+40&=32p-288\\
p&=\dfrac{328}{31}
\end{aligned}
\begin{aligned}
U_4&=2p-18\\
ar^3&=2\left(\dfrac{328}{31}\right)-18\\
a(8)&=\dfrac{656}31-18\\
8a&=\dfrac{98}{31}\\
a&=\dfrac{49}{124}
\end{aligned}
\begin{aligned}
S_n&=\dfrac{a\left(r^n-1\right)}{r-1}\\
S_4&=\dfrac{\left(\dfrac{49}{124}\right)\left(2^4-1\right)}{2-1}\\
&=\dfrac{\left(\dfrac{49}{124}\right)\left(16-1\right)}1\\
&=\left(\dfrac{49}{124}\right)\left(15\right)\\
&=\dfrac{735}{124}\\
&=5{,}93
\end{aligned}
Jadi, jumlah 4 suku pertama deret tesebut adalah \dfrac{735}{124} .
JAWAB: TIDAK ADA
JAWAB: TIDAK ADA
No.
Suku pertama dan suku kedua suatu deret geometri berturut-turut adalah p2 dan px. Jika suku ke lima deret tersebut adalah p18 maka x = ...1 2
4
6 8
ALTERNATIF PENYELESAIAN
U1 = a = p2,
U2 = px,
U5 = p18 \begin{aligned} r&=\dfrac{U_2}{U_1}\\[3.5pt] &=\dfrac{p^x}{p^2}\\[3.5pt] &=p^{x-2} \end{aligned}
U2 = px,
U5 = p18 \begin{aligned} r&=\dfrac{U_2}{U_1}\\[3.5pt] &=\dfrac{p^x}{p^2}\\[3.5pt] &=p^{x-2} \end{aligned}
\begin{aligned}
U_5&=p^{18}\\
ar^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{\boxed{6}}
\end{aligned}
Jadi, x = 6.
JAWAB: D
JAWAB: D
No.
Jika persamaan 2x2 + x + k mempunyai akar-akar x1 dan x2. Jika x1, x2 danALTERNATIF PENYELESAIAN
\begin{aligned}
x_1+x_2&=-\dfrac{b}a\\
&=-\dfrac12\\
x_1&=-x_2-\dfrac12
\end{aligned}
\begin{aligned}
{x_2}^2&=x_1\cdot\dfrac12\\
{x_2}^2&=\left(-x_2-\dfrac12\right)\cdot\dfrac12\\
{x_2}^2&=-\dfrac12x_2-\dfrac14\\
4{x_2}^2&=-2x_2-1\\
4{x_2}^2+2x_2+1&=0\\
x_2&=\dfrac{-2\pm\sqrt{2^2-4(4)(1)}}{2(4)}\\
&=\dfrac{-2\pm\sqrt{4-16}}8\\
&=\dfrac{-2\pm\sqrt{-12}}8\\
&=\dfrac{-2\pm2i\sqrt3}8\\
&=\dfrac{-1\pm i\sqrt3}4
\end{aligned}
\begin{aligned}
x_1&=-x_2-\dfrac12\\
&=-\left(\dfrac{-1\pm i\sqrt3}4 \right)-\dfrac12\\
&=\dfrac{1\mp i\sqrt3}4-\dfrac24\\
&=\dfrac{-1\mp i\sqrt3}4
\end{aligned}
\begin{aligned}
r&=\dfrac{x_2}{x_1}\\
&=\dfrac{\dfrac{-1\pm i\sqrt3}4 }{\dfrac{-1\mp i\sqrt3}4}\\
&=\dfrac{-1\pm i\sqrt3}{-1\mp i\sqrt3}{\color{red}\cdot\dfrac{-1\pm i\sqrt3}{-1\pm i\sqrt3}}\\
&=\dfrac{1\mp2i\sqrt3\mp3}{1+3}\\
&=\dfrac{1\mp2i\sqrt3\mp3}4
\end{aligned}
\begin{aligned}
U_4&=\dfrac12\cdot r\\
&=\dfrac12\cdot \dfrac{1\mp2i\sqrt3\mp3}4\\
&=\dfrac{1\mp2i\sqrt3\mp3}8
\end{aligned}
U_4=\dfrac{1-2i\sqrt3-3}8 \begin{aligned} U_4&=\dfrac{-2-2i\sqrt3}8\\ &=\boxed{\boxed{\dfrac{-1-i\sqrt3}4}} \end{aligned}U_4=\dfrac{1+2i\sqrt3+3}8 \begin{aligned} U_4&=\dfrac{4+2i\sqrt3}8\\ &=\boxed{\boxed{\dfrac{2+i\sqrt3}4}} \end{aligned}
Jadi, suku ke empat deret tersebut adalah \dfrac{-1-i\sqrt3}4 atau \dfrac{2+i\sqrt3}4 .
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