Question

Kuis +50:
{¼(4p³-9p²), (p-2)³, p²} membentuk deret aritmatika
Jika jumlah 6 suku pertamanya adalah -72, tentukan:

Jumlah 3 suku pertama deret aritmatika tersebut!
Jawab = [tex]\left[....,\frac{...\pm...\sqrt{...}}{...}\right][/tex]

Answer 1

Jumlah 3 suku pertama deret aritmatika tersebut adalah:[tex]\left[\:\bf648\,,\ \dfrac{-3969\pm1377\sqrt{65}}{1024}\:\right][/tex] PembahasanDeret Aritmetika{¼(4p³–9p²), (p–2)³, p²} membentuk deret aritmatika.⇒ a = ¼(4p³–9p²)⇒ b = (p–2)³ – ¼(4p³–9p²)Jika jumlah 6 suku pertamanya adalah –72, maka:[tex]\begin{aligned}S_6&=\frac{6}{2}(2a+(6-1)b)\\&=3(2a+5b)\\&=6a+15b\end{aligned}[/tex][tex]\begin{aligned}{\Rightarrow\ }-72&=6\left(\frac{1}{4}\left(4p^3-9p^2\right)\right)\\&{\quad}+15\left((p-2)^3-\frac{1}{4}\left(4p^3-9p^2\right)\right)\\{\Rightarrow\ }-24&=2\left(\frac{1}{4}\left(4p^3-9p^2\right)\right)\\&{\quad}+5\left((p-2)^3-\frac{1}{4}\left(4p^3-9p^2\right)\right)\\&=\left(\frac{2}{4}-\frac{5}{4}\right)\left(4p^3-9p^2\right)+5(p-2)^3\\&=\frac{-3\left(4p^3-9p^2\right)+20(p-2)^3}{4}\end{aligned}[/tex][tex]\begin{aligned}{\Rightarrow\ }-96&=-3\left(4p^3-9p^2\right)+20(p-2)^3\\&=-12p^3+27p^2+20\left(p^3-6p^2+12p-8\right)\\&=(-12+20)p^3+(27-120)p^2+240p-160\\&=8p^3-93p^2+240p-160\\{\Rightarrow\ }0&=8p^3-93p^2+240p-64\end{aligned}[/tex]Faktor dari koefisien [tex]p^3[/tex] (4) = 1, 2, 4Faktor dari koefisien [tex]p^0[/tex] (64) = 1, 2, 4, 8, 16, 32, 64Dari [tex]{}\pm \dfrac{1, 2, 4, 8, 16, 32, 64}{1, 2, 4}[/tex]  ditemukan akar p = 8/1 = 8.Maka, faktorkan terhadap [tex](p-8)[/tex].[tex]\begin{aligned}0&=(p-8)\left(8p^2\right)-29p^2+240p-64\\&=(p-8)\left(8p^2-29p\right)+8p-64\\&=(p-8)\left(8p^2-29p+8\right)\\\therefore\ &p_1={\bf8}\,,\ 8p^2-29p+8=0\end{aligned}[/tex][tex]\begin{aligned}&8p^2-29p+8=0\\&{\Rightarrow\ }8p^2-29p=-8\\&{\Rightarrow\ }p^2-\frac{29}{8}p=-1\\&{\Rightarrow\ }p^2-2\cdot\frac{29}{16}p+\left(\frac{29}{16}\right)^2=-1+\left(\frac{29}{16}\right)^2\\&{\Rightarrow\ }\left(p-\frac{29}{16}\right)^2=-1+\frac{841}{256}=\frac{585}{256}\\&{\Rightarrow\ }p_{2,3}-\frac{29}{16}={}\pm\sqrt{\frac{585}{256}}={}\pm\sqrt{\frac{65\cdot9}{256}}\\&{\Rightarrow\ }p_{2,3}-\frac{29}{16}={}\pm\frac{3\sqrt{65}}{16}\end{aligned}[/tex][tex]\begin{aligned}&{\therefore\ \ }p_{2,3}=\bf\frac{29\pm3\sqrt{65}}{16}\end{aligned}[/tex]Jumlah 3 suku pertama sebuah deret aritmetika sama dengan 3 kali suku kedua, karena:[tex]\begin{aligned}S_3&=\frac{3}{2}\bigl[\:2U_1+(3-1)b\:\bigr]\\&=3U_1+\frac{3}{\cancel{2}}\cdot\cancel{2}b\\&=3U_1+3b\\&=3(U_1+b)\\&=3U_2\end{aligned}[/tex]Oleh karena itu, untuk deret aritmetika ini:[tex]S_3=3(p-2)^3[/tex]Untuk [tex]p_1={\bf8}[/tex] :[tex]\begin{aligned}S_{3a}&=3(8-2)^3\\&=3\cdot6^3\\&=3\cdot216\\\therefore\ S_{3a}&=\bf648\end{aligned}[/tex]Untuk [tex]p_{2,3}=\bf\dfrac{29\pm3\sqrt{65}}{16}[/tex] :[tex]\begin{aligned}S_{3b}&=3\left(\frac{29\pm3\sqrt{65}}{16}-2\right)^3\\&=3\left(\frac{29\pm3\sqrt{65}-32}{16}\right)^3\\&=\frac{3}{16^3}\left(-3\pm3\sqrt{65}\right)^3\\&=\frac{3}{16^3}\left(3\left(-1\pm\sqrt{65}\right)\right)^3\\&=\frac{3\cdot{3}^3}{16^3}\left(-1\pm\sqrt{65}\right)^3\\&=\frac{81}{2^{12}}\left(-1\pm\sqrt{65}\right)\left(-1\pm\sqrt{65}\right)^2\\&=\frac{81}{2^{12}}\left(-1\pm\sqrt{65}\right)\left(1+65\mp2\sqrt{65}\right)\end{aligned}[/tex][tex]\begin{aligned}S_{3b}&=\frac{81}{2^{12}}\left(-1\pm\sqrt{65}\right)\left(66\mp2\sqrt{65}\right)\\&=\frac{81}{2^{11}}\left(-1\pm\sqrt{65}\right)\left(33\mp\sqrt{65}\right)\\&=\frac{81}{2^{11}}\left(-33\pm33\sqrt{65}-\left(\mp\sqrt{65}\right)-65\right)\\&=\frac{81}{2^{11}}\left(-98\pm34\sqrt{65}\right)\\&=\frac{81}{2^{10}}\left(-49\pm17\sqrt{65}\right)\\&=\bf\frac{-3969\pm1377\sqrt{65}}{1024}\end{aligned}[/tex] KESIMPULAN∴  Jumlah 3 suku pertama deret aritmatika tersebut adalah:[tex]\left[\:\bf648\,,\ \dfrac{-3969\pm1377\sqrt{65}}{1024}\:\right][/tex]