9Nomor 9PembahasanJumlah deret tak hingga memiliki nilai 94\frac{9}{4}49. Dengan suku pertama u1=au_1 = au1=a dan rasio r=−1ar = -\frac{1}{a}r=−a1. Jika a>0a > 0a>0, tentukan nilai dari 3u6−u53u_6 - u_53u6−u5.000227\frac{2}{27}272−227-\frac{2}{27}−272127\frac{1}{27}271−127-\frac{1}{27}−271PembahasanJumlah deret tak hingga memiliki nilai 94\frac{9}{4}49 atau S∞=94S_\infty = \frac{9}{4}S∞=49. Dengan u1=au_1 = au1=a dan r=−1ar = -\frac{1}{a}r=−a1. S∞=u11−rS_\infty = \frac{u_1}{1 - r}S∞=1−ru1a1−(−1a)=94\frac{a}{1 - \left(-\frac{1}{a}\right)} = \frac{9}{4}1−(−a1)a=49a1+1a=94\frac{a}{1 + \frac{1}{a}} = \frac{9}{4}1+a1a=49aa+1a=94\frac{a}{\frac{a + 1}{a}} = \frac{9}{4}aa+1a=49a⋅aa+1=94a \cdot \frac{a}{a + 1} = \frac{9}{4}a⋅a+1a=49a2a+1=94\frac{a^2}{a + 1} = \frac{9}{4}a+1a2=494a2=9(a+1)4a^2 = 9(a + 1)4a2=9(a+1)4a2=9a+94a^2 = 9a + 94a2=9a+94a2−9a−9=04a^2 - 9a - 9 = 04a2−9a−9=0 Faktorkan (4a+3)(a−3)=0(4a + 3)(a - 3) = 0(4a+3)(a−3)=0 a=−34∪a=3a = -\frac{3}{4} \cup a = 3a=−43∪a=3 Karena nilai a>0a > 0a>0 maka yang memenuhi adalah a=3a = 3a=3. Sehingga r=−1a=−13r = -\frac{1}{a} = -\frac{1}{3}r=−a1=−31. Menentukan nilai dari 3u6−u53u_6 - u_53u6−u5 u6=ar5u_6 = ar^5u6=ar5u5=ar4u_5 = ar^4u5=ar4=3ar5−ar4= 3ar^5 - ar^4=3ar5−ar4=ar4(3r−1)= ar^4(3r - 1)=ar4(3r−1)=3(−13)4(3(−13)−1)= 3\left(-\frac{1}{3}\right)^4 \left(3\left(-\frac{1}{3}\right) - 1\right)=3(−31)4(3(−31)−1)=3⋅181(−1−1)= 3 \cdot \frac{1}{81} (-1 - 1)=3⋅811(−1−1)=3⋅181(−2)= 3 \cdot \frac{1}{81} (-2)=3⋅811(−2)=127(−2)= \frac{1}{27} (-2)=271(−2)=−227= -\frac{2}{27}=−272
