Jawab: A

$x^2-(2p+4)x+(3p+4)=0$

Barisan geometri $x_1,p,x_2$

$$\begin{array}{rl}& \frac{p}{x_1}=\frac{x_2}{p}\\ \\ & p^2=x_1.x_2\\ \\ & p^2=3p+4\\ \\ & p^2-3p-4=0\\ \\ & (p-4)(p+1)=0\\ \\ & p=4\text{ atau }p=-1\end{array}$$

Karena $x_1$ dan $x_2$ adalah bilangan bulat, maka $x_1=1$ dan $x_2=1$

Deret geometri adalah $1,-1,1$

$$\begin{array}{rl}& U_{12}=a.r^{n-1}\\ \\ & U_{12}=1.(-1{)}^{12-1}\\ \\ & U_{12}=1.(-1{)}^{11}\\ \\ & U_{12}=1.-1\\ \\ & U_{12}=-1\end{array}$$

Jawab: D

Menentukan jumlah $1+3+5+7+9+11+13+15+17+\dots +193+195+197$

Menentukan jumlah $-3-9-15-\dots -195$

Maka $9801-2.(3267)=3267$

Jawab: A

$$\begin{array}{rl}\frac{3}{2}+\frac{3}{4}+\frac{9}{8}+\frac{15}{16}+\dots & =m\\ \\ \text{masing-masing suku dikurangi 1 sebanyak 10 suku}& \\ \\ \frac{3}{2}-1+\frac{3}{4}-1+\frac{9}{8}-1+\frac{15}{16}-1+\dots & =m-10\\ \\ \frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\dots & =m-10\dots (1)\end{array}$$

Bentuk ${\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{4}}+{\displaystyle \frac{1}{8}}-{\displaystyle \frac{1}{16}}+\dots$ merupakan deret geometri dengan $a={\displaystyle \frac{1}{2}}$ dan $r=-{\displaystyle \frac{1}{2}}$

$$\begin{array}{rl}{S}_{10}& =\frac{a[1-r^n]}{1-r}\\ \\ S_{10}& =\frac{{\displaystyle \frac{1}{2}}[1-{(-{\displaystyle \frac{1}{2}})}^{10}]}{1-(-{\displaystyle \frac{1}{2}})}\\ \\ S_{10}& =\frac{{\displaystyle \frac{1}{2}}[1-2^{-10}]}{{\displaystyle \frac{3}{2}}}\\ \\ S_{10}& =\frac{1}{3}[1-2^{-10}]\end{array}$$

$$\begin{array}{rl}& m-10=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+\dots \dots (1)\\ \\ & m-10=\frac{1}{3}[1-2^{-10}]\\ \\ & m=10+\frac{1}{3}[1-2^{-10}]\end{array}$$

Jawab: (1) dan (3)

Barisan aritmetika $\sqrt{y^2+2y+1},{\displaystyle \frac{y^2+3y-1}{3}},y-1$

$$\begin{array}{rl}{U}_2-U_1& =U_3-U_2\\ \\ \frac{y^2+3y-1}{3}-\sqrt{y^2+2y+1}& =y-1-\frac{y^2+3y-1}{3}\\ \\ 2.\frac{y^2+3y-1}{3}+1-y& =\sqrt{y^2+2y+1}\\ \\ \frac{2y^2+6y-2}{3}+\frac{3}{3}-\frac{3y}{3}& =\sqrt{(y+1{)}^2}\\ \\ \frac{2y^2+3y+1}{3}& =|y+1|\\ \end{array}$$

Ada dua kemungkinan penyelesaian:

Nilai y yang mungkin adalah $-2,-1,1$

Jawab: B

$$\begin{array}{rl}& x=\frac{1}{2013}-\frac{2}{2013}+\frac{3}{2013}-\frac{4}{2013}+\dots +\frac{2011}{2013}-\frac{2012}{2013}\\ \\ & x=\frac{1}{2013}+\frac{3}{2013}+\dots +\frac{2011}{2013}-\frac{2}{2013}-\frac{4}{2013}-\dots -\frac{2012}{2013}\end{array}$$

Deret aritmetika ${\displaystyle \frac{1}{2013}}+{\displaystyle \frac{3}{2013}}+\dots +{\displaystyle \frac{2011}{2013}}$

Deret aritmetika $-{\displaystyle \frac{2}{2013}}-{\displaystyle \frac{4}{2013}}-\dots -{\displaystyle \frac{2012}{2013}}$

$$\begin{array}{rl}& S_{\text{total}}=503\left(\frac{2012}{2013}\right)+503(-\frac{2014}{2013})\\ \\ & S_{\text{total}}=503(\frac{2012}{2013}-\frac{2014}{2013})\\ \\ & S_{\text{total}}=503(-\frac{2}{2013})\\ \\ & S_{\text{total}}=-\frac{1006}{2013}\end{array}$$

Jawab: C

$$\begin{array}{rl}\frac{a_3-a_2}{b_5-b_3}& =\frac{b}{2b^{\prime }}\\ \\ \frac{a_3-a_2}{b_5-b_3}& =\frac{{\displaystyle \frac{y-x}{4}}}{2.{\displaystyle \frac{y-x}{6}}}\\ \\ \frac{a_3-a_2}{b_5-b_3}& =\frac{{\displaystyle \frac{y-x}{4}}}{{\displaystyle \frac{y-x}{3}}}\\ \\ \frac{a_3-a_2}{b_5-b_3}& =\frac{y-x}{4}.\frac{3}{y-x}\\ \\ \frac{a_3-a_2}{b_5-b_3}& =\frac{3}{4}\end{array}$$

Jawab: A

$s_n={\displaystyle \frac{1}{2^n}}+{\displaystyle \frac{1}{3^n}}+{\displaystyle \frac{1}{4^n}}+\dots$

$$\begin{array}{rl}& s_2={\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{3^2}}+{\displaystyle \frac{1}{4^2}}+\dots \\ \\ & s_3={\displaystyle \frac{1}{2^3}}+{\displaystyle \frac{1}{3^3}}+{\displaystyle \frac{1}{4^3}}+\dots \\ \\ & s_4={\displaystyle \frac{1}{2^4}}+{\displaystyle \frac{1}{3^4}}+{\displaystyle \frac{1}{4^4}}+\dots (+)\\ \\ \\ & s_2+s_3+s_4+\dots ={\displaystyle \frac{1}{2^2}}+{\displaystyle \frac{1}{2^3}}+{\displaystyle \frac{1}{2^4}}+\dots +{\displaystyle \frac{1}{3^2}}+{\displaystyle \frac{1}{3^3}}+{\displaystyle \frac{1}{3^4}}+\dots +{\displaystyle \frac{1}{4^2}}+{\displaystyle \frac{1}{4^3}}+{\displaystyle \frac{1}{4^4}}+\dots \end{array}$$

$$\begin{array}{rl}& S_{\text{total}}=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\dots \\ \\ & S_{\text{total}}=\sum _1^{\sim }\frac{1}{n.(n+1)}\\ \\ & S_{\text{total}}=\sum _1^{\sim }(\frac{1}{n}-\frac{1}{n+1})\end{array}$$

Metode sum of difference

$$\begin{array}{rl}& S_1=\frac{1}{1}\cancel{-\frac{1}{2}}\\ \\ & S_2=\cancel{\frac{1}{2}}\cancel{-\frac{1}{3}}\\ \\ & S_3=\cancel{\frac{1}{3}}\cancel{-\frac{1}{4}}\\ \\ & \dots \\ \\ & S_n=\cancel{\frac{1}{n}}-\frac{1}{n+1}(+)\\ \\ & \\ & S_{\text{total}}=1-\frac{1}{n+1}\\ \\ & S_{\text{total}}=1-\frac{1}{\sim }\\ \\ & S_{\text{total}}=1-0\\ \\ & S_{\text{total}}=1\end{array}$$

Jawab: C

Suku pertama

$$\begin{array}{rl}& a=2+4+6+\dots +2n\\ \\ & a=\frac{n}{2}[2+2n]\\ \\ & a=n+n^2\end{array}$$

Jumlah deret

$$\begin{array}{rl}& S_n=\frac{n}{2}[2a+(n-1).b]\\ \\ & S_n=\frac{n}{2}[2(n+n^2)+(n-1).n]\\ \\ & S_n=\frac{n}{2}[2n+2n^2+n^2-n]\\ \\ & S_n=\frac{n}{2}[3n^2+n]\\ \\ & S_n=\frac{3}{2}{n}^3+\frac{1}{2}{n}^2\end{array}$$

Jawab: D

Barisan aritmetika $\log a^3{b}^7,\log a^5{b}^{12},\log a^8{b}^{15}$

$$\begin{array}{rl}b& =U_2-U_1\\ \\ b& =\log a^5{b}^{12}-\log a^3{b}^7\\ \\ b& =\log \left(\frac{a^5{b}^{12}}{a^3{b}^7}\right)\\ \\ b& =\log a^2{b}^5\end{array}$$

Suku ke-12

$$\begin{array}{rl}{U}_n& =a+(n-1).b\\ \\ U_{12}& =\log a^3{b}^7+(12-1).\log a^2{b}^5\\ \\ U_{12}& =\log a^3{b}^7+11.\log a^2{b}^5\\ \\ U_{12}& =\log a^3{b}^7+\log {\left(a^2{b}^5\right)}^{11}\\ \\ U_{12}& =\log a^3{b}^7+\log a^{22}{b}^{55}\\ \\ U_{12}& =\log (a^3{b}^7.a^{22}{b}^{55})\\ \\ U_{12}& =\log a^{25}{b}^{62}\end{array}$$

$m=25,n=62$

$2m+n=112$

Jawab: A

Jawab: D

$$\begin{array}{rl}m& =1+2.2+3.2^2+4.2^3+\dots +2015.2^{2014}+2016.2^{2015}\\ \\ 2m& =1.2+2.2^2+3.2^3+4.2^4+\dots +2015.2^{2015}+2016.2^{2016}(-)\\ \\ \\ -m& =1+2+2^2+2^3+\dots +2^{2015}-2016.2^{2016}\dots (1)\end{array}$$

Bentuk $2+2^2+2^3+\dots +2^{2015}$ merupakan deret geometri

$a=2,r=2,n=2015$

$$\begin{array}{rl}{S}_n& =\frac{a(r^n-1)}{r-1}\\ \\ S_{2015}& =\frac{2(2^{2015}-1)}{2-1}\\ \\ S_{2015}& =2(2^{2015}-1)\\ \\ S_{2015}& =2^{2016}-2\end{array}$$

$$\begin{array}{rl}& -m=1+2+2^2+2^3+\dots +2^{2015}-2016.2^{2016}\dots (1)\\ \\ & -m=1+2^{2016}-2-2016.2^{2016}\\ \\ & m=2016.2^{2016}-2^{2016}+1\\ \\ & m=2015.2^{2016}+1\end{array}$$

Jawab: E

Barisan $4,8,14,22,32,\dots$ merupakan barisan aritmetika tingkat 2

$$\begin{array}{rl}& U_n=a+(n-1).b+(n-1)(n-2).{\displaystyle \frac{c}{2!}}\\ \\ & U_{20}=4+(20-1).4+(20-1)(20-2).{\displaystyle \frac{2}{2!}}\\ \\ & U_{20}=4+19.4+19.18.1\\ \\ & U_{20}=422\end{array}$$

Jawab: C

Misalkan deret aritmetika tersebut adalah

$a-4b,a-3b,a-2b,a-b,a,a+b,a+2b,a+2b,a+3b,a+4b$

Jumlah bilangan adalah 153

$$\begin{array}{rl}& a-4b+a-3b+a-2b+a-b+a+a+b+a+2b+a+2b+a+3b+a+4b=153\\ \\ & 9a=153\\ \\ & a=17\end{array}$$

Sisipan

$a-4b,a-3\frac{1}{2}b,a-3b,a-2\frac{1}{2}b,a-2b,\dots$

Banyaknya sisipan adalah 8 bilangan

$a-3\frac{1}{2}b,a-2\frac{1}{2}b,a-1\frac{1}{2}b,a-\frac{1}{2}b,a+\frac{1}{2}b,a+1\frac{1}{2}b,a+2\frac{1}{2}b,a+3\frac{1}{2}b$

Jumlah seluruh sisipan

$$\begin{array}{rl}& a-3\frac{1}{2}b+a-2\frac{1}{2}b+a-1\frac{1}{2}b+a-\frac{1}{2}b+a+\frac{1}{2}b+a+1\frac{1}{2}b+a+2\frac{1}{2}b+a+3\frac{1}{2}b\\ \\ & 8a\\ \\ & 136\end{array}$$

Jumlah deret baru

$153+136=289$

Jawab: A

Barisan aritmetika $a^2-bc,b^2-ac,c^2-ab$

$$\begin{array}{rl}{U}_2-U_1& =U_3-U_2\\ \\ b^2-ac-(a^2-bc)& =c^2-ab-(b^2-ac)\\ \\ b^2-ac-a^2+bc& =c^2-ab-b^2+ac\\ \\ b^2-a^2+bc-ac& =c^2-b^2+ac-ab\\ \\ (b-a)(b+a)+c(b-a)& =(c-b)(c+b)+a(c-b)\\ \\ (b-a)(b+a+c)& =(c-b)(c+b+a)\\ \\ (b-a)\cancel{(b+a+c)}& =(c-b)\cancel{(c+b+a)}\\ \\ b-a& =c-b\\ \\ 2b& =a+c\end{array}$$

$$\begin{array}{rl}& \frac{a+c}{b}=\frac{2b}{b}\\ \\ & \frac{a+c}{b}=2\end{array}$$