Jump to heading🔗 Module 5-03 Geometric Sequence

If in the sequence $\left\{a_n\right\},\frac{a_{n+1}}{a_n}=q(constant)(n\in N_{+})$, called the sequence $\left\{a_n\right\}$ for Geometric sequence, q is the common ratio.

Remark: If two elements are known, need to know determine a common ratio $\frac{a_n}{a_m}=q^{n-m}$.

${\mathbf{a}}_{\mathbf{n}}\mathbf{=}{\mathbf{a}}_{\mathbf{1}}{\mathbf{q}}^{\mathbf{n}\mathbf{-}\mathbf{1}}$

$\begin{array}{l}\frac{a_{n+1}}{a_n}=q\\ \frac{\cancel{a_2}}{a_1}=q,\frac{\cancel{a_3}}{\cancel{a_2}}=q,\dots ,\frac{a_n}{\cancel{a_{n-1}}}=q\\ \frac{a_n}{a_1}=q^{n-1}\\ a_n=a_1{q}^{n-1}\end{array}$

${\mathbf{a}}_{\mathbf{n}}\mathbf{=}{\mathbf{a}}_{\mathbf{k}}{\mathbf{q}}^{\mathbf{n}\mathbf{-}\mathbf{k}}$

$\begin{array}{l}\frac{a_{n+1}}{a_n}=q\\ \frac{\cancel{a_2}}{a_k}=q,\frac{\cancel{a_3}}{\cancel{a_2}}=q,\dots ,\frac{a_n}{\cancel{a_{n-1}}}=q\\ \frac{a_n}{a_k}=q^{n-k}\\ a_n=a_k{q}^{n-k}\\ a_n=a_1{q}^{n-1}& k=1\end{array}$

${\mathbf{a}}_{\mathbf{n}}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{q}}{\mathbf{q}}^{\mathbf{n}}$

$\begin{array}{l}{a}_n=a_1{q}^{n-1}\\ a_n=a_1\frac{q^n}{q}\\ a_n=\frac{a_1}{q}{q}^n\end{array}$

${\mathbf{S}}_{\mathbf{n}}\mathbf{=}{\mathbf{na}}_{\mathbf{1}}$

$\begin{array}{ll}{S}_n=a_1+a_1+a_1+\cdots +a_n& \text{Constant sequence}\\ S_n=n{a}_1\end{array}$

${\mathbf{S}}_{\mathbf{n}}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{1}}\mathbf{-}{\mathbf{a}}_{\mathbf{n}}\mathbf{q}}{\mathbf{1}\mathbf{-}\mathbf{q}}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{1}}\mathbf{-}{\mathbf{a}}_{\mathbf{n}\mathbf{+}\mathbf{1}}}{\mathbf{1}\mathbf{-}\mathbf{q}}$

$\begin{array}{l}{S}_n=a_1+a_2+\cdots +a_{n-1}+a_n\\ q{S}_n=q(a_1+a_2+\cdots +a_{n-1}+a_n)\\ q{S}_n=a_2+a_3+\cdots +a_n+a_{n}q\\ (S_n=a_1+a_2+\cdots +a_{n-1}+a_n)-(q{S}_n=a_2+a_3+\cdots +a_n+a_{n}q)& \text{Displaced subtraction}\\ (1-q)S_n=a_1-a_{n}q\\ S_n=\frac{a_1-a_{n}q}{1-q}\vee S_n=\frac{a_1-a_{n+1}}{1-q}\end{array}$

${\mathbf{S}}_{\mathbf{n}}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{1}}\mathbf{(}\mathbf{1}\mathbf{-}{\mathbf{q}}^{\mathbf{n}}\mathbf{)}}{\mathbf{1}\mathbf{-}\mathbf{q}}$

$\begin{array}{l}{S}_n=\frac{a_1-a_{n}q}{1-q}\\ a_n=a_1{q}^{n-1}\\ S_n=\frac{a_1-a_1{q}^n}{1-q}\\ S_n=\frac{a_1(1-q^n)}{1-q}\vee S_n=\frac{a_1}{1-q}(1-q^n)\end{array}$

1. If $m+n=k+t\boxed{\text{Arithmetic sequence}}$ then $a_m{a}_n=a_k{a}_t\boxed{\text{Geometric sequence}}$.$\begin{array}{l}{a}_3·a_9=a_5·a_7\\ a_3·q^2=a_5\\ a_7·q^2=a_9\\ a_3·a_9=a_5·a_7=a_6^2\end{array}$

2. $S_n$ is the sum of the first n terms of a geometric sequence, then $S_n,S_{2n}-S_n,S_{3n}-S_{2n},\dots$ are still geometric sequences$\boxed{\text{Segment summation}}$, and their common ratio is $q^n$.
   $\boxed{{\displaystyle \begin{array}{l}\underset{S_3}{\underbrace{a_1{a}_2{a}_3}}\underset{S_6-S_3}{\underbrace{a_4{a}_5{a}_6}}\underset{S_9-S_6}{\underbrace{a_7{a}_8{a}_9}}\dots \\ \frac{a_4+a_5+a_6}{a_1+a_2+a_3}=\frac{(a_1+a_2+a_3)q^3}{a_1+a_2+a_3}=q^3\end{array}}}$

   $\boxed{{\displaystyle \begin{array}{l}{S}_n=\frac{a_1(1-q^n)}{1-q}\to \frac{S_m}{S_n}=\frac{1-q^m}{1-q^n}\\ \text{Special case}{S}_m=S_{2n}.\\ \frac{S_{2n}}{S_n}=\frac{1-q^{2n}}{1-q^n}=\frac{(1-q^n)(1+q^n)}{1-q^n}=1+q^n\\ \text{Proof common ratio}.\\ \frac{S_{2n}-S_n}{S_n}=\frac{S_{2n}}{S_n}-1=1+q^n-1=q^n\end{array}}}$

3. If $\left|q\right|<1$, then the sum of all terms in the geometric sequence is $S=\underset{n\to \mathrm{\infty }}{lim}{S}_n=\frac{a_1}{1-q}$.
   $\begin{array}{l}\left|q\right|<1⟹n\to \mathrm{\infty }⟹q^n\to 0\\ {\left(\frac{1}{3}\right)}^{100}\approx 0\\ S_n=\frac{a_{1}0}{1-q}=\frac{a_1}{1-q}\end{array}$

• If three numbers $a,b,c$ form a geometric sequence, then b is called the geometric mean of a and c, that is $ac=b^2$.
  • $b=\pm \sqrt{ac}$
  • $a,c$ same sign operators

Jump to heading🔗 $\boxed{\text{28}}$If $2,2^x-1,2^x+3$ form a geometric sequence, that is $x=?$.

$\begin{array}{llllll}(\text{A}){\log }_{2}5& (\text{B}){\log }_{2}6& (\text{C}){\log }_{2}7& (\text{D})3& (\text{E})4& \end{array}$

• $\mathbf{2}\mathbf{,}{\mathbf{2}}^{\mathbf{x}}\mathbf{-}\mathbf{1}\mathbf{,}{\mathbf{2}}^{\mathbf{x}}\mathbf{+}\mathbf{3}$ form a geometric sequence; it means $\mathbf{a}\mathbf{+}\mathbf{c}\mathbf{=}{\mathbf{b}}^{\mathbf{2}}$

  $\begin{array}{l}2\times (2^x+3)=(2^x-1{)}^2\\ 2(t+3)=(t-1{)}^2& t=2^x(t>0)\\ 2(t+3)=t^2-2t+1& (a-b{)}^2=a^2-2ab+b^2\\ 2t+6=t^2-2t+1\\ t^2-4t-5=0\\ (t-5)(t+1)=0\\ t=5\vee -1& t\ne -1\\ t=5\\ 2^x=5\\ x={\log }_{2}5& b^c=a⟹{\log }_{b}a=c\end{array}$

${\mathbf{a}}_{\mathbf{n}}\mathbf{=}{\mathbf{a}}_{\mathbf{1}}{\mathbf{q}}^{\mathbf{n}\mathbf{-}\mathbf{1}}\mathbf{=}{\mathbf{a}}_{\mathbf{k}}{\mathbf{q}}^{\mathbf{n}\mathbf{-}\mathbf{k}}\mathbf{=}\frac{{\mathbf{a}}_{\mathbf{1}}}{\mathbf{q}}{\mathbf{q}}^{\mathbf{n}}$

• No element in a geometric sequence can be 0, and the common ratio can't be 0.
  • $a_n=\text{constants}\times \text{exponents}$
  • $q=\text{base}$
  • $a_n=\text{constants}\text{Special case }q=1$

Jump to heading🔗 $\boxed{\text{29}}$Following there are $?$that can be used as general term in geometric sequence.

$\begin{array}{lllll}(\text{1})a_n=n^3& (\text{2})a_n=3^n& (\text{3})a_n=\frac{1}{3}& (\text{4})a_n=\frac{2_n}{3}& \\ (\text{5})a_n=3^{-n}& (\text{6})a_n=(-1{)}^n& (\text{7})a_n=2^n-1& \\ (\text{A})2& (\text{B})3& (\text{C})4& (\text{D})5& (\text{E})6& \end{array}$

• Currently, know the Expressions can use characterization analysis $a_n=\text{Constant}\times \text{Exponent}$

  $\begin{array}{ll}(\text{1})a_n=n^3& \text{This is a power function}❌\\ (\text{2})a_n=3^n& \boxed{{\displaystyle 1\times 3^n}}q=3✅\\ (\text{3})a_n=\frac{1}{3}& \boxed{{\displaystyle \frac{1}{3}\times 1^{n-1}}}q=1✅\\ (\text{4})a_n=\frac{2_n}{3}& \boxed{{\displaystyle \frac{1}{3}\times 2_n}}q=2✅\\ (\text{5})a_n=3^{-n}& \boxed{{\displaystyle 1\times (\frac{1}{3}{)}^n}}q=\frac{1}{3}✅\\ (\text{6})a_n=(-1{)}^n& \boxed{{\displaystyle 1\times (-1{)}^n}}q=-1✅\\ (\text{7})a_n=2^n-1& \text{There is a constant term -1 after the exponent}❌\end{array}$

Jump to heading🔗 $\boxed{\text{30}}$If $\left\{a_n\right\}$ is a geometric sequence, among the following four statements, the number of correct statements is $?$.

$\begin{array}{lll}(\text{1})\text{The sequence}\left\{a_n^2\right\}\text{is a geometric sequence.}& (\text{2})\text{The sequence}\left\{a_{2n}\right\}\text{is a geometric sequence.}& \\ (\text{3})\text{The sequence}\left\{\frac{1}{a_n}\right\}\text{is a geometric sequence.}& (\text{4})\text{The sequence}\left\{|a_n|\right\}\text{is a geometric sequence.}& \end{array}$

$\begin{array}{llllll}(\text{A})0& (\text{B})1& (\text{C})2& (\text{D})3& (\text{E})4& \end{array}$

• Currently, don't know the Expressions, can use Geometric sequence definition analysis $\frac{a_n+1}{a_n}=q$

  $\begin{array}{ll}(\text{1})\left\{a_n^2\right\}& \boxed{{\displaystyle \frac{a_{n+1}^2}{a_n^2}=(\frac{a_{n+1}}{a_n}{)}^2=q^2}}✅\\ (\text{2})\left\{a_{2n}\right\}& \boxed{{\displaystyle \frac{a_{2(n+1)}}{a_{2n}}=q^2}}✅\\ (\text{3})\left\{\frac{1}{a_n}\right\}& \boxed{{\displaystyle \frac{\frac{1}{a_{n+1}}}{\frac{1}{a_n}}=\frac{a_n}{a_{n+1}}=\frac{1}{q}}}✅\\ (\text{4})\left\{|a_n|\right\}& \boxed{{\displaystyle \frac{|a_{n+1}|}{|a_n|}=\left|\frac{a_{n+1}}{a_n}\right|=|q|}}✅\end{array}$

Jump to heading🔗 $\boxed{\text{31}}$In the geometric sequence $\left\{a_n\right\}$, if $a_4{a}_7=-512,a_3+a_8=124$ and the is common ratio is $q\in \mathbb{Z}$, then $a_{10}=?$.

$\begin{array}{llllll}(\text{A})124& (\text{B})64& (\text{C})512& (\text{D})-124& (\text{E})-512& \end{array}$

• According to the characteristics of Vieta's formulas, rearrange the equation

  $\begin{array}{l}{x}^2-124x-512=0\{\begin{array}{l}{a}_4{a}_7=a_3{a}_8=-512\\ a_3+a_8=124\end{array}\\ (1+4)(1-128)=0\\ x=-4\vee 128\\ q\in \mathbb{z}\{\begin{array}{l}{a}_3=-4\\ a_8=128\end{array}\\ \frac{a_8}{a_3}=\frac{128}{-4}=-32=q^{\frac{\log 32}{\log 2}}=q^5\\ q^5=-32⟹\sqrt[5]{-32}\\ q=-2\\ a_{10}=a_8{q}^2=128\times (-2{)}^2=512\end{array}$

Jump to heading🔗 $\boxed{\text{32}}$In the known geometric sequence $\left\{a_n\right\}$, if $a_3+a_9=130,a_3-a_9=-126$, then common ratio $q=?$.

$\begin{array}{llllll}(\text{A})2\vee -2& (\text{B})2& (\text{C})3& (\text{D})-3& (\text{E})-2& \end{array}$

$\begin{array}{l}{a}_3+a_9+a_3-a_9=130-126\{\begin{array}{l}{a}_3+a_9=130\\ a_3-a_9=-126\end{array}\\ 2a_3=4\\ a_3=\frac{4}{2}=2\\ a_3+a_9=130\\ 2+a_9=130\\ a_9=130-2=128\\ \frac{a_9}{a_3}=\frac{128}{2}=64=q^{\frac{\log 64}{\log 2}}=q^6\\ q^6=64⟹\pm \sqrt[6]{64}\\ q=\pm 2\end{array}$

${\mathbf{S}}_{\mathbf{n}}\mathbf{=}\{\begin{array}{ll}n{a}_1& q=1\\ \frac{a_1(1-q^n)}{1-q}=\frac{a_1-a_{n}q}{1-q}=\frac{a_1-a_{n+1}}{1-q}& q\ne 1\end{array}$

• $q=1$
  • $S_n=n{a}_1\text{linear function}$
• $q\ne 1$
  • $S_n=\frac{a_1}{1-q}\times (1-q^n)=k(1-q^n)=\boxed{{\displaystyle k-k{q}^n}}$
• $S_n=a+b{q}^n$
  • $a+b\ne 0\text{it is not }{S}_n$
    • Form $a_2$ onwards it is still a geometric sequence.
  • $a+b=0\text{it is }{S}_n$
    • $S_n=3-3q^n$ ✅$k-k=0$
    • $S_n=5-3q^n$ ❌$k-k\ne 0$

Jump to heading🔗 $\boxed{\text{33}}$Following there are $?$ that can be used as a sum of the first n terms of a geometric sequence.

$\begin{array}{lllll}(\text{1})S_n=\frac{1}{3}& (\text{2})S_n=2n& (\text{3})S_n=2n-1& (\text{4})S_n=2^n& \\ (\text{5})S_n=2^n-1& (\text{6})S_n=2^n+1& (\text{7})S_n=3(2^n-1)& \\ (\text{A})2& (\text{B})3& (\text{C})4& (\text{D})5& (\text{E})6& \end{array}$

• Currently, know the Expressions can use characterization analysis

  $S_n=\{\begin{array}{ll}{S}_n=n{a}_1& q=1\\ S_n=k-k{q}^n& q\ne 1\end{array}$

  $\begin{array}{ll}(\text{1})S_n=\frac{1}{3}& S_n\text{cannot be a constant, but }{a}_n\text{can be a constant}❌\\ (\text{2})S_n=2n& \boxed{{\displaystyle n2}}q=1✅\\ (\text{3})S_n=2n-1& 2n-1\text{is not the }{q}^n\text{exponential}❌\\ (\text{4})S_n=2^n& \text{There is no constant term after }{2}^n❌\\ (\text{5})S_n=2^n-1& \boxed{{\displaystyle 1-1\cdot 2^n}}q=2✅\\ (\text{6})S_n=2^n+1& \text{Does not satisfy }k-k=0❌\\ (\text{7})S_n=3(2^n-1)& \boxed{{\displaystyle 3-3\cdot 2^n}}q=2✅\end{array}$

Jump to heading🔗 $\boxed{\text{34}}$It is known that $S_n$ is the sum of the first n terms of the geometric sequence $\left\{a_n\right\}$, if $S_2+S_5=2S_8$, then common ratio $q=?$.

$\begin{array}{llllll}(\text{A})1\vee -2& (\text{B})2& (\text{C})1\vee -\frac{\sqrt[3]{4}}{2}& (\text{D})-\frac{\sqrt[3]{4}}{2}& (\text{E})-2\vee -\frac{\sqrt[3]{4}}{2}& \end{array}$

• $q=1$

  $\begin{array}{l}2a_1+5a_1=2\times 8a_1\\ 7a_1=16a_1\\ 0=9a_1\\ a_1=\frac{0}{9}=0& \text{No any element in a geometric sequence can be 0}❌\end{array}$

• $q\ne 1$

  $\begin{array}{l}\frac{a_1(1-q^2)}{1-q}+\frac{a_1(1-q^5)}{1-q}=\frac{2a_1(1-q^8)}{1-q}\\ 1-q^2+1-q^5=2(1-q^8)\\ q^2+q^5=2q^8\\ \frac{q^2+q^5}{q^2}=\frac{2q^8}{q^2}& q\text{is not 0, divide both sides by }{q}^2,\text{similar to }{q}^{\frac{{\log }_{10}x}{{\log }_{10}2}}\\ \frac{q^2}{q^2}+\frac{q^2+q^3}{q^2}=\frac{2(q^2+q^6)}{q^2}\\ 1+q^3=2q^6\\ 2t^2-t-1=0& t=q^3\\ (1t-1)(2t+1)=0\\ t=1\vee -\frac{1}{2}& t\ne 1\text{if }t=1\text{then }q=1\\ q^3=-\frac{1}{2}\\ q=-\frac{1}{\sqrt[3]{2}}=-\frac{\sqrt[3]{4}}{\sqrt[3]{8}}=-\frac{\sqrt[3]{4}}{2}\end{array}$

• If $k\in {\mathbb{z}}_{\mathbb{+}},m+n=k+t$, then $a_m\cdot a_n=a_k\cdot a_t$.

Jump to heading🔗 $\boxed{\text{35}}$In the geometric sequence $\left\{a_n\right\}$, $a_3,a_8$ are the two roots of the equation $3^x+2x-18=0$, then $a_4{a}_7=?$.

$\begin{array}{llllll}(\text{A})-9& (\text{B})-8& (\text{C})-6& (\text{D})6& (\text{E})8& \end{array}$

$\begin{array}{l}{a}_4{a}_7=a_3{a}_8=\frac{-18}{3}=-6\end{array}$

Jump to heading🔗 $\boxed{\text{36}}$If the geometric sequence a satisfies $a_2{a}_4+2a_3{a}_5+a_2{a}_8=25$ and $a_1>0$, then $a_3+a_5=?$.

$\begin{array}{llllll}(\text{A})8& (\text{B})5& (\text{C})2& (\text{D})-2& (\text{E})-5& \end{array}$

$\begin{array}{l}{a}_3^2+2a_3{a}_5+a_5^2=25\\ (a_3+a_5{)}^2=25\\ \sqrt{(a_3+a_5{)}^2}=\sqrt{25}\\ a_3+a_5=\pm 5\\ a_3+a_5=5& a_1>0\text{so }{a}_1{q}^{n-1}>0\end{array}$

• If $S_n$ is the sum of the first n terms of a geometric sequence, then $S_n,S_{2n}-S_n,S_{3n}-S_{2n},\dots$ are still geometric sequences$\boxed{\text{Segment summation}}$, and their common ratio is $q^n$.

Jump to heading🔗 $\boxed{\text{37}}$In the geometric sequence $\left\{a_n\right\}$, knew $S_n=36,S_{2n}=54$, then $S_{3n}=?$.

$\begin{array}{llllll}(\text{A})63& (\text{B})68& (\text{C})76& (\text{D})89& (\text{E})92& \end{array}$

$\begin{array}{l}\underset{36}{\underbrace{S_n}}\underset{54-36=18}{\underbrace{S_{2n}-S_n}}\underset{\frac{18}{2}=9}{\underbrace{S_{3n}-S_{2n}}}\dots \\ S_3=S_{2n}+9\vee \bcancel{S_n}+\cancel{S_{2n}}-\bcancel{S_n}+S_{3n}-\cancel{S_{2n}}\\ S_3=54+9\vee 36+18+9\\ S_3=63\end{array}$

Jump to heading🔗 $\boxed{\text{38}}$It is known that $S_n$ is the sum of the first n terms of the geometric sequence $\left\{a_n\right\}$, if $S_4=30,S_8=150$, then common ratio $q=?$.

$\begin{array}{llllll}(\text{A})\pm 2& (\text{B})\sqrt{2}& (\text{C})\pm \sqrt{2}& (\text{D})\pm \frac{1}{2}& (\text{E})-\sqrt{2}& \end{array}$

• $\boxed{\text{1}}$All indexs in $S_{\text{index}}$ are even numbers

  $\begin{array}{ll}\underset{30}{\underbrace{S_4}}\underset{150-30=120}{\underbrace{S_8-S_4}}\dots & q^4\\ \frac{120}{30}=4\\ q^4=4\\ \sqrt{q^{2^2}}=\sqrt{2^2}\\ q^2=2\\ q=\pm \sqrt{2}\end{array}$

• $\boxed{\text{2}}$All indexs in $S_{\text{index}}$ aren't even numbers

  $\begin{array}{l}\frac{S_m}{S_n}=\frac{1-q^m}{1-q^n}\\ \frac{S_8}{S_4}=1+q^4& S_m=S_{2n}\\ \frac{150}{30}=5\\ 5=1+q^4\\ q^4=4\\ \sqrt{q^{2^2}}=\sqrt{2^2}\\ q^2=2\\ q=\pm \sqrt{2}\end{array}$