mathematics-class-11-unit-10-chapter-01-arithmetic-geometric-harmonic-progre-l-1-2-n7wowjkq9py

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Introduction </primary>
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- Let $\left(a_n\right)_{n \geqslant 1}$ be a sequence. We say
	
- **1)** $ \{(a_n)}_{n \geqslant 1}$ is in arithmetic progression (AP)

- $ \text{if} \ \exists \ \text{a,d such that}$  $ \{\forall n \geqslant 1}$,  
- $a_n=a+(n-1) d \ \ \ $  ($d \rightarrow$ common difference)
- **2)** $\left(a_n\right)_{n \geqslant 1}$ is in geometric progression (GP)
- if $ \ \exists \ a, r$ such that $\forall n \geqslant 1$,  
- $a_n=a r^{n-1} \ \ \ $ ($r \rightarrow$ common ratio)

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<footer style = "font-size: 18px"> $\rightarrow$ Arithmetic geometric & harmonic progressions lecture -1 $\rightarrow$ <span style = "color:blue"> Introduction </span> $\rightarrow$ Introduction $\rightarrow$ Notes </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Facts case-II </primary>
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- $a, b, c$ are in both GP  and HP

- $ a=\frac{b}{r}, c=b r $

- $a c=b^2$

- $ a=\frac{1}{p-d}, b=\frac{1}{p}, c=\frac{1}{p+d} $

- $ a c=\frac{1}{p^2-d^2}$

- $\Rightarrow  b p=1  \Rightarrow b^2 p^2=1$

- $ b^2 p^2-b^2 d^2=1 \ \Rightarrow b^2 d^2=0 $

- $\quad b \neq 0 \Rightarrow d=0$

- $ a=b=c$

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<footer style = "font-size: 18px">Facts case-I $\rightarrow$ Facts case-I $\rightarrow$ <span style = "color:blue"> Facts case-II </span> $\rightarrow$ Facts case-III $\rightarrow$ Case III </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Facts case-III </primary>
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- $a, b, c$ are in both AP and  H P 
- $a=b-d_1$
- $c=b+d_1$

- $a=\frac{1}{p-d_2}$
- $ b = \frac{1}{p} \Rightarrow bp=1$

- $c=\frac{1}{p+d_2}$

- $ \left(b-d_1\right)\left(p-d_2\right)=1 $

- $ \Rightarrow b p-p d_1-b d_2+d_1 d_2=1$

- $ \Rightarrow p d_1+b d_2=d_1 d_2$

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<footer style = "font-size: 18px">Facts case-I $\rightarrow$ Facts case-II $\rightarrow$ <span style = "color:blue"> Facts case-III </span> $\rightarrow$ Case III $\rightarrow$ Question 1 </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Case III </primary>
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- $ \left(b+d_1\right)\left(p+d_2\right)=1$
- $ \Rightarrow  \underbrace{bp}_{= 1} +p d_1+b d_2+d_1 d_2=1$
- $ \Rightarrow p d_1+b d_2=-d_1 d_2 $
- $ d_1 d_2=-d_1 d_2$
- $ \Rightarrow  d_1 d_2=0$
- $ \Rightarrow  d_1=0 \text { or } d_2=0$
	
- $a=b=c$

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<footer style = "font-size: 18px">Facts case-II $\rightarrow$ Facts case-III $\rightarrow$ <span style = "color:blue"> Case III </span> $\rightarrow$ Question 1 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Question 1 </primary>
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- For any three positive real numbers a, b, c,

- $9(25 a^2+b^2)+25\left(c^2-3 a c\right)=15 b(3 a+c)$. Then

- (1) $ a, b, c\ \text{are in arithmetic progression.}$

- (2) $b, c, a \text{ are in arithmetic progression.}$

- (3) $c, a, b \text{ are in arithmetic progression.}$

- (4) $\text{ none of the above statements are true.}$

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<footer style = "font-size: 18px">Facts case-III $\rightarrow$ Case III $\rightarrow$ <span style = "color:blue"> Question 1 </span> $\rightarrow$ Solution $\rightarrow$ Question 2 </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \underbrace{225 a^2}_ {(15 a)^2} +$ $ \underbrace{9 b^2}_ {(3 b)^2} +$  $ \underbrace{25 c^2}_ {(5 c)^2} - $ $\underbrace{75 a c}_ {(15 a)(5 c)}-$ $\underbrace{45 a b}_ {(15 a)(3 b)}$ $-\underbrace{(15 b c)}_{(3b)(5 c)}=0$

- $\frac{1}{2}\\{\{(15 a-3 b)^2}+(3 b-5 c)^2+(15 a-5 c)^2\\} = 0 $

- are non-negative.

- $15 a-3 b=0 \ \Rightarrow \ b=5 a $

- $3 b-5 c=0 \ \Rightarrow \ 3 b=5 c$

- $15 a-5 c=0 \ \Rightarrow \ c=3 a $
- $ a+b $
- $ =\frac{c}{3}+\frac{5 c}{3}$
- $ =2 c$
- $b, c, a$ are in AP

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<footer style = "font-size: 18px">Case III $\rightarrow$ Question 1 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Question 2 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $  d\rightarrow \text { common difference }$

- $ a_r=a_1+(r-1) d$

- $ a_1 + a_5+a_9+a_{13}+\cdots+a_{49}=416$

- $ a_1+\left(a_1 +4 d\right)+\left(a_1 +8 d\right)+\cdots+\left(a_1 +48 d\right)=416$

- $ 13 a_1+(0+\underbrace{4+8+\cdots+48)}_{12 \text { terms }} d=416$
	
- $ 4(1+2+3+\cdots+12)$
	
- $ =4 \frac{12 \cdot 13}{2} \\\\ =13 \cdot 24$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Question 2 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \sum_{r=1}^{17} a_r^2=140 \mathrm{~m}$       
	
- $ a_r=a_1+(r-1) d $    
- $ =8+(r-1) $        
- $=7+r$
	
- $ \Rightarrow \sum_{r=1}^{17}(7+r)^2=140 m $        
	
- $ \Rightarrow \quad \underbrace{49 \sum_ {r=1}^{17} 1}_ {49 \times 17}$	$ + \underbrace{14 \sum_ {r=1}^{17} r}_ {14 \times \frac{17 \times 18}{2}}$ $+ \underbrace{\sum_ {r=1}^{17}  r^2}_{\frac{17 \times 18 \times 35}{6}} =140 m$
	
- $ m=\frac{4760}{140}=34 $

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Question 3 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Question 3 </primary>
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- For distinct $a, b, c$, if $\log (a+c), \log (a-c), \log (a-2 b+c)$ are in arithmetic progression then

- (1) $a, b, c$ are in arithmetic progression.

- (2) $a b, a c, b c$ are in arithmetic progression.

- (3) $a, b, c$ are in geometric progression.

- (4) $a, b, c$ are in harmonic progression.

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Question 3 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $  \frac{a b+b c}{2}=a c \quad \frac{1}{a},$ 
	
- $\frac{1}{b}, \frac{1}{c} \text { are in AP } $
	
- $   \Rightarrow a b+b c=2 a c$

- $\Rightarrow b=\frac{2 a c}{a+c} \quad $
	
- $\Rightarrow \frac{1}{1 / b}=\frac{2}{1 / c+1 / a}$
	
- $\Rightarrow \frac{1}{a}+\frac{1}{c}=\frac{2}{b}$
	
- $\Rightarrow a, b, c \text { are in HP}$

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<footer style = "font-size: 18px">Question 3 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Question 4 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $a_{10}  =a_1+9 d \quad d \rightarrow \text { common difference }$

- $\Rightarrow 3  =2+9 d $

- $\Rightarrow d  =\frac{1}{9}$

- $h_1, h_2, \ldots, h_{10}$ in HP

- $\Rightarrow  \frac{1}{h_1}, \frac{1}{h_2}, \cdots, \frac{1}{h_{10}} \text { in AP }$

- $ \frac{1}{h_{10}}=\frac{1}{h_1}+9 C \quad C \rightarrow \text { common difference} $

- $ 9C=\frac{1}{3}-\frac{1}{2}=-\frac{1}{6}$

- $ \Rightarrow  C=-\frac{1}{54}$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Question 4 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Question 5 </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Question 5 </primary>
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- Find three numbers $a, b, c$ between 2 and 18 such that

- a. their sum is 25 ,

- b. the terms $2, a, b$ are consecutive terms of an arithmetic progression,

- c. the terms $b, c, 18$ are consecutive terms of a geometric progression.  
  
- **Solution:**

- $ 2<a, b, c<18$

- $ a+b+c=25$
- $ 2, a, b \rightarrow A P$  
- $\frac{2+b}{2}=a \Rightarrow b=2(a-1)$

- $ b, c, 18 \rightarrow   G P \quad\quad  18 b=c^2$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Question 5 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \Rightarrow c^2=36(a-1)$

- $ \Rightarrow c=6 \sqrt{a-1}$

- $a+2(a-1)+6 \sqrt{a-1}=25 \\\\ $
- $\Rightarrow  3 a+6 \sqrt{a-1}=27 \\\ $
- $\Rightarrow  a-9=-2 \sqrt{a-1} \\\ $
- $ a^2-18 a+81=4 a-4 \\\ $
- $\Rightarrow  a^2-22 a+85=0 \\\ $
- $\Rightarrow  (a-5)(a-17)=0 \\\ $
- $ a=5 \text { or } 17$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Question 5 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Question 6 </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- **Case -I**

- $a=5, b=8, c=12$  
- $b=2(a-1)$  
- $c=6 \sqrt{a-1}$

- $2,5,8 \rightarrow$ are in AP with the common difference $3.$

- $8,12,18 \rightarrow$ are in GP with the common ratio $\frac{3}{2}$  
  
- **Case -II**

- $
\begin{aligned}
& a=17 \\
& b=32
\end{aligned} $

- Not possible

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<footer style = "font-size: 18px">Question 5 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Question 6 $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Question 6 </primary>
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- Let $\alpha, \beta$ be the solutions of $x^2-x+p=0$ and $\gamma, \delta$ be the solutions of $x^2-4 x+q=0$. If $\alpha, \beta, \gamma, \delta$ are in geometric progression, then integral values of $p$ and $q$ respectively, are

- (1) $-2,-32$

- (2) $-2,3$

- (3) $-6,3$

- (4) $-6,-32$  
  
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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Question 6 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Question 7 </primary>
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- Let $\alpha, \beta$ be the solutions of the quadratic equation $a x^2+b x+c=0$. Denote $\Delta=b^2-4 a c$. If $\alpha+\beta, \alpha^2+\beta^2, \alpha^3+\beta^3$ are in geometric progression, then which of the following statements are definitely true?

- (1) $\Delta=0$

- (2) $b \Delta=0$

- (3) $c \Delta=0$

- (4) $\Delta \neq 0$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Question 7 </span> $\rightarrow$ Solution $\rightarrow$ Solution </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ \alpha+\beta=-\frac{b}{a} \quad \quad\quad\quad  a \neq 0$

- $ \alpha \beta=\frac{c}{a} $  
  
- $ (\alpha+\beta)  \left(\alpha^3+\beta^3\right)=\left(\alpha^2+\beta^2\right)^2$
	
- $ \alpha^3+\beta^3  =(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)$
	
- $ =-\frac{b^3}{a^3}+\frac{3 b c}{a^2}=\frac{3 a b c-b^3}{a^3}$
	
- $\alpha^2+\beta^2  =(\alpha+\beta)^2-2 \alpha \beta$
	
- $ =\frac{b^2}{a^2}-\frac{2 c}{a}=\frac{b^2-2 a c}{a^2}$

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<footer style = "font-size: 18px">Solution $\rightarrow$ Question 7 $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Solution $\rightarrow$ Thank you </footer>

### <Success style="font-size:25px"> Arithmetic Geometric Harmonic Progre L-1 </Success>
### <primary style="font-size:24px"> Solution </primary>
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- $ -\frac{b}{a}\left(\frac{3 a b c-b^3}{a^3}\right)=\frac{\left(b^2-2 a c\right)^2}{a^4}$
	
- $ \Rightarrow  -3 a b^2 c+b^4=b^4-4 a b^2 c+4 a^2 c^2$
	
- $ \Rightarrow  a b^2 c=4 a^2 c^2$
	
- $ \Rightarrow  a c \underbrace{\left(b^2-4 a c\right)}_{\Delta}=0$
	
- $  a c \Delta=0 \ \ \text { as } a \neq 0$

- $ \Rightarrow c \Delta =0$

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<footer style = "font-size: 18px">Question 7 $\rightarrow$ Solution $\rightarrow$ <span style = "color:blue"> Solution </span> $\rightarrow$ Thank you $\rightarrow$  </footer>