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)

3) $\left(a_n\right)_{n \geqslant 1}$ is in harmonic progression(HP)

if $\ \exists \ a, d$ such that $\forall n \geqslant 1$,

$\frac{1}{a_n}=a+(n-1) d$

$\text { i.e. } a_n=\frac{1}{a+(n-1) d}$

1) If $a, b, c$ are in AP, then $a=b-d$,

$c=b+d$, where

$d \rightarrow$ common difference

2) If $a, b, c$ are in $G P$,

then $a=\frac{b}{r}$,

$c=b r$, where

$r \rightarrow$ common ratio

$a, b, c$ are in $A P$ and as well as in GP

$a=b-d$

$c=b+d$

$a c=(b-d)(b+d)$

$=b^2-d^2$

$a, b, c$ are in both GP and HP

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

$a c=b^2$

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

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

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

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

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

$a=b=c$

$a, b, c$ are in both AP and H P

$a=b-d_1$

$c=b+d_1$

$a=\frac{1}{p-d_2}$

$b = \frac{1}{p} \Rightarrow bp=1$

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

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

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

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

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

$\Rightarrow \underbrace{bp}_{= 1} +p d_1+b d_2+d_1 d_2=1$

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

$d_1 d_2=-d_1 d_2$

$\Rightarrow d_1 d_2=0$

$\Rightarrow d_1=0 \text { or } d_2=0$

$a=b=c$

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

$\underbrace{225 a^2}_ {(15 a)^2} +$ $\underbrace{9 b^2}_ {(3 b)^2} +$ $\underbrace{25 c^2}_ {(5 c)^2} -$ $\underbrace{75 a c}_ {(15 a)(5 c)}-$ $\underbrace{45 a b}_ {(15 a)(3 b)}$ $-\underbrace{(15 b c)}_{(3b)(5 c)}=0$

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

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

Let $a_1 ,a_2, \ldots, a_{49}$ be in arithmetic progression such that $\sum_{k=0}^{12} a_{4 k+1}=416$

and $a_9+a_{43}=66$.

If $a_1^2+a_2^2+\cdots+a_{17}^2=140 m$, then $m$ is equal to:

(1) $34$

(2) $68$

(3) $33$

(4) $66$

$d\rightarrow \text { common difference }$

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

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

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

$13 a_1+(0+\underbrace{4+8+\cdots+48)}_{12 \text { terms }} d=416$

$4(1+2+3+\cdots+12)$

$=4 \frac{12 \cdot 13}{2} \\ =13 \cdot 24$

$13 a_1+13 \cdot 24 d=416$

$\Rightarrow 13\left(a_1+24 d\right)=416$

$a_1+24 d=32$

$a_9+a_{43}=66$

$\Rightarrow a_1+8 d+a_1+42 d=66$

$\Rightarrow 2 a_1+50 d=66$

$a_1+25 d=33$

$d=1, a_1=33-25=8$

$\sum_{r=1}^{17} a_r^2=140 \mathrm{~m}$

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

$=8+(r-1)$

$=7+r$

$\Rightarrow \sum_{r=1}^{17}(7+r)^2=140 m$

$\Rightarrow \quad \underbrace{49 \sum_ {r=1}^{17} 1}_ {49 \times 17}$ $+ \underbrace{14 \sum_ {r=1}^{17} r}_ {14 \times \frac{17 \times 18}{2}}$ $+ \underbrace{\sum_ {r=1}^{17} r^2}_{\frac{17 \times 18 \times 35}{6}} =140 m$

$m=\frac{4760}{140}=34$

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.

$\frac{\log (a+c)+\log (a-2 b+c)}{2}=\log (a-c)$

$\Rightarrow \log (a+c)(a-2 b+c)=\log (a-c)^2$

$(a+c)(a-2 b+c)=(a-c)^2$

$\Rightarrow a^2+c a-2 a b-2 b c+a c+c^2 = a^2-2 a c+c^2$

$\Rightarrow 2(a b+b c)=4 a c$

$\frac{a b+b c}{2}=a c$

$a b, a c, b c$ are in AP

$\frac{a b+b c}{2}=a c \quad \frac{1}{a},$

$\frac{1}{b}, \frac{1}{c} \text { are in AP }$

$\Rightarrow a b+b c=2 a c$

$\Rightarrow b=\frac{2 a c}{a+c} \quad$

$\Rightarrow \frac{1}{1 / b}=\frac{2}{1 / c+1 / a}$

$\Rightarrow \frac{1}{a}+\frac{1}{c}=\frac{2}{b}$

$\Rightarrow a, b, c \text { are in HP}$

Let $a_1, a_2, \cdots, a_{10}$ be in arithmetic progression and $h_1, h_2, \cdots, h_{10}$ be in harmonic progression. If $a_1=h_1=2$ and $a_{10}=h_{10}=3$, then $a_4 h_7$ is

(1) $2$

(2) $3$

(3) $5$

(4) $6$

$a_{10} =a_1+9 d \quad d \rightarrow \text { common difference }$

$\Rightarrow 3 =2+9 d$

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

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

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

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

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

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

$a_4 =a_1+3 d$

$=2+\frac{3}{9}=\frac{7}{3}$

$\frac{1}{h_7} =\frac{1}{h_1}+6 c=\frac{1}{2}-\frac{1}{9}$

$\Rightarrow h_7 =\frac{18}{7}$

$a_4 \cdot h_7 =\frac{7}{3} \cdot \frac{18}{7}=6$

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:

$2<a, b, c<18$

$a+b+c=25$

$2, a, b \rightarrow A P$

$\frac{2+b}{2}=a \Rightarrow b=2(a-1)$

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

$\Rightarrow c^2=36(a-1)$

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

$a+2(a-1)+6 \sqrt{a-1}=25 \\$

$\Rightarrow 3 a+6 \sqrt{a-1}=27 \\$

$\Rightarrow a-9=-2 \sqrt{a-1} \\$

$a^2-18 a+81=4 a-4 \\$

$\Rightarrow a^2-22 a+85=0 \\$

$\Rightarrow (a-5)(a-17)=0 \\$

$a=5 \text { or } 17$

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

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) $-6,-32$

$\alpha+\beta=1, \quad \alpha \beta=p=a^2 r$

$\gamma+\delta=4, \quad \gamma \delta=q=a^2 r^5$

$\alpha, \beta, \gamma, \delta \rightarrow$ are in a GP

$[\alpha=a, \quad \beta= a r , \quad \gamma= a r^2, \quad \delta= a r^3 ]$

$a(1+r)=1$

$a r^2(1+r)=4$

$r^2=4 \Rightarrow r= \pm 2$

$a=\frac{1}{1+r}= \frac{1}{3}, \text { when } r=2$

$\quad\quad\quad\quad -1 \text { when } r=-2$

$p=a^2 r=-2 \quad \quad \mid r=-2, a=-1$

$q=a^2 r^5=-32$

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) $\Delta \neq 0$

$\alpha+\beta=-\frac{b}{a} \quad \quad\quad\quad a \neq 0$

$\alpha \beta=\frac{c}{a}$

$(\alpha+\beta) \left(\alpha^3+\beta^3\right)=\left(\alpha^2+\beta^2\right)^2$

$\alpha^3+\beta^3 =(\alpha+\beta)^3-3 \alpha \beta(\alpha+\beta)$

$=-\frac{b^3}{a^3}+\frac{3 b c}{a^2}=\frac{3 a b c-b^3}{a^3}$

$\alpha^2+\beta^2 =(\alpha+\beta)^2-2 \alpha \beta$

$=\frac{b^2}{a^2}-\frac{2 c}{a}=\frac{b^2-2 a c}{a^2}$

$-\frac{b}{a}\left(\frac{3 a b c-b^3}{a^3}\right)=\frac{\left(b^2-2 a c\right)^2}{a^4}$

$\Rightarrow -3 a b^2 c+b^4=b^4-4 a b^2 c+4 a^2 c^2$

$\Rightarrow a b^2 c=4 a^2 c^2$

$\Rightarrow a c \underbrace{\left(b^2-4 a c\right)}_{\Delta}=0$

$a c \Delta=0 \ \ \text { as } a \neq 0$

$\Rightarrow c \Delta =0$