Suppose $a, b, c$ are in arithmetic progression and $a^2, b^2, c^2$ are in geometric progression.

If $a<b<c$ and $a+b+c=3 / 2$, then the value of $a$ is

(1)$1 / 2 \sqrt{2}$

(2)$1 / 2 \sqrt{3}$

(3)$1 / 2-1 / \sqrt{2}$

(4)$1 / 2-1 / \sqrt{3}$

$a, b, c \rightarrow A P$

$a=p-r$

$b=p \text { for some } p, p$

$c=p+r$

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

$\Rightarrow p-r+p+p+r= \frac{3}{2}$

$\Rightarrow p= \frac{1}{2}$

$a=p-r =\frac{1}{2}-r$

$a^2, b^2, c^2 \rightarrow \text{G P}$

$(p-r)^2, p^2,(p+r)^2 \rightarrow \text{G P}$

$(p-r)^2(p+r)^2=p^4$

$\left(p^2-r^2\right)^2=p^4$

$\Rightarrow p^4-2 p^2 r^2+r^4=p^4$

$\Rightarrow r^2\left(r^2-2 p^2\right)=0$

$r \neq 0$

$r^2=2 p^2$

$r^2=\frac{1}{2} \ \ \Rightarrow r= \pm \frac{1}{\sqrt{2}}$

$a<b<c$

$\quad \downarrow$

$\quad \frac{1}{2}$

$\quad a = \frac{1}{2} + \frac{1}{\sqrt{2}} \ \text { not possible }$

$\quad a = \frac{1}{2} - \frac{1}{\sqrt{2}}$

Suppose the sum of the first $n$ terms of an arithmetic progression is $\mathrm{cn}^2$. Then the sum of the square of these $n$ terms is

(1)$n\left(4 n^2-1\right) c^2 / 6$

(2) $n\left(4 n^2+1\right) c^2 / 6$

(3) $n\left(4 n^2-1\right) c^2 / 3$

(4)$n\left(4 n^2+1\right) c^2 / 3$

The sum of first $(n-1)$ terms

$\quad=c(n-1)^2$

$n-t h \text { term }=c n^2-c(n-1)^2=c(2 n-1)$

call $r-$ th term $a_r$

$\sum_{r=1}^n a_r^2 =c^2 \sum_{r=1}^n(2 r-1)^2$

$=4 c^2 \sum_{r=1}^n r^2-4 c^2 \sum_{r=1}^n r+c^2 \sum_{r=1}^n 1$

$=\frac{4 c^2 n(n+1)(2 n+1)}{6}-4 c^2 \frac{n(n+1)}{2}+c^2 n$

$=\frac{c^2 n}{3}{({(2 n+2)(2 n+1)-6 n-6+ 3 })}$

$=\frac{c^2 n}{3}{({4n^2+4n+2n+2-6 n-3 })}$

$=\frac{c^2 n}{3}{({4n^2-1})}$

Let $b_i>1$ for $1 \leq i \leq 101$. Let $\log b_1, \log b_2, \ldots, \log b_{101}$ be in arithmetic progression with common difference $\log 2$. Suppose $a_1, a_2, \ldots, a_{101}$ are in arithmetic progression such that $a_1=b_1$ and $a_{51}=b_{51}$. If $s=a_1+\cdots+a_{51}$ and $t=b_1+\cdots+b_{51}$, then

(1) $s>t$ and $a_{101}>b_{101}$

(2) $s>t$ and $a_{101}<b_{101}$

(3) $s<t$ and $a_{101}>b_{101}$

(4) $s<t$ and $a_{101}<b_{101}$

$\log b_2 =\log b_1+\log 2$

$=\log 2 b_1 \ \Rightarrow b_2=2 b_1$

$\log b_3 =\log b_2+\log 2$

$\ =\log 2 b_1+\log 2$

$\ =\log 2^2 b_1$

$\ b_3 =2^2 b_1$

$\ b_i =2^{i-1} b_1 \forall 2 \leqslant i \leqslant 101$

$t =b_1+b_2+b_3+\cdots+b_{51}$

$=b_1+2 b_1+2^2 b_1+\cdots+2^{50} b_1$

$=b_1(\underbrace{1+2+2^2+\cdots+2^{50}}_{\downarrow})$

$\quad \quad \quad \quad ({2^{51}-1})$

$=b_1\left(2^{51}-1\right)$

$s=a_1+a_2+a_3+\cdots+a_{51}$

$=a_1+\left(a_1+d\right)+\left(a_1+2 d\right)+\cdots+\left(a_1+50 d\right)$

d = common difference of ${a_1, \ldots, a_{101}}$

$=51 a_1+d(\underbrace{1+2+\cdots+50}_{\downarrow})$

$\quad \quad \quad \quad \quad \quad \quad \quad \frac{50 \times 51}{2}$

$a_{51}=b_{51}$

$a_1=b_1$

$a_{51}=a_1+50 d$

$\Rightarrow b_{51}=b_1+50 d$

$\Rightarrow d=\frac{b_{51}-b_1}{50}=\frac{\left(2^{50}-1\right) b_1}{50}$

Let $a_1, a_2, a_3, \ldots$ be in harmonic progression with $a_1=5$ and $a_{20}=25$.

Then the least positive integer $n$ for which $a_n<0$ is

$(1)22$

$(2)23$

$(3)24$

$(4)25$

$a_n=\frac{1}{b+(n-1) d} \text { for some b,d }$

$a_1=\frac{1}{b}$

$b=\frac{1}{5}$

$a_{20}=\frac{1}{b+19 d} \\ \Rightarrow b+19 d=\frac{1}{a_{20}}=\frac{1}{25}$

$\ b=\frac{1}{5}$

$\Rightarrow d=-\frac{4}{25 \times 19}$

$a_n < 0$

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

$b+(n-1) d<0$

$\frac{1}{5}+(n-1)\left(-\frac{4}{25 \times 19}\right)<0$

$\Rightarrow 1+(n-1)\left(-\frac{4}{5 \times 19}\right)<0$

$\Rightarrow(n-1)>\frac{5 \times 19}{4} \Rightarrow n>{\frac{5 \times 19}{4}+1}=\frac{99}{4}$

$n \geqslant 25$

Suppose four distinct numbers $a_1, a_2, a_3, a_4$ are in geometric progression. Let $b_1=a_1$ and $b_i=b_{i-1}+a_i$ for $i=2,3,4$.

Statement 1: The numbers $b_1, b_2, b_3, b_4$ are neither in arithmetic progression, nor in geometric progression.

Statement 2: The numbers $b_1, b_2, b_3, b_4$ are in harmonic progression.

(1) Statement 1 is false, Statement 2 is true.

(2) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation of Statement 1.

(3) Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation of Statement 1.

(4) Statement 1 is true, Statement 2 is false.

$b_ 1=a_1$

$b_ 2=b_ 1+a_ 2=a_ 1+a_2$

$b_ 3=b_ 2+a_ 3=a_ 1+a_ 2+a_3$

$b_ 4=b_ 3+a_ 4=a_ 1+a_ 2+a_ 3+a_4$

$b_ 2-b_ 1=a_ 2$

$b_ 3-b_ 2=a_ 3, \ \ a_ 2 \neq a_3$

$b_ 2-b_ 1 \neq b_ 3-b_ 2 \ \Rightarrow 2 b_ 2 \neq b_ 1+b_3$

$b_ 1, b_ 2, b_3 \text { are not in AP }$

$a_ i=a_ 1 r^{i-1} \ \ \ \text{for} \ i=2,3,4$
$\quad \quad \downarrow$

${\text{common ratio of} \ a_ 1, a_ 2, a_ 3, a_4}$

Note

$a_1 \neq 0, \ \ r \neq 0$

$b_2 =a_1+a_1 r=a_1(1+r)$

$b_3 =a_1+a_1 r+a_1 r^2=a_1\left(1+r+r^2\right)$

$b_4 =a_1+a_1 r+a_1 r^2+a_1 r^3 =a_1\left(1+r+r^2+r^3\right)$

$b_1 b_3 =b_2^2\left(\text { if } b_1, b_2, b_3\right. \text { are in GP) }$

$a_1 \cdot a_1\left(1+r+r^2\right)=a_1^2(1+r)^2$

$\quad \text { or } a_1 \neq 0,$

$\left(1+r+r^2\right)=\left(1+2 r+r^2\right)$

For $r \neq 0$ does not hold.

$b_1 b_3 \neq b_2^2$

$b_1, b_2, b_3, b_4$ are not in GP

If $b_1, b_2, b_3, b_4$ are in HP, then

$\quad \frac{1}{b_1}+\frac{1}{b_3}=\frac{2}{b_2}$

$\text { i.e. } \frac{1}{a_1}+\frac{1}{a_1\left(1+r+r^2\right)}=\frac{2}{a_1(1+r)}$

$\text { i.e. } \quad \frac{1+r+r^2+1}{1+r+r^2}=\frac{2}{1+r}$

$\text { i.e. } \quad(1+r)\left(2+r+r^2\right)=2\left(1+r+r^2\right)$

$\text { i.e. } \quad 2+r+r^2+2 r+r^2+r^3 =2+2 r+2 r^2$

ie. $\quad r\left(1+r^2\right)=0$

i.e. $r=0$ or $r= \pm i$

Recall $\quad r \neq 0$

if $r= \pm i$

$b_4=a_1\left(1+r+r^2+r^3\right)=0$

$b_1, b_2, b_3, b_4$ can not be in HP

$a_ 2=a_1+d$

$S_ p = \sum_ {i=1}^p a_ i$

$=p a_1+(1+2+\cdots+(p-1)) d$

$=3 p+\frac{p(p-1)}{2} d$

$=\frac{p}{2} (6+(p-1) d)$

$\frac{S_m}{S_n}=\frac{S_{5 n}}{S_n}=\frac{\frac{5 n}{2}{6+(5 n-1) d}}{\frac{n}{2}{6+(n-1) d}}$

$=\frac{30+(25 n-5) d}{6+(n-1) d}$

$30+(25 n-5) d=c\{6+(n-1) d\}$

$\Rightarrow 30+25 n d-5 d=6 c+c n d-c d$

$\Rightarrow(25-c) n d=6 c-c d-30+5 d$

$(25-c)d=0$

${d=0}$

$\quad {\downarrow}$

$a_2=a_1+0 =3$

$\text {or}$

$\quad c=25$

$\quad \quad \downarrow$

$\frac{30+25 n d-5 d}{6+n d-d}=25$

$\Rightarrow 30+25 n d-5 d=150 +25 n d -25 d$

$\Rightarrow 20 d=150-30=120$

$\Rightarrow \quad d=6$

$\begin{aligned} a_2 & =a_1+6 =9\end{aligned}$

$a=\frac{b}{r} \quad \frac{b}{a} \in \mathbb{Z}$

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

$\therefore r \in \mathbb{Z}_{>0}$

$\frac{b}{r}-2 b+b r=6$

$\Rightarrow b-2 b r+b r^2=6 r$

$\Rightarrow b\left(r^2-2 r+1\right)=6 r$

$\Rightarrow \frac{b}{r}(r-1)^2=6$

$a(r-1)^2=6 \ \ a, r \in \mathbb{Z}>0$

$r=2, a=6$

$\frac{a^2+a-14}{a+1} \\ =\frac{36+6-14}{7} \\ = \frac{28}{7} \\ = 4$

Solution:

$d \rightarrow$ common difference

$a_r \rightarrow$ r-th term

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

$\forall r \geqslant 1$

$a_r \in \mathbb{Z}_{>0} \quad \forall r \geqslant 1$

$d \in \mathbb{Z}_{>0}$

$\frac{\sum_{i=1}^7 a_i}{\sum_{i=1}^{11} a_i}=\frac{6}{11}$

$\sum_{i=1}^7 a_i=7 a_1+{ \underbrace{(1+2+\cdots+6)} } d$

$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \frac{6 \cdot 7}{2}$

$=7(a_1+3 d)$

$\sum_{i=1}^{11} a_i=11 a_1+(1+2+\cdots+10) d$

$\ \ =11 a_1+\frac{10 \cdot 11}{2} d$

$\ \ =11\left(a_1+5 d\right)$

$\frac{7\left(a_1+3 d\right)}{11\left(a_1+5 d\right)}=\frac{6}{11}$

$\Rightarrow 7 a_1+21 d=6 a_1+30 d$

$\Rightarrow \ a_1=9 d$

$130<a_7<140$

$\quad \quad \quad \downarrow$

$\quad \quad a_1+6 d$

$\quad \quad \quad \downarrow$

$\quad \quad \quad 15 d$

$\frac{130}{15}<d<\frac{140}{15} \quad d=9$

$\frac{26}{3}<d<\frac{28}{3}$

$9-\frac{1}{3}<d<9+\frac{1}{3}$