Maths M.C.Q (1 Marks)

We have,

$\frac{2^2+4^2+6^2+\ldots+(2 n)^2}{1^2+3^2+5^2+\ldots+(2 n-1)^2} < 1.01$

$=\frac{\Sigma 4 n^2}{\sum\left(4 n^2-4 n+1\right)} < 1.01$

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

$= \frac{4 \frac{n(n+1)(2 n+1)}{6}}{\frac{n(2 n+1)(2 n-1)}{3}}<101=\frac{2(n+1)}{2 n-1} < \frac{101}{100}$

$= 200 n+200 < 202 n-101$

$\Rightarrow 2 n > 301 \Rightarrow n > \frac{301}{2}=150.5$

$\therefore n > 151$

Let

$S_n=\left(1^2-1+1\right)(1 !)+ \left(2^2-2+1\right)(2 !)$

$+\ldots+\left(n^2-n+1\right)(n !)$

Here, $T_r=\left(r^2-r+1\right)(r !)$

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

$T_r=\left(r^2-1\right) r !-(r-2) r !$

$T_r=(r-1)(r+1) !-(r-2) r !$

$S_n= T_1+T_2+T_3+\ldots+T_n$
$S_n= (1-0)+(3 !-0)+(2 \cdot 4 !-1 \cdot 3 !)$

$\quad\quad+(3 \cdot 5 !-2 \cdot 4 !)+(4 \cdot 6 !)-3 \cdot 5 ! \ldots$

$\quad\quad+[(n-1)(n+1) !-(n-2)(n !)]$

$S_n= 1+(n-1)(n+1) !$

$S_n=(n-1)(n+1) !+1$

We have,

$\frac{1^2+2^2+3^2+4^2+\ldots+n^2}{1+2+3+4+\ldots+n}$

$=\frac{\frac{n(n+1)(2 n+1)}{6}}{n(n+1)}$

$=\frac{2 n+1}{2}=k(l \text { let })$

$\Rightarrow \quad n=\frac{3 k-1}{2}$

$\text { Now, } 1 \leq \frac{3 k-1}{2} \leq 100$

$\Rightarrow \quad 2 \leq 3 k-1 \leq 200$

$\Rightarrow \quad 2+1 \leq 3 k \leq 200+1$

$\Rightarrow \quad 3 \leq 3 k \leq 201$

$\Rightarrow \quad 1 \leq k \leq 201$

$\Rightarrow \quad 1 \leq k \leq 67$

Number of odd integer $=34$

Let $P(x)=1+x^2+x^4+\ldots+x^{2010}$

${c}P(x)=\frac{1-x^{2012}}{1-x^2}$

$P(x)=\frac{\left(1-x^{1006}\right)\left(1+x^{1006}\right)}{(1-x)(1+x)}$$P(x)=\left(1+x^{1006}\right)\left(\frac{1-x^{503}}{1-x}\right)\left(\frac{1+x^{503}}{1+x}\right)$

$\left.P(x)=\left(1+x^{1006}\right) 1+x+x^2+x^3+\ldots+x^{502}\right)$

$\left(1-x+x^2-x^3+\ldots+x^{502}\right)$

Thus, $P(x)$ is divisible by $1+x+x^2+\ldots x^{n-1}$

If $n-1=502 \Rightarrow n=503$

Let the sides of triangle are

$a, a r, a r^2 . \quad[\because$ sides of triangle in $GP ]$

Case I $r > 1$

We know sum of two sides is greater than third side.

$\therefore a+a r>a r^2 \Rightarrow r^2-r-1<0$

$\begin{array}{ll} \Rightarrow & r=\frac{1 \pm \sqrt{5}}{2} \\\Rightarrow & 1 < r < \frac{\sqrt{5}+1}{2}, r > 1\end{array}$

Case $II$ $0 < r < 1$

$\therefore a r^2+a r > a \Rightarrow r^2+r-1 > 0$

$\Rightarrow \quad r=\frac{-1 \pm \sqrt{5}}{2}$

$\Rightarrow \quad 1 > r > \frac{\sqrt{5}-1}{2}, 0 < r < 1$

$\therefore \quad r \in\left(\frac{\sqrt{5}-1, \sqrt{5}+1}{2}, \frac{2}{2}\right)$

If all trucks had packets of $1000\,g$, then total weight is

$1000(1+2 \left.+2^2+\ldots+2^9\right)$

$=1000\left(2^{10}-1\right)$

$=1000(1024-1)=1023000$

We have given total weight is $1022870$.

$\therefore 1022870 < 1023000$

Extra amount of weight

$=1023000-1022870$

$=130$

$130 =2^7+2^1=128+2$

$\therefore$ The trucks have the lighter bags are $2,8$

We have, $a_1, a_2, a_3, \ldots, a_n$ in an $AP$ and $a_1, a_2, a_4, a_8$ in GP.

Let $a_1=a, a_2=a+d, a _3=a+2 d$ and $a_1$, $a_2, a_4, a_8$ are in $GP$.

Let common ratio is $r$.

$\therefore \quad a_1=a, a_2=a r, a_4=a r^2, a_8=a r^3$

$\therefore a+ d =a r, a+3 d=a r^2, a+7 d=a r^3$

$\Rightarrow \quad 2 d =a r(r-1), 4 d=a r^2(r-1)$

$\Rightarrow \quad r=2$

We have, $2^n < k < 2^{n+1}, k \in N$

Number of integer between $2^n$ and $2^{n+1}$ is i.e $k=2^{n+1}-2^n-1$

First term $=2^n+1$

Last term $=2^{n+1}-1$

$\therefore S_n=\frac{2^{n+1}-2^n-1}{2}\left[2^n+1+2^{n+1}-1\right]$

$S_n=\frac{2^{n+1}-2^n-1}{2}\left(2^n\right)(1+2)$

$S_n=\frac{\left(2^n-1\right)\left(2^n\right) \cdot 3}{2}$

But $S_n=9 m, m \in I$

$\therefore \quad \frac{\left(2^n-1\right) 2^n \cdot 3}{2}=9 m$

$\Rightarrow \quad\left(2^n-1\right) 2^{n-1}=3\,m$

$\Rightarrow \quad 2^n\left(2^n-1\right)=6\,m$

It is possible when, $n$ is even.

We have,

$P(x)=1+x+x^2+x^3+x^4+x^5$

$P(x)=\frac{1-x^6}{1-x}$

${\left[\because a+a r+a r^2+\ldots+a r^n=\frac{a\left(-r^n\right)}{1-r}\right]}$

It has $5$ roots let $\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5$ they are $6$ th roots of unity

Now,

$P\left(x^{12}\right)=1+x^{12}+x^{24}+x^{36}+x^{48}+x^{60}$

$\therefore P\left(x^{12}\right) =P(x) \cdot Q(x)+R(x)$

Here, $R(x)$ is a polynomial of maximum degree $4 .$

Put $x=\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5$; we get

$R\left(\alpha_1\right)=6=R\left(\alpha_2\right) =R\left(\alpha_3\right)$

$=R\left(\alpha_4\right)=R\left(\alpha_5\right)$

$\therefore R(x)-6=0$ has 6 roots, which contradicts that $R(x)$ is maximum of degree $4 .$

$\therefore$ So it is an identity.

$\because \quad R(x)=6$

Let $a, b, c$ be the sides of a triangle.

$\therefore \quad a^2+b^2 \geq 2 a b \quad[\because AM \geq GM ]$

Similarly, $b^2+c^2 \geq 2 b c$

$c^2+a^2 \geq 2 a c$

$\Rightarrow 2\left(a^2+b^2+c^2\right) \geq 2(a b+b c+c a)$

$\Rightarrow \quad \frac{a^2+b^2+c^2}{a b+b c+a c} \geq 1$

$t \geq 1 \quad\left[\because \frac{a^2+b^2+c^2}{a b+b c+a c}=t\right]$

$\because a, b, c$ be the side of a triangle.

$\begin{aligned} a+b>c \\ & \quad|a|>|c-b| \\ a \end{aligned}$

$\Rightarrow \quad a^2 > (c-b)^2$

$\Rightarrow \quad a^2 > c^2+b^2-2 b c$

$\Rightarrow \quad b^2+c^2-a^2<2 b c$

Similarly, $a^2+b^2-c^2 < 2 a b$

and $\quad c^2+a^2-b^2 < 2 a c$

On adding Eqs.$(i), (ii)$ and $(iii)$, we get

$a^2+b^2+c^2 < 2(a b+b c+c a)$

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

$\therefore \quad t < 2$

$\therefore \quad\{x \in R \mid 1 \leq x < 2\}$

If we fix $j=1$, then $i$ range, we get $1+2+3+\ldots+(n-1)$

if we fix $j=2$, then $i$ range, we get $1+2+3+\ldots+(n-2)$

Similarly, we fix $j=n-1$, then $i$ range, We gret

if we put them all together we get a total of ( $n-1)$ numbers $1^{\prime}$ s,$(n-2)$ number of $2^{\prime} s . \ldots .1$ number of $(n-1)^{\prime}$ sthen sum equals

$\sum \limits_{k=1}^{n-1} k(n-k)=n \sum \limits_{k=1}^{n-1} k-\sum \limits_{k=1}^{n-1} k^2$

$=\frac{n(n)(n-1) \quad n(n-1)(2 n-1)}{2}$

$\left.=\frac{n(n-1)\left[n-\frac{2 n-1}{2}[n\right.}{3}\right]$

$=\frac{n(n-1)(n+1)}{2 \cdot 3}={ }^{n+1} C_3$

$ 2 \mathrm{a}+19 \mathrm{~d}=79$      $.............(1)$

$ \mathrm{~S}_{10}=\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}]=145 $

$ 2 \mathrm{a}+9 \mathrm{~d}=29$        $................(2)$   

From $(1)$ and $(2)$ a $=-8, d=5$

$ S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d] $

$ =\frac{15}{2}[-16+70]-\frac{5}{2}[-16+20] $

$ =405-10 $

$ =395$

$ p=A, q=A R, r=A R^2$

$ A x^2+A R x-A R^2=0$

$ x^2+R x-R^2=0 < \beta $

$ \because \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} $

$ \therefore \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \Rightarrow \frac{-R}{-R^2}=\frac{3}{4} \Rightarrow R=\frac{4}{3} $

$ (\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta=R^2-4\left(-R^2\right)=5\left(\frac{16}{9}\right) $

$ =80 / 9$

$ \frac{a+b+c}{3}=8 \rightarrow b=8, a+c=16 $

$ 64=(a+1)(19-a)=19+18 a-a^2 $

$ a^2-18 a-45=0 \rightarrow(a-15)(a+3)=0,(a>10) $

$ a=15, c=1, b=8 $

$ \left((a b c)^{1 / 3}\right)^3=a b c=120$

(sum of infinite terms of A.G.P $=\frac{\mathrm{a}}{1-\mathrm{r}}+\frac{\mathrm{dr}}{(1-\mathrm{r})^2}$ )

$\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9$

$ 17 \mathrm{~m}=25 \mathrm{~m}-4(1+2 \ldots 24) $

$ 8 \mathrm{~m}=\frac{4 \cdot 24 \cdot 25}{2}=150$

$=7\left(a+a r^2+a r^4 \ldots+a r^{62}\right) $

$\Rightarrow \frac{a\left(1-r^{64}\right)}{1-r}=\frac{7 a\left(1-r^{64}\right)}{1-r^2}$

$r=6$

$ \quad \geq 2 \quad \geq 2 $

$ \mathrm{k} \geq 10$

$ \mathrm{A}_{\mathrm{k}}=-\mathrm{kd}[2 \mathrm{a}+(2 \mathrm{k}-1) \mathrm{d}] $

$ \mathrm{A}_3=-153 $

$ \Rightarrow 153=13 \mathrm{~d}[2 \mathrm{a}+5 \mathrm{~d}] $

$ 51=\mathrm{d}[2 \mathrm{a}+5 \mathrm{~d}] $

$ \mathrm{A}_5=-435 $

$ 435=5 \mathrm{~d}[2 \mathrm{a}+9 \mathrm{~d}] $

$ 87=\mathrm{d}[2 \mathrm{a}+9 \mathrm{~d}] $

$ (2)-(1) $

$ 36=4 \mathrm{~d}^2$

$ \mathrm{~d}=3, \mathrm{a}=1 $

$ \mathrm{a}_{17}-\mathrm{A}_7=49-[-7.3[2+39]]=910$

$ \frac{10}{2}[2 \mathrm{a}+(10-1) \mathrm{d}]=390 $

$ \Rightarrow 2 \mathrm{a}+9 \mathrm{~d}=78 $               $......(1)$

$ \frac{\mathrm{t}_{10}}{\mathrm{t}_5}=\frac{15}{7} \Rightarrow \frac{\mathrm{a}+9 \mathrm{~d}}{\mathrm{a}+4 \mathrm{~d}}=\frac{15}{7} \Rightarrow 8 \mathrm{a}=3 \mathrm{~d} $                        $......(2)$

$ \text { From }(1) \&(2) \quad \mathrm{a}=3 \& \mathrm{~d}=8 $

$ \mathrm{~S}_{15}-\mathrm{S}_5=\frac{15}{2}(6+14 \times 8)-\frac{5}{2}(6+4 \times 8) $

$ =\frac{15 \times 118-5 \times 38}{2}=790$

$2,5,8,11, \ldots ., 404$

$\operatorname{LCM}(4,3)=12$

$11,23,35, \ldots . \text { let }(403)$

$403=11+(n-1) \times 12$

$\frac{392}{12}=n-1$

$33 \cdot 66=n$

$n=33$

$\operatorname{Sum} \frac{33}{2}(22+32 \times 12)$

$=6699$

${n}{2}(6+(n-1) 4)=3 n+2 n^2-2 n $

$ =2 n^2+n $

$ \sum_{k=1}^n S_k=2 \sum_{k=1}^n K^2+\sum_{k=1}^n K $

$ =2 \cdot \frac{n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} $

$=n(n+1)\left[\frac{2 n+1}{3}+\frac{1}{2}\right]$

$=\frac{n(n+1)(4 n+5)}{6} $

Rightarrow $40<\frac{6}{n(n+1)} \sum_{k=1}^n S_k<42 $

$ 40<4 n+5<42 $

$ 35<4 n<37 $

$ n=9$

$\therefore \mathrm{b}^2=\mathrm{ac}$

Also

$\log _{\circ}\left(\frac{a}{2 b}\right), \log _{\circ}\left(\frac{2 b}{3 c}\right), \log _{\circ}\left(\frac{3 c}{a}\right)$ are in $A.P.$

$\left(\frac{2 b}{3 \mathrm{c}}\right)^2=\frac{\mathrm{a}}{2 \mathrm{~b}} \times \frac{3 \mathrm{c}}{\mathrm{a}} $

$ \frac{\mathrm{b}}{\mathrm{c}}=\frac{3}{2}$

Putting in eq. $(i)$ $b^2=a \times \frac{2 b}{3}$

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

$ a: b: c=9: 6: 4$

This is $A.P$. with common difference

$ d_1=-1+\frac{1}{4}=-\frac{3}{4} $

$ -129 \frac{1}{4}, \ldots \ldots \ldots . . ., 19 \frac{1}{4}, 20$

This is also $A.P.$ $a=-129 \frac{1}{4}$ and $d=\frac{3}{4}$

Required term $=$

$ -129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right)$

$ =-129-\frac{1}{4}+15-\frac{3}{4}=-115$

$\mathrm{T}_{25}=4+(25-1) 5=4+120=124$

$3,6,9,12, \ldots$, up to $37^{\text {th }}$ term

$\mathrm{T}_{37}=3+(37-1) 3=3+108=111$

Common difference of $\mathrm{I}^{\text {st }}$ series $\mathrm{d}_{\mathrm{l}}=5$

Common difference of $\mathrm{In}^{\text {nd }}$ series $\mathrm{d}_2=3$

First common term $=9$, and

their common difference $=15\left(\mathrm{LCM}\right.$ of $\mathrm{d}_1$ and $\mathrm{d}_2$ )

then common terms are

$9,24,39,54,69,84,99$

$ \mathrm{a}+\mathrm{ar}+\mathrm{ar}^2+\infty=57 $

$ \frac{\mathrm{a}}{1-\mathrm{r}}=57 \ldots \ldots \ldots \ldots .(\mathrm{I}) $

$ \sum_{\mathrm{n}=0}^{\infty} \mathrm{a}^3 \mathrm{r}^{3 \mathrm{n}}=9747 $

$ \mathrm{a}^3+\mathrm{a}^3 \cdot \mathrm{r}^3+\mathrm{a}^3 \cdot \mathrm{r}^6+\ldots \ldots \ldots \infty=9746 $

$ \frac{\mathrm{a}^3}{1-\mathrm{r}^3}=9746 \ldots \ldots \ldots \ldots(\mathrm{II}) $

$ \frac{(\mathrm{I})^3}{(\mathrm{II})} \Rightarrow \frac{\mathrm{a}^3}{\frac{(1-\mathrm{r})^3}{\mathrm{a}^3}}=\frac{57^3}{9717}=19 $

$ \text { On solving, } \mathrm{r}=\frac{2}{3} \text { and } \mathrm{r}=\frac{3}{2}(\text { rejected) } $

$ \mathrm{a}=19 $

$ \therefore \mathrm{a}+18 \mathrm{r}=19+18 \times \frac{2}{3}=31$

$ \mathrm{ar}+\mathrm{ar}^5=\frac{70}{3} $

$ \mathrm{~T}_3 \cdot \mathrm{T}_5=49 $

$ \mathrm{ar}^2 \cdot \mathrm{ar}^4=49 $

$ \mathrm{a}^2 \mathrm{r}^6=49 $

$ \mathrm{ar}^3=+7, \mathrm{a}=\frac{7}{\mathrm{r}^3} $

$ \mathrm{ar}\left(1+\mathrm{r}^4\right)=\frac{70}{3} $

$ \frac{7}{\mathrm{r}^2}\left(1+\mathrm{r}^4\right)=\frac{70}{3}, \mathrm{r}^2=\mathrm{t} $

$ \frac{1}{\mathrm{t}}\left(1+\mathrm{t}^2\right)=\frac{10}{3} $

$ 3 \mathrm{t}^2-10 \mathrm{t}+3=0 $

$ \mathrm{t}=3, \frac{1}{3}$

Increasing $G.P$. $\mathrm{r}^2=3, \mathrm{r}=\sqrt{3}$

$ \mathrm{T}_4+\mathrm{T}_6+\mathrm{T}_8 $

$ =\mathrm{ar}^3+\mathrm{ar}^5+\mathrm{ar}^7 $

$ =\mathrm{ar}^3\left(1+\mathrm{r}^2+\mathrm{r}^4\right) $

$ =7(1+3+9)=91$

$ f(\theta)=1+\frac{2 \cos ^2 \theta}{\sin ^4 \theta+\cos ^2 \theta} $

$ f(\theta)=\frac{2 \cos ^2 \theta}{\cos ^4 \theta-\cos ^2 \theta+1}+1 $

$ f(\theta)=\frac{2}{\cos ^2 \theta+\sec ^2 \theta-1}+1 $

$ \left.f(\theta)\right|_{\min .}=1 $

$ f(\theta)_{\max }=3 $

$ S=\frac{64}{1-1 / 3}=96$

Area of second $\Delta=\frac{\sqrt{3} a^2}{4} \frac{a^2}{4}=\frac{\sqrt{3} a^2}{16}$

Area of third $\Delta=\frac{\sqrt{3} \mathrm{a}^2}{64}$

sum of area $=\frac{\sqrt{3} a^2}{4}\left(1+\frac{1}{4}+\frac{1}{16} \ldots\right)$

$\mathrm{Q}=\frac{\sqrt{3} \mathrm{a}^2}{4} \frac{1}{\frac{3}{4}}=\frac{\mathrm{a}^2}{\sqrt{3}}$

perimeter of $1^{\text {st }} \Delta=3 \mathrm{a}$

perimeter of $2^{\text {nd }} \Delta=\frac{3 a}{2}$

perimeter of $3^{\text {rd }} \Delta=\frac{3 \mathrm{a}}{4}$

$ \mathrm{P}=3 \mathrm{a}\left(1+\frac{1}{2}+\frac{1}{4}+\ldots\right) $

$ \mathrm{P}=3 \mathrm{a} \cdot 2=6 \mathrm{a} $

$ \mathrm{a}=\frac{\mathrm{P}}{6} $

$ \mathrm{Q}=\frac{1}{\sqrt{3}} \cdot \frac{\mathrm{P}^2}{36} $

$ \mathrm{P}^2=36 \sqrt{3} \mathrm{Q}$

Sum of any two sides $>$ third side

$ a+a r>a r^2, a+a r^2>a r, a r+a r^2>a $

$ r^2-r-1<0 $

$ r \in\left(\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right) $

$ r^2-r+1>0$                    $............(1)$

always true

$ \mathrm{r}^2+\mathrm{r}-1>0 $

$ \mathrm{r} \in\left(-\infty,-\frac{1-\sqrt{5}}{2}\right) \cup\left(\frac{-1+\sqrt{5}}{2}, \infty\right)$               $...............(2)$

Taking intersection of $(1)$, $(2)$

$\mathrm{r} \in\left(\frac{-1+\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right)$

As $\mathrm{r}>1$

$ r \in\left(1, \frac{1+\sqrt{5}}{2}\right) $

$ {[r]=1[-r]=-2} $

$ 3[r]+[-r]=1$

$ \mathrm{f}(\mathrm{x})=\mathrm{a}+\mathrm{b}-2 \mathrm{c}+\mathrm{b}+\mathrm{c}-2 \mathrm{a}+\mathrm{c}+\mathrm{a}-2 \mathrm{~b}=0$

$ \mathrm{f}(1)=0 $

$\therefore \alpha \cdot 1=\frac{\mathrm{c}+\mathrm{a}-2 \mathrm{~b}}{\mathrm{a}+\mathrm{b}-2 \mathrm{c}} $

$ \alpha=\frac{\mathrm{c}+\mathrm{a}-2 \mathrm{~b}}{\mathrm{a}+\mathrm{b}-2 \mathrm{c}} $

$ \text { If, }-1<\alpha<0 $

$ -1<\frac{\mathrm{c}+\mathrm{a}-2 \mathrm{~b}}{\mathrm{a}+\mathrm{b}-2 \mathrm{c}}<0$

$\mathrm{b}+\mathrm{c}<2 \mathrm{a} \text { and } \mathrm{b}>\frac{\mathrm{a}+\mathrm{c}}{2}$
therefore, $\mathrm{b}$ cannot be G.M. between $\mathrm{a}$ and $\mathrm{c}$.
If, $0<\alpha<1$
$0<\frac{\mathrm{c}+\mathrm{a}-2 \mathrm{~b}}{\mathrm{a}+\mathrm{b}-2 \mathrm{c}}<1$
$\mathrm{b}>\mathrm{c}$ and $\mathrm{b}<\frac{\mathrm{a}+\mathrm{c}}{2}$
Therefore, $\mathrm{b}$ may be the $G.M.$ between $\mathrm{a}$ and $\mathrm{c}$.

$ t_{11}  =a r^{10}=a\left(r^2\right)^5=a \cdot\left(\frac{b}{a}\right)^5 $

$2^{\text {nd }} \text { G.P. }  \Rightarrow T_1=a, T_5=a r^4=b $

$\Rightarrow r^4  =\left(\frac{b}{a}\right) \Rightarrow r=\left(\frac{b}{a}\right)^{1 / 4} $

$ T_p  =a r^{p-1}=a\left(\frac{b}{a}\right)^{\frac{p-1}{4}} $

$ t_{11}  =T_p \Rightarrow a\left(\frac{b}{a}\right)^5=a\left(\frac{b}{a}\right)^{\frac{p-1}{4}} $

$ \Rightarrow 5  =\frac{p-1}{4} \Rightarrow p=21$

$ {\left[\frac{1}{2+\sqrt{2}}, \frac{1}{2-\sqrt{2}}\right]} $

$ \frac{\alpha}{\beta}=\frac{a+b}{2 \sqrt{a b}}=\frac{1}{2}\left(\sqrt{\frac{a}{b}}+\sqrt{\frac{b}{a}}\right) $

$ =\frac{1}{2}\left(\sqrt{\frac{2-\sqrt{2}}{2+\sqrt{2}}}+\sqrt{\frac{2+\sqrt{2}}{2-\sqrt{2}}}\right) $

$ =\frac{(2-\sqrt{2})+(2+\sqrt{2})}{2 \times \sqrt{2}}=\sqrt{2}$

$ p=a+7 d \quad \ldots \text { (ii) } $

$ q=a+12 d \quad \ldots \text { (iii) } $

$ p-2=d $                    $((ii)-(i))$

$ q-p=5 d $                 $((iii)-(ii))$

$ q-p=5(p-2) $

$ q=6 p-10 $

$ p^2=2(6 p-10) $

$ p^2-12 p+20=0 $

$ p=10,2 $

$ p=10 ; q=50 $

$ d=8 $

$ a=-46 $

$ 2,10,50,250,1250 $

$ a^4=a+(n-1) d $

$ 1250=-46+(n-1) 8 $

$ n=163$

$3, \mathrm{a}-1, \mathrm{~b}+1, \mathrm{c}+9 \rightarrow \mathrm{G} . \mathrm{P} \Rightarrow 3,2+\mathrm{d}, 4+2 \mathrm{~d}, 12+3 \mathrm{~d}$

$\mathrm{a}=3+\mathrm{d}$      $(2+d)^2=3(4+2 d)$

$\mathrm{b}=3+2{~d}$         ${~d}=4,-2$

${c}=3+3{~d}$

$\text { If } \mathrm{d}=4 \quad \text { G.P } \Rightarrow 3,6,12,24$

$\mathrm{a}=7$

$\mathrm{~b}=11$

$\mathrm{c}=15$

$\frac{a+b+c}{3}=11$

$ (1+7 d)^2=(1+d)(1+43 d) $

$ 1+49 d^2+14 d=1+44 d+43 d^2 $

$ 6 d^2-30 d=0 $

$ d=5 $

$ S_{20}=\frac{20}{2}[2 \times 1+(20-1) \times 5] $

$=10[2+95] $

$=970$

$ 2 a_r=a_{r+1}+a_{r+2} $

$2 a r^{n-1}=a r^n+a r^{n-1} $

$ \frac{2}{r}=1+r $

$ r^2+r-2=0$

Hence, we get, $r=-2($ as $r \neq 1)$

So, $\mathrm{S}_{20}-\mathrm{S}_{18}=$ (Sum upto $20$ terms) - (Sum upto

$18$   terms )$=\mathrm{T}_{19}+\mathrm{T}_{20} $

$ \mathrm{~T}_{19}+\mathrm{T}_{20}=\mathrm{ar}^{18}(1+\mathrm{r})$

Putting the values $\mathrm{a}=\frac{1}{8}$ and $\mathrm{r}=-2$;

we get $T_{19}+T_{20}=-2^{15}$

$\Rightarrow $ $ \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) $ $ -\left\{\left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\ldots+\frac{1}{2023}\right. $ $ \left.-\frac{1}{2024}-2\left(\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2022}\right)\right\}=\frac{1}{2024} $

$\Rightarrow $ $ \left(\frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}\right) $ $ -\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{2023}\right) $ $ \frac{1}{2024}+\left(\frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{1011}\right)=\frac{1}{2024} $

$\Rightarrow $ $ \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012} $ = $ \frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023} $

$\Rightarrow $ $ \alpha=1011$

$ \frac{1}{d}\left[\frac{(1+d)-1}{1 \cdot(1+d)}+\frac{(1+2 d)-(1-d)}{(1+d)(1+2 d)}\right]+\ldots  . . . . $ $ \frac{(1+10 d)-(1+9 d)}{(1+9 d)(1+10 d)}=5 $

$ \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+ . . .\right. $ $ \left.\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5 $

$ \frac{1}{d}\left[1-\frac{1}{(1+10 d)}\right]=5 $

$ \frac{10 d}{1+10 d}=5 d $

$ 50 d=5$

$ \text { General term }=\frac{3 n^2-3 n+4}{2} $

$ \mathrm{T}_{10}=\frac{3(100)-3(10)+4}{2} $

$=137 $

$ 10 \text { terms with c.d. }=3$

sum $ =\frac{10}{2}(2(137)+9(3))$

$ =1505$

$ \mathrm{S}=1+2+4+\ldots \ldots $

$ \mathrm{Tn}=1+1+2+3+\ldots \ldots+\left(\mathrm{T}_{\mathrm{n}}-\mathrm{T}_{\mathrm{n}-1}\right) $

$ \mathrm{T}_{\mathrm{n}}=1+\left(\frac{\mathrm{n}-1}{2}\right)[2+(\mathrm{n}-2) \times 1] $

$ \mathrm{T}_{\mathrm{n}}=1+1+\frac{\mathrm{n}(\mathrm{n}-1)}{2} $

$ \mathrm{n}=100 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{100 \times 99}{2}=4950+1 $

$ \mathrm{n}=101 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{101 \times 100}{2}=5050+1=5051 $

$ \mathrm{n}=102 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{102 \times 101}{2}=5151+1=5152 $

$ \mathrm{n}=103 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{103 \times 102}{2}=5254 $

$ \mathrm{n}=104 \quad \mathrm{~T}_{\mathrm{n}}=1+\frac{104 \times 103}{2}=5357$

$(1+x) S=(1+x)^2+\ldots \ldots . . \quad 59(1+x)^{60}+60(1+x)^{61} $

$-x S=\frac{(1+x)(1+x)^{60}-1}{x}-60(1+x)^{61}$

Put $\mathrm{x}=60$

$-60 \mathrm{~S}=\frac{61\left((61)^{60}-1\right)}{60}-60(61)^{61}$

on solving $3660$

$ \mathrm{T}_2=3 \cdot \mathrm{T}_1+6^2 $

$ \mathrm{~T}_2=3 \cdot 6+6^2 $  ................($1$)

$ \mathrm{~T}_3=3 \mathrm{~T}_2+6^3 $

$ \mathrm{~T}_3=3 \mathrm{~T}_2+6^3 $

$ \mathrm{~T}_3=3\left(3 \cdot 6+6^2\right)+6^3 $

$ \mathrm{~T}_3=3^2 \cdot 6+3 \cdot 6^2+6^3 \quad \ldots(2) $ ...............($2$)

$ \mathrm{T}_{\mathrm{r}}=3^{\mathrm{r}-1} \cdot 6+3^{\mathrm{r}-2} \cdot 6^2+\ldots+6^{\mathrm{r}} $

$ \mathrm{T}_{\mathrm{r}}=3^{\mathrm{r}-1} \cdot 6\left[1+\frac{6}{3}+\left(\frac{6}{3}\right)^2+\ldots+\left(\frac{6}{3}\right)^{\mathrm{r}-1}\right] $

$ \mathrm{T}_{\mathrm{r}}=3^{\mathrm{r}-1} \cdot 6\left(1+2+2^2+\ldots+2^{\mathrm{r}-1}\right) $

$ \mathrm{T}_{\mathrm{r}}=6 \cdot 3^{\mathrm{r}-1} 1 \cdot \frac{\left(1-2^{\mathrm{r}}\right)}{(-1)} $

$ \mathrm{T}_{\mathrm{r}}=6 \cdot 3^{\mathrm{r}-1} \cdot\left(2^{\mathrm{r}}-1\right) $

$ \mathrm{T}_{\mathrm{r}}=\frac{6 \cdot 3^{\mathrm{r}}}{3} \cdot\left(2^{\mathrm{r}}-1\right)$

$ \mathrm{T}_{\mathrm{r}}=2 \cdot\left(6^{\mathrm{T}}-3^{\mathrm{r}}\right) $

$ \mathrm{S}_{\mathrm{n}}=2 \Sigma\left(6^{\mathrm{r}}-3^{\mathrm{r}}\right) $

$ \mathrm{S}_{\mathrm{n}}=2 \cdot\left[\frac{6 \cdot\left(6^{\mathrm{n}}-1\right)}{5}-\frac{3 \cdot\left(3^{\mathrm{n}}-1\right)}{2}\right] $

$ \mathrm{S}_{\mathrm{n}}=2\left[\frac{12\left(6^{\mathrm{n}}-1\right)-15\left(3^{\mathrm{n}}-1\right)}{10}\right] $

$ \mathrm{S}_{\mathrm{n}}=\frac{3}{5}\left[4 \cdot 6^4-5 \cdot 3^{\mathrm{n}}+1\right] $

$ \therefore \mathrm{n}^2-12 \mathrm{n}+39=3 $

$ \mathrm{n}^2-12 \mathrm{n}+36=0 $

$ \mathrm{n}=6$

$ \text { Put } \frac{x}{\sqrt{3}}=t \text {, where } x=\sqrt{3}-\sqrt{2} $

$ S=1+\frac{t}{2}+\frac{t^2}{6}+\frac{t^3}{12}+\frac{t^4}{20}+\ldots $

$ S=1+t\left(1-\frac{1}{2}\right)+t^2\left(\frac{1}{2}-\frac{1}{3}\right)+t^3\left(\frac{1}{3}-\frac{1}{4}\right)+t^4\left(\frac{1}{4}-\frac{1}{5}\right) $

$ S=\left(1+t+\frac{t^2}{2}+\frac{t^3}{3}+\frac{t^3}{4}+\ldots\right)-\left(\frac{t}{2}+\frac{t^2}{3}+\frac{t^3}{4}+\frac{t^4}{5}+\ldots\right) $

$ S=\left(t+\frac{t^2}{2}+\ldots\right)-\frac{1}{t}\left(t+\frac{t^2}{2}+\frac{t^3}{3}+\ldots\right)+2 $

$ S=2+\left(1-\frac{1}{t}\right)(-\log (1-t))=\left(\frac{1}{t}-1\right) \log (1-t)+2 $

$ S=2+\left(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}-1\right) \log \left(1-\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}}\right) $

$ S=2+\left(\frac{\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right) \log e \frac{\sqrt{2}}{\sqrt{3}} $

$ S=2+\frac{(\sqrt{6}+2)}{2} \log e \frac{2}{3}=2+\left(\sqrt{\frac{3}{2}}+1\right) \log e \frac{2}{3} $

$ a=2, b=3 $

$ 11 a+18 b=11 \times 2+18 \times 3=76$

$ =\frac{\sum_{\mathrm{r}=1}^{100}\left(\mathrm{r}^3+2 \mathrm{r}^2+\mathrm{r}\right)}{\sum_{\mathrm{r}=1}^{100}\left(\mathrm{r}^3+\mathrm{r}^2\right)}=\frac{\left(\frac{\mathrm{n}(\mathrm{n}+1)^2}{2}\right)+\frac{2 \mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2}}{\left(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\right)^2+\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}} $

$ =\frac{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}+\frac{2}{3} \cdot(2 \mathrm{n}+1)+1\right]}{\frac{\mathrm{n}(\mathrm{n}+1)}{2}\left[\frac{\mathrm{n}(\mathrm{n}+1)}{2}+\frac{(2 \mathrm{n}+1)}{3}\right]} ; \text { Put } \mathrm{n}=100 $

$ \frac{\frac{100(101)}{2}+\frac{2}{3}(201)+1}{\frac{100 \times 101}{2}+\frac{201}{3}}=\frac{5185}{5117}=\frac{305}{301}$

$\mathrm{T}_{\mathrm{r}}=\frac{\mathrm{r}}{1-3 \mathrm{r}^2+\mathrm{r}^4}$

$\mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\mathrm{r}^4-2 \mathrm{r}^2+1-\mathrm{r}^2}$

$\mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-1\right)^2-\mathrm{r}^2} $

$\mathrm{~T}_{\mathrm{r}}=\frac{\mathrm{r}}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)}$

$\mathrm{T}_{\mathrm{r}}=\frac{\frac{1}{2}\left[\left(\mathrm{r}^2+\mathrm{r}-1\right)-\left(\mathrm{r}^2-\mathrm{r}-1\right)\right]}{\left(\mathrm{r}^2-\mathrm{r}-1\right)\left(\mathrm{r}^2+\mathrm{r}-1\right)}$

$=\frac{1}{2}\left[\frac{1}{\mathrm{r}^2-\mathrm{r}-1}-\frac{1}{\mathrm{r}^2+\mathrm{r}-1}\right]$

Sum of $10$ terms,

$\sum_{\mathrm{r}=1}^{10} \mathrm{~T}_{\mathrm{r}}=\frac{1}{2}\left[\frac{1}{-1}-\frac{1}{109}\right]=\frac{-55}{109}$

$t_n=a^2+b n+c$

$ 1=a+b+c$

$ 4=4 a+2 b+c $

$ 8=9 a+3 b+c$

On solving we get, $\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{3}{2}, \mathrm{c}=-1$

$ \alpha=\sum_{n=1}^{10}\left(\frac{n^2}{2}+\frac{3 n}{2}-1\right)^2 $

$ 4 \alpha=\sum_{n=1}^{10}\left(n^2+3 n-2\right)^2, \beta=\sum_{n=1}^{10} n^4 $

$ 4 \alpha-\beta=\sum_{n=1}^{10}\left(6 n^3+5 n^2-12 n+4\right)=55(353)+40$

$ \Rightarrow \mathrm{f}(\mathrm{x})=\mathrm{kx} $

$ \Rightarrow \mathrm{f}(1)=1 $

$ \Rightarrow \mathrm{k}=1 $

$ \mathrm{f}(\mathrm{x})=\mathrm{x}$

Now

$ \sum_{\mathrm{k}=1}^{2022} \mathrm{f}(\lambda+\mathrm{k}) \leq(2022)^2 $

$ \Rightarrow \sum_{\mathrm{k}=1}^{2022}(\lambda+\mathrm{k}) \leq(2022)^2 $

$ \Rightarrow 2022 \lambda+\frac{2022 \times 2023}{2} \leq(2022)^2$

$ \Rightarrow \lambda \leq 2022-\frac{2023}{2} $

$ \Rightarrow \lambda \leq 1010.5$

$\therefore$ largest natural no. $\lambda$ is $1010$ .

$ \frac{\sqrt{1}-\sqrt{2}}{-1}+\frac{\sqrt{2}-\sqrt{3}}{-1} \ldots \frac{\sqrt{99}-\sqrt{100}}{-1}=\mathrm{m}$

$ \sqrt{100}-1=\mathrm{m} \Rightarrow \mathrm{m}=9 $

$ \frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\ldots \frac{1}{99 \cdot 100}=\mathrm{n} $

$ \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3} \ldots \frac{1}{99}-\frac{1}{100}=\mathrm{n} $

$ 1-\frac{1}{100}=\mathrm{n} $

$ \frac{99}{100}=\mathrm{n} $

$ (\mathrm{m}, \mathrm{n})=\left(9, \frac{99}{100}\right) $

$ \Rightarrow 11(9)-100\left(\frac{99}{100}\right) $

$ =99-99=0$

Ans. option ($4$) $11 x-100 y=0$

$ \Rightarrow 2 \sec ^2 \theta-5 \sec \theta-3=0 $

$ \Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0 $

$ \Rightarrow \sec \theta=-\frac{1}{2}, $

$ \Rightarrow \cos \theta=-2, \frac{1}{3} $

$ \Rightarrow \cos \theta=\frac{1}{3}$

For $7$ solutions $n=13$

$ \text { So, } \sum_{\mathrm{k}=1}^{13} \frac{\mathrm{k}}{2^{\mathrm{k}}}=\mathrm{S} \text { (say) } $

$ \mathrm{S}=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^j}+\ldots .+\frac{13}{2^{13}} $

$ \frac{1}{2} \mathrm{~S}=\frac{1}{2^2}+\frac{1}{2^3}+\ldots . .+\frac{12}{2^{13}}+\frac{13}{2^{14}} $

$ \Rightarrow \frac{\mathrm{S}}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow \mathrm{S}=2 \cdot\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}}$