Sequence and Series

\(\begin{array}{rlrl}
d=3 / 2 & a_{10}=\alpha =a+9 d \\
a_{11} & =\beta=a+10 d \\
3_{11}=\frac{11}{2}(2 a+10 d) & =88 \\
(a+5 d) & =8 \\
a+\frac{15}{2} & =8 \Rightarrow a=1 / 2
\end{array}\)

\(\begin{gathered}
\alpha=\frac{1}{2}+9 \times \frac{3}{2}=14 \\
\beta=\frac{1}{2}+10 \times \frac{3}{2}=\frac{31}{2} \\
3 x^2-p x+q=0 \\
\alpha+\beta=\frac{p}{3} \\
\frac{14+81}{2}=\frac{p}{3} \Rightarrow \frac{28+31}{2}=\frac{p}{3} \\
\frac{59}{2}=\frac{p}{3} \Rightarrow p=\frac{59 \times 3}{2}=\frac{177}{2}
\end{gathered}\)

\(\begin{aligned}
& \alpha \beta=q / 3 \\
& 14 \times \frac{31}{2}=\frac{q}{3} \\
& 217 \times 3=q \\
& q=651 \\
& q-2 p \\
& 651-2\left(\frac{177}{2}\right)=651-177=474
\end{aligned}\)

Let $a=\frac{4}{7}, b=\frac{1}{3}$

Therefore we get

$S=\left(a+b\right)+\left(a^2+ab+b^2\right)+\left(a^3+a^{2}b+a{b}^2+b^3\right)+...\infty$

Multiply numerator and denominator by \((a – b)\) $=\frac{4}{7}-\frac{1}{3}=\frac{5}{21}$

$S=\frac{1}{a-b}\left[\left(a^2-b^2\right)+\left(a^3-b^3\right)+\left(a^4-b^4\right)+\dots .\infty \right]$

$S=\frac{1}{a-b}\left[\frac{a^2}{1-a}-\frac{b^2}{1-b}\right]$

$=\frac{21}{5}\left[\frac{\frac{16}{49}}{1-\frac{4}{7}}-\frac{\frac{1}{9}}{1-\frac{1}{3}}\right]$

$=\frac{21}{5}\left[\frac{16}{21}-\frac{1}{6}\right]=\frac{21}{5}\left[\frac{96-21}{21·6}\right]$

$=\frac{75}{5·6}=\frac{15}{6}=\frac{5}{2}$

From the given condition, we can say that,

$$\begin{gathered} AM =2\times GM \\ \frac{a+b}{2}=2\sqrt{ab} \\ a+b = 4\sqrt{ab} \end{gathered}$$

$$\begin{gathered} \text{we know,} \\ {\left(a-b\right)}^2={\left(a+b\right)}^2-4ab \\ =16ab-4ab=12ab \\ a-b = 2\sqrt{3}\sqrt{ab} \end{gathered}$$

$$\begin{gathered} \text{ (Taking }+ve\text{ sign only as }a>b\text{ ) } \\ \therefore \frac{a+b}{a-b}=\frac{4\sqrt{ab}}{2\sqrt{3}\sqrt{ab}}=\frac{2}{\sqrt{3}} \\ \text{ By componendo and dividendo, } \\ \frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}} \end{gathered}$$

\( \mathrm{T}_{\mathrm{r}}=\frac{1^3+2^3+3^3+\ldots +\ldots r^3}{1+3+5+\ldots .+(2 \mathrm{r}-1)}=\frac{\left(\frac{\mathrm{r}(\mathrm{r}+1)}{2}\right)^2}{\mathrm{r}^2} \)
\( =\frac{\mathrm{r}^2+2 \mathrm{r}+1}{4} \)
\( \mathrm{~S}_{\mathrm{n}}=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{~T}_{\mathrm{r}} \)
\( \mathrm{~S}_{\mathrm{n}}=\frac{1}{4} \sum_{\mathrm{r}=1}^{\mathrm{n}} \left(\mathrm{r}^2+2 \mathrm{r}+1\right) \)
\( =\frac{1}{4}\left[\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n}\right] \)
\( \mathrm{S}_8=\frac{1}{4}\left[\frac{8 \times 9 \times 17}{6}+8 \times 9+8\right]\)
\( =\frac{1}{4}[204+72+8]=71\)

$$\begin{gathered} 2a_{n+2}=5a_{n+1}-3a_n \\ \Rightarrow 2t^2-5t+3=0 \\ \Rightarrow t=1,\frac{3}{2} \\ {a}_n=A·(1{)}^n+B·{\left(\frac{3}{2}\right)}^n \\ {a}_0=3, a_1=4 \\ then A = 1 \& B =2 \\ Now, a_{n }=1+2{\left(\frac{3}{2}\right)}^n \\ {a}_{100}=1+2{\left(\frac{3}{2}\right)}^{100} \\ {\left(\frac{3}{2}\right)}^{100}=\frac{a_{100}-1}{2} \\ Now \sum _{k=1}^{100}{a}_{k }=S_{100}=100+6\left({\left(\frac{3}{2}\right)}^{100}-1\right) \\ =100+6\left(\frac{a_{100}-1}{2}-1\right) \\ =3a_{100}+91 \end{gathered}$$

\(S_n=3 n^2+5 n\)

\(T_n=S_n-S_{n-1}\)

\(=3\left(n^2-(n-1)^2\right)+5(n-(n-1))\)

\(=3(2 n-1)+5\)

\(=6 n+2\)

\(\sum_{n=1}^{10}(6 n+2)^2=\sum 36 n^2+\sum 4+\sum 24 n\)

\(=36 \times \frac{10 \times 11 \times 21}{6}+4 \times 10+24 \frac{10 \times 11}{2}\)

\(=15220\)

Let the side of the original equilateral triangle be \(a\).

When the middle points of the sides are joined, the side of the new equilateral triangle becomes half of the previous side.

So, the sides of the triangles formed are \(\frac{a}{2},\frac{a}{4},\frac{a}{8},\ldots\)

Now, perimeter of an equilateral triangle of side \(s\) is \(3s\).

Therefore, \(P=3\cdot\frac{a}{2}+3\cdot\frac{a}{4}+3\cdot\frac{a}{8}+\cdots\)

\(P=3a\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots\right)\)

\(P=3a\cdot 1\)

\(P=3a\)

Area of an equilateral triangle of side \(s\) is \(\frac{\sqrt{3}}{4}s^2\).

Therefore, \(Q=\frac{\sqrt{3}}{4}\left(\frac{a}{2}\right)^2+\frac{\sqrt{3}}{4}\left(\frac{a}{4}\right)^2+\frac{\sqrt{3}}{4}\left(\frac{a}{8}\right)^2+\cdots\)

\(Q=\frac{\sqrt{3}}{4}a^2\left(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+\cdots\right)\)

\(Q=\frac{\sqrt{3}}{4}a^2\cdot \frac{\frac{1}{4}}{1-\frac{1}{4}}\)

\(Q=\frac{\sqrt{3}}{4}a^2\cdot \frac{1}{3}\)

\(Q=\frac{\sqrt{3}a^2}{12}\)

Now, \(P^2=(3a)^2=9a^2\)

Also, \(36\sqrt{3}Q=36\sqrt{3}\cdot \frac{\sqrt{3}a^2}{12}\)

\(=36\cdot \frac{3a^2}{12}\)

\(=9a^2\)

Thus, \(P^2=36\sqrt{3}Q\)

$$\begin{gathered} {\text{Given, T}}_r \text{represents terms of A.P.} \\ {\mathrm{T}}_{\mathrm{m}}=\frac{1}{25},{ \mathrm{T}}_{25}=\frac{1}{20},20\sum _{\mathrm{r}=1}^{25}{ \mathrm{T}}_{\mathrm{r}}=13 \\ 20\sum _{r=1}^{25}{T}_r=20\left[\frac{25}{2}\left[a+\frac{1}{20}\right]\right]=13 \\ \Rightarrow a=\frac{1}{20\times 25}=\frac{1}{500} \\ \because T_{25}=a+24d=\frac{1}{20} \\ \Rightarrow d=\frac{1}{20\times 25}=\frac{1}{500} \\ \text{Now}, \\ {T}_m=a+(m-1)d=\frac{1}{25} \\ =\frac{1}{500}m=\frac{1}{25} \\ \Rightarrow m=20 \\ \text{ Now, }5m\sum _{r=m}^{2m}{T}_r=5\times 20\left[\sum _{r=20}^{40}{T}_r\right] \\ =100\left[\frac{40}{2}(2a+39d)-\frac{19}{2}(2a+18d)\right] \\ \text{but a=d (we already find above)} \\ 5m\sum _{r=m}^{2m}{T}_r=100\left[\frac{40}{2}\times 41d-\frac{19}{2}\times 20d\right] \\ =126 \end{gathered}$$

Let the first term of the G.P. be \(a\) and the common ratio be \(r\).

Given

\(a r+a r^3+a r^5=21 \Rightarrow a r\left(1+r^2+r^4\right)=21 \ldots (1)\)

and \(a r^7+a r^9+a r^{11}=15309 \Rightarrow a r^7\left(1+r^2+r^4\right)=15309 \ldots (2)\)

Divide Equation 2 by Equation 1

\(\frac{a r^7\left(1+r^2+r^4\right)}{a r\left(1+r^2+r^4\right)}=\frac{15309}{21}\)

\(r^6=729\)

\(r=3\)

from eq (1)

\(\begin{aligned} & 3 a(1+9+81)=21 \\ & 3 a(91)=21 \Rightarrow 273 a=21 \\ & a=\frac{21}{273}=\frac{1}{13}\end{aligned}\)

Now

\(\begin{aligned} & S_9=\frac{\frac{1}{13}\left(3^9-1\right)}{3-1} \\ & S_9=\frac{1}{13} \cdot \frac{19683-1}{2} \\ & S_9=\frac{19682}{26} \\ & S_9=757\end{aligned}\)

$$\begin{gathered} S=\frac{6}{3^{26}}+\frac{10\cdot 1}{3^{25}}+\frac{10\cdot 2}{3^{24}}+\frac{10\cdot 2^2}{3^{23}}+\mathrm{...}+\frac{10\cdot 2^{24}}{3} \\ =\frac{6}{3^{26}}+\frac{10}{3^{25}}\left[1+6+6^2+\mathrm{...}+6^{24}\right] \end{gathered}$$

$=\frac{6}{3^{26}}+\frac{10}{3^{25}}\left[\frac{{\left(6\right)}^{25}-1}{6-1}\right]$

$=\frac{6}{3^{26}}+\frac{10}{3^{25}}\left[\frac{6^{25}-1}{5}\right]$

$=\frac{2}{3^{25}}+2\left[2^{25}-\frac{1}{3^{25}}\right]$

$=2^{26}$

Given the sum of all the terms is 7 times the sum of the odd terms of the G.P.

\(a+a r+a r^2+a r^3+\ldots+a r^{63}=7\left(a+a r^2+a r^4 \ldots+a r^{62}\right)\)

\(\Rightarrow \frac{a\left(1-r^{64}\right)}{1-r}=\frac{7a\left(1-\left(r^2\right)^{32}\right)}{1-r^2}\)

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

$$\begin{gathered} \Rightarrow 1+r=7 \\ \Rightarrow r=6 \end{gathered}$$

Given the sum of all the terms is 7 times the sum of the odd terms of the G.P.

\(a+a r+a r^2+a r^3+\ldots+a r^{63}=7\left(a+a r^2+a r^4 \ldots+a r^{62}\right)\)

\(\Rightarrow \frac{a\left(1-r^{64}\right)}{1-r}=\frac{7a\left(1-\left(r^2\right)^{32}\right)}{1-r^2}\)

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

$$\begin{gathered} \Rightarrow 1+r=7 \\ \Rightarrow r=6 \end{gathered}$$

Let \(\alpha, \beta, \gamma, \delta\) be \(a, a r, a r^2, a r^3\), respectively.
For \(x^2-x+p=0\),

sum of roots \(=1\),

so \(a+a r=a(1+r)=1\)

For \(x^2-4 x+q=0\),

sum of roots \(=4\),

so \(a r^2+a r^3=a r^2(1+r)=4\)

Dividing,

\(\frac{a r^2(1+r)}{a(1+r)}=\frac{4}{1}\)

\(r^2=4\)

So, \(r=2\) or \(r=-2\).
If \(r=2\), then \(a(3)=1 \Rightarrow a=\frac{1}{3}\),

so \(p=\alpha \beta=a \cdot a r=a^2 r=\frac{2}{9}\), not an integer.

Hence \(r=-2\).
Now \(a(1-2)=1 \Rightarrow-a=1 \Rightarrow a=-1\).
So,

\(p=\alpha \beta=a \cdot a r=a^2 r=-2\)

Also,

\(q=\gamma \delta=a r^2 \cdot a r^3=a^2 r^5=(-2)^5=-32\)

\(|p+q|=|-2-32|=34\)

Let four terms are $a-3d,a-d,a+d,a+3d$

Where $d=\frac{l}{2}$

$\because a_1+a_2+a_3+a_4=48$

$\Rightarrow 4a=48\Rightarrow a=12$

$a_1{a}_2{a}_3{a}_4+{\mathcal{l}}^4=361$

$\Rightarrow \left(a^2-9 d^2\right)\left(a^2-d^2\right)+16 d^4=361$

$\Rightarrow \left(144-9 d^2\right)\left(144-d^2\right)+16 d^4=361$

$\Rightarrow 25 d^4-1440 d^2+{\left(144\right)}^2=361$

$\Rightarrow {\left(5 d^2-144\right)}^2={19}^2$

$\Rightarrow 5 d^2-144=19, -19$

$\Rightarrow d^2=\frac{163}{5}, 25$

$\Rightarrow d=\sqrt{\frac{163}{5}}, 5$

$\therefore l=2\sqrt{\frac{163}{5}}, 10$

Common difference is an integer therefore \(l=10\)

Largest term \(= 12 + 15 = 27\)

We have \(\left(n^2-2 n+2\right) x^2-3 x+\left(n^2-2 n+2\right)^2=0\)

Since \(n^2-2 n+2=(n-1)^2+1\)

Sum of roots \(=\frac{3}{(n-1)^2+1}\)

Since the denominator is minimum when \((n-1)^2=0\)

Therefore \(\beta=3\)

Product of roots \(=(n-1)^2+1\)

So the minimum value of the product is clearly

\(\alpha=1\)

Now, first term \(a=\alpha=1\)

Common ratio \(r=\frac{\alpha}{\beta}=\frac{1}{3}\)

Now, \(S_6=1 \cdot \frac{1-\left(\frac{1}{3}\right)^6}{1-\frac{1}{3}}\)

\(S_6=\frac{1092}{729}=\frac{364}{243}\)

Given, \(S_{20}=\frac{20}{2}[2 a+19 d]=790\)

\(2 a+19 d=79\)\(\ldots(1)\)

and, \(S_{10}=\frac{10}{2}[2 a+9 d]=145\)

\(2 a+9 d=29\)\(\ldots(2)\)

From (1) and (2)

\(a=-8, d=5\)

Now, \(S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d]\)

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

\(=405-10\)

\(=395\)

Given, \(S_{20}=\frac{20}{2}[2 a+19 d]=790\)

\(2 a+19 d=79\)\(\ldots(1)\)

and, \(S_{10}=\frac{10}{2}[2 a+9 d]=145\)

\(2 a+9 d=29\)\(\ldots(2)\)

From (1) and (2)

\(a=-8, d=5\)

Now, \(S_{15}-S_5=\frac{15}{2}[2 a+14 d]-\frac{5}{2}[2 a+4 d]\)

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

\(=405-10\)

\(=395\)

Let

\(S=5+\frac{1}{7}(5+\alpha)+\frac{1}{72}(5+2 \alpha)+\ldots \infty \quad \ldots(i)\)

On multiply by \(\frac{1}{7}\), we get

\(\frac{1}{7} S=\frac{1}{7}(5)+\frac{1}{7^2}(5+\alpha)+\ldots \infty \ldots(i i)\)

Subtracting equation (ii) from (i), we get

\(\begin{aligned} & \frac{6}{7}(S)=5+\frac{1}{7} \alpha\left(\frac{1}{1-\frac{1}{7}}\right) \\ & 6=5+\frac{\alpha}{6} \\ & \Rightarrow \alpha=6\end{aligned}\)

$$\begin{gathered} S=\frac{6}{3^{26}}+\frac{10\cdot 1}{3^{25}}+\frac{10\cdot 2}{3^{24}}+\frac{10\cdot 2^2}{3^{23}}+\mathrm{...}+\frac{10\cdot 2^{24}}{3} \\ =\frac{6}{3^{26}}+\frac{10}{3^{25}}\left[1+6+6^2+\mathrm{...}+6^{24}\right] \end{gathered}$$

$=\frac{6}{3^{26}}+\frac{10}{3^{25}}\left[\frac{{\left(6\right)}^{25}-1}{6-1}\right]$

$=\frac{6}{3^{26}}+\frac{10}{3^{25}}\left[\frac{6^{25}-1}{5}\right]$

$=\frac{2}{3^{25}}+2\left[2^{25}-\frac{1}{3^{25}}\right]$

$=2^{26}$

\( \log _e a, \log _e b, \log _e c \) are in A.P.
\(\therefore \mathrm{b}^2=\mathrm{ac} \quad \ldots .(i)\)

Also \(\log _e\left(\frac{a}{2 b}\right), \log _c\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right)\) are in A.P.

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

Putting in eq. (i)

\( b^2=a \times \frac{2 b}{3} \)
\(\frac{a}{b}=\frac{3}{2} \)
\( a: b: c=9: 6: 4\)

\( \log _e a, \log _e b, \log _e c \) are in A.P.
\(\therefore \mathrm{b}^2=\mathrm{ac} \quad \ldots .(i)\)

Also \(\log _e\left(\frac{a}{2 b}\right), \log _c\left(\frac{2 b}{3 c}\right), \log _e\left(\frac{3 c}{a}\right)\) are in A.P.

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

Putting in eq. (i)

\( b^2=a \times \frac{2 b}{3} \)
\(\frac{a}{b}=\frac{3}{2} \)
\( a: b: c=9: 6: 4\)

\(S=1^3-2^3+3^3-\ldots+15^3\)

\(=\left(1^3+2^3+\ldots+15^3\right)-2\left(2^3+4^3+\ldots+14^3\right)\)

\(=\left(1^3+2^3+\ldots+15^3\right)-2^4\left(1^3+2^3+\ldots+7^3\right)\)

\(=\left[\frac{15 \times 16}{2}\right]^2-16\left[\frac{7 \times 8}{2}\right]^2\)

\(=14400-16\times 784\)

\(=14400-12544\)

\(=1856\)

\(\begin{aligned}
& T_n=S_{n-} S_{n-1} \\
& =\frac{(2 n-1)(2 n+1)(2 n+3)(2 n+5)-(2 n-3)(2 n-1)(2 n+1)(2 n+3)}{64} \\
& T_n=\frac{(2 n-1)(2 n+1)(2 n+3)}{8} \\
& \frac{1}{T_n}=\frac{8}{(2 n-1)(2 n+1)(2 n+3)} \\
& \frac{1}{T_n}=2\left(\frac{1}{(2 n-1)(2 n+1)}-\frac{1}{(2 n-1)(2 n+3)}\right) \\
& \sum_{r=1}^n \frac{1}{T_r}=2\left(\frac{1}{1 \times 3}-\frac{1}{(2 n-1)(2 n+3)}\right) \\
& \operatorname{Lim}_{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{T_r}=\frac{2}{3}
\end{aligned}\)

\(\sum_{n=0}^{\infty} a r^n=57\)

\(\frac{a}{1-r}=57\) \(\ldots(i)\)

\(\sum_{n=0}^{\infty} a^3 r^{3 n}=9747\)

\(\frac{\mathrm{a}^3}{1-\mathrm{r}^3}=9747\) \(\ldots (ii)\)

from (i) and (ii)

\(\frac{\frac{a^3}{(1-r)^3}}{\frac{a^3}{1-r^3}}=\frac{57^3}{9747}=19\)

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

\(a=19\)

\(\therefore a+18 r=19+18 \times \frac{2}{3}=31\)

From the given condition, we can say that,

$$\begin{gathered} AM =2\times GM \\ \frac{a+b}{2}=2\sqrt{ab} \\ a+b = 4\sqrt{ab} \end{gathered}$$

$$\begin{gathered} \text{we know,} \\ {\left(a-b\right)}^2={\left(a+b\right)}^2-4ab \\ =16ab-4ab=12ab \\ a-b = 2\sqrt{3}\sqrt{ab} \end{gathered}$$

$$\begin{gathered} \text{ (Taking }+ve\text{ sign only as }a>b\text{ ) } \\ \therefore \frac{a+b}{a-b}=\frac{4\sqrt{ab}}{2\sqrt{3}\sqrt{ab}}=\frac{2}{\sqrt{3}} \\ \text{ By componendo and dividendo, } \\ \frac{a}{b}=\frac{2+\sqrt{3}}{2-\sqrt{3}} \end{gathered}$$

First A.P.

Set \(\mathrm{A}=\{1,6,11,16 \ldots .101\) terms \(\}\)

Second A.P.
Set \(\mathrm{B}=\{9,16 \ldots . .71\) terms \(\}\)
\(\mathrm{D}=\mathrm{L} . \mathrm{C} . \mathrm{M}\left\{\mathrm{d}_1, \mathrm{~d}_2\right\}=35\)
\(1^{\text {st }}\) Common term is \(16\)
\(16+(\mathrm{n}-1) 35 \leq 499\)
\(\mathrm{n} \leq 14.8\)
\(\Rightarrow \mathrm{n}=14\)
\(A \cap B=\{16,51,86,121,156,191,226,261,296,331,366,401,436,471\}\)
Terms divisible by \(3=\{51,156,261,366,471\}\)
\(=5\) terms

$$\begin{gathered} \text{Given, }{S}_{40}=1030\text{ and }{S}_{12}=57 \\ \mathrm{Let} \mathrm{a} \& \mathrm{d} \mathrm{are} \mathrm{first} \mathrm{term} \mathrm{and} \mathrm{common} \mathrm{diff}\text{erence} \mathrm{of} \mathrm{an} \mathrm{AP}. \\ \text{According to question,} \\ {S}_{\mathit{40}}\mathit{=}\frac{\mathit{40}}{\mathit{2}}\mathit{[}\mathit{2}a\mathit{+}\mathit{39}\mathit{ }d\mathit{]}\mathit{=}\mathit{1030}........\left(1\right) \\ {S}_{\mathit{12}}\mathit{=}\frac{\mathit{12}}{\mathit{2}}\mathit{[}\mathit{2}a\mathit{+}\mathit{11}\mathit{ }d\mathit{]}\mathit{=}\mathit{57}..........\left(2\right) \\ \text{solve equation (1) and (2)\hspace{0.17em}, we get} \\ a\mathit{=}\mathit{-}\frac{\mathit{7}}{\mathit{2}}\mathit{ }\mathit{ }d\mathit{=}\frac{\mathit{3}}{\mathit{2}} \\ \mathit{\therefore }{S}_{\mathit{30}}\mathit{-}{S}_{\mathit{10}}\mathit{=}\frac{\mathit{30}}{\mathit{2}}\mathit{[}\mathit{2}a\mathit{+}\mathit{29}\mathit{ }d\mathit{]}\mathit{-}\frac{\mathit{10}}{\mathit{2}}\mathit{[}\mathit{2}a\mathit{+}\mathit{9}\mathit{ }d\mathit{]} \\ \mathit{=}\mathit{20}a+\mathit{390}\mathit{ }d\mathit{=}\mathit{515} \end{gathered}$$

Let the common difference of A.P. be \(d\) and the common ratio of G.P. be \(r\)

Since \(a_1=g_1=1, a_2=1+d, a_3=1+2 d, g_2=r, g_3=r^2\)

\(a_2+g_2=1 \Rightarrow 1+d+r=1 \Rightarrow d+r=0 \Rightarrow d=-r\)

\(a_3+g_3=4 \Rightarrow 1+2 d+r^2=4\)

\(\Rightarrow 1-2 r+r^2=4\)

\(\Rightarrow r^2-2 r-3=0\)

\(\Rightarrow (r-3)(r+1)=0\)

\(\Rightarrow r=3,-1\)

Since the G.P. is increasing and \(g_1=1, r=3\), so \(d=-3\)

\( \therefore a_{10}=1+9 d=1+9(-3)=-26 \)

\( \mathrm{g}_5=1 . \mathrm{r}^4=3^4=81 \)

\(\therefore \mathrm{a}_{10}+\mathrm{g}_5=55\)

Let the common difference of the A.P. be \(d\).

Since the first term of the A.P. is \(1\), the \(n\)-th term is \(a_n=1+(n-1)d\)

So, \(a_2=1+d\)

\(a_8=1+7d\)

\(a_{44}=1+43d\)

Given that \(a_2,a_8,a_{44}\) are the first, second and third terms of a G.P.

Therefore, \((a_8)^2=a_2a_{44}\)

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

\(1+14d+49d^2=1+44d+43d^2\)

\(6d^2-30d=0\)

\(6d(d-5)=0\)

So, \(d=0\) or \(d=5\)

Since the A.P. is non-constant, \(d\neq 0\).

Therefore, \(d=5\)

Now, the sum of first \(20\) terms is

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

\(S_{20}=10(2+95)\)

\(S_{20}=10\cdot 97\)

\(S_{20}=970\)

Let the common difference be \(d\) and the last term be \(l\)

Given first term \(a=\dfrac{10}{3}\) and number of terms \(n=30\)

So \(l=a+29d=\dfrac{10}{3}+29d\)

Sum of the A.P. is \(S_{30}=\dfrac{30}{2}(a+l)=15\left(\dfrac{10}{3}+l\right)\)

Given \(S_{30}=l^3\),

hence \(15\left(\dfrac{10}{3}+l\right)=l^3\)

So \(50+15l=l^3\)

\(l^3-15l-50=0\)

$\left(l-5\right)\left(l^2+5l+10\right)=0$

The discriminant of the second factor is \(25-40=-15<0\); therefore, it gives no real values.

Thus the last term is \(l=5\)

Now, \(5=\dfrac{10}{3}+29d\),

\(29d=\dfrac{5}{3}\).

Therefore \(d=\dfrac{5}{87}\)

General term

\(\mathrm{T}_{\mathrm{n}}=\frac{1}{\mathrm{n}}\left(1^2+2^2+\ldots+\mathrm{n}^2\right)\)

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

\(=\frac{1}{6}\left[2 \mathrm{n}^2+3 \mathrm{n}+1\right]\)

Now,

\(\mathrm{S}_{\mathrm{n}}=\Sigma \mathrm{T}_{\mathrm{n}}=\frac{1}{6}\left[2 \Sigma \mathrm{n}^2+3 \Sigma \mathrm{n}+\Sigma 1\right]\)

\(=\frac{1}{6}\left[2 \cdot \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n}\right]\)

Hence,

\(\mathrm{S}_{10}=\frac{1}{6}\left[\frac{2 \times 10 \times 11 \times 21}{6}+\frac{3}{2} \times 10 \times 11+10\right]\)

\(=\frac{10}{6}\left[77+\frac{33}{2}+1\right]=\frac{315}{2}\)

$$\begin{gathered} T_n=S_{n-}{S}_{n-1} \\ =\frac{(2n-1)(2n+1)(2n+3)(2n+5)}{64} \\ -\frac{(2n-3)(2n-1)(2n+1)(2n+3)}{64} \\ \Rightarrow T_n=\frac{(2n-1)(2n+1)(2n+3)}{8} \\ \Rightarrow \frac{1}{T_n}=\frac{8}{(2n-1)(2n+1)(2n+3)} \\ \Rightarrow \frac{1}{T_n}=2\left(\frac{1}{(2n-1)(2n+1)}-\frac{1}{(2n-1)(2n+3)}\right) \\ \underset{n\to \infty }{\lim }\sum _{ r=1}^n\frac{1}{T_r} \\ =\underset{n\to \infty }{\lim }2\left[\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\dots \right] \\ \Rightarrow \underset{n\to \infty }{\lim } \sum _{r=1}^n\frac{1}{T_r}=2\left(\frac{1}{1\times 3}-\frac{1}{(2n-1)(2n+3)}\right) \\ \Rightarrow \underset{n\to \infty }{\lim }\sum _{r=1}^n\frac{1}{T_r}=\frac{2}{3} \end{gathered}$$

Let the \(A P\) be

\(
a, a+d, a+2 d, \ldots, a+(2 n-1) d
\)

Now given that

\(
\begin{aligned}
& (a+d)+(a+3 d)+\ldots+(a+(2 n-1) d)=40 \\
& n a+n^2 d=40...(1)
\end{aligned}
\)

Also \(a+(a+2 d)+(a+4 d)+\ldots+(a+(2 n-2) d)=55\)

\(\begin{aligned}
&n a+d n(n-1)=55...(2)\\
&\text { Also } a-(a+(2 n-1) d)=27...(3)\\
&\begin{aligned}
& -(2 n-1) d=27 \\
& d=\frac{-27}{2 n-1}
\end{aligned}\\
&\begin{aligned}
& (2)-(1) \\
& d n(n-1)-n^2 d=15 \\
& d\left[n^2-n-n^2\right]=15 \\
& \left(\frac{-27}{2 n-1}\right)(-n)=15 \\
& 27 n=30 n-15 \\
& 15=3 n \\
& n=5
\end{aligned}
\end{aligned}\)

Given: $f\left(x+y\right)=f\left(x\right)\cdot f\left(y\right)\Rightarrow f\left(x\right)=a^x$

$\because f\left(1\right)=7\Rightarrow a^1=7$

So $f\left(x\right)=7^x$

Now $g\left(x+y\right)=g\left(xy\right)($put $y=1)$

$\Rightarrow g\left(x+1\right)=g\left(x\right)$

So $g\left(1\right)=g\left(2\right)=g\left(3\right)=\dots =g\left(n\right)=1$

Given ${\sum }_{\text{x}=1}^{\text{n}}\frac{\text{f}\left(\text{x}\right)}{\text{g}\left(\text{x}\right)}=19607$

$\Rightarrow {\sum }_{x=1}^n\frac{7^x}{1}=19607$

$\Rightarrow 7\left(\frac{7^{\text{n}}-1}{7-1}\right)=19607$

$\Rightarrow 7^n-1=\frac{6}{7}\times 19607$

$\Rightarrow 7^{\text{n}}=16807\Rightarrow \text{n}=5$

$a=3$, let common difference $=d$.

$\frac{4}{2}[2\times 3+(4-1\left)d\right]=\frac{4}{5\times 2}[2\times (3+4d)+(4-1\left)d\right]$
$5(6+3 d)=6+8 d+3 d$
$30+15 d=6+11 d$
$4 d=-24$
$d=-6$

$S_{20}=\frac{20}{2}\left[2\times 3+19\times -6\right]=-1080$

$$\begin{gathered} \text{Given:} e^{5(\ln x{)}^2+3}=x^8 \\ \text{Taking log on both sides:} \\ \Rightarrow ln e^{5(l\mathrm{nx}{)}^2+3}={\mathrm{lnx}}^8 \\ \Rightarrow 5(\ln x{)}^2+3=8\ln x \\ (\text{Let} \ln x=t) \\ \Rightarrow 5t^2-8t+3=0 \\ {\mathrm{t}}_1+{\mathrm{t}}_2=\frac{8}{5} \\ \Rightarrow \ln x_1+\ln x_2=\frac{8}{5} \\ \Rightarrow \ln x_1{\mathrm{x}}_2=\frac{8}{5} \\ \Rightarrow x_1{\mathrm{x}}_2={\mathrm{e}}^{8/5} \end{gathered}$$

$$\begin{gathered} \frac{\mathrm{n}}{2}(2\mathrm{a}+(\mathrm{n}-1)6°)=(\mathrm{n}-2)·180° \\ \text{ an }+3n^2-3n=(n-2)·180 ...\left(1\right) \\ \text{ Now according to question ,} \\ a+(n-1)6°=219° \\ \Rightarrow \mathrm{a}=225-6\mathrm{n} ...\left(2\right) \\ \mathrm{Putting} \mathrm{value} \mathrm{of} a \mathrm{from} \mathrm{equation} \left(2\right) \mathrm{in} \left(1\right) \mathrm{We} \mathrm{get} \\ \left(225n-6n^2\right)+3n^2-3n=180n-360 \\ {n}^2-14n-120=0 \\ n = 20, -6 \left(rejected\right) \end{gathered}$$

Let common difference of A.P.'s are $d_1$ and $d_2$

$\therefore d_1=13+d_2$

$$\begin{gathered} b_{31}=-277 \\ \Rightarrow b_1+30 d_2=-277 ...\left(1\right) \end{gathered}$$

$$\begin{gathered} b_{43}=-385 \\ \begin{array}{c}\Rightarrow b_1+42 d_2=-385\end{array} ...\left(2\right) \end{gathered}$$

By \((2) – (1)\)

$12d_2=-108$

$\Rightarrow d_2=-9$

$\therefore d_1=4$

Now $a_{78}=327$

$\Rightarrow a_1+77 d_1=327$

$\Rightarrow a_1+308=327$

$\Rightarrow a_1=19$

Given \(a_2-a_1=d=\frac{-3}{4}, a_{\mathrm{n}}=\frac{1}{4} a_1\)

Now, $S_n=\frac{n}{2}\left[a_1+a_n\right]=\frac{525}{2}$

$\Rightarrow \frac{n}{2}\left[a_1+\frac{a_1}{4}\right]=\frac{525}{2}$

$\Rightarrow \frac{5a_{1}n}{4}=525$

$\Rightarrow a_{1}n=420$

\(\because a_n=a_1+(n-1)\left(\frac{-3}{4}\right)\)

\(\Rightarrow \frac{-3}{4} a_1=\left(\frac{-3}{4}\right)(n-1) \Rightarrow a_1=n-1\)

$\therefore n(n-1)=420$

$\Rightarrow n^2-n-420=0$

$\Rightarrow (n-21)(n+20)=0$

$\Rightarrow n=21, a_1=20$

Hence, $\sum _{i=1}^{17}{a}_i=\frac{17}{2}\left[2a_1+16 d\right]$

$=\frac{17}{2}\left[40+16\left(\frac{-3}{4}\right)\right]$

$=\frac{17}{2}[40-12]$

$=17\times 14=238$

\(p x^2+q x-r=0\)

Let \(p=\frac{a}{r_1}, q=a, r=a r_1\)

and \(\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}\)

\(\frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4}\)

\(\begin{aligned} & \Rightarrow \frac{-\frac{q}{p}}{-\frac{r}{p}}=\frac{3}{4} \\ & \Rightarrow \frac{q}{r}=\frac{3}{4} \\ & \Rightarrow \frac{1}{r_1}=\frac{3}{4} \\ & \Rightarrow r_1=\frac{4}{3} \\ & (\alpha-\beta)^2=\alpha^2+\beta^2-2 \alpha \beta \\ & =(\alpha+\beta)^2-4 \alpha \beta \\ & =\left(\frac{-q}{p}\right)^2-4\left(\frac{-r}{p}\right) \\ & =\frac{q^2}{p^2}+\frac{4 r}{p} \\ & =r_1^2+4 r_1^2\end{aligned}\)

\(\begin{aligned} & =5 r_1^2 \\ & =5\left(\frac{4}{3}\right)^2 \\ & =\frac{80}{9}\end{aligned}\)

$$\begin{gathered} \text{ Let }\mathrm{P}=1+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+\frac{35}{6^4}+\dots .\dots \text{ (i) } \\ \Rightarrow \frac{P}{6}=\frac{1}{6}+\frac{5}{6^2}+\frac{12}{6^3}+\frac{22}{6^4}+\dots ...\text{ (ii) } \\ \text{Subtracting eqn }\left(ii\right)\text{ from eqn} \left(i\right), \text{we get} \end{gathered}$$

$$\begin{gathered} P-\frac{P}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+\frac{13}{6^4}+\dots \\ \Rightarrow \frac{5P}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+\frac{13}{6^4}+\dots .\text{...(iii) } \\ \Rightarrow \frac{5P}{6^2}=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+\frac{10}{6^4}+\dots ...\text{ (iv) } \end{gathered}$$

$$\begin{gathered} \text{Subtracting eqn }\left(ii\right)\text{ from }\left(i\right)\text{, we get} \\ \frac{5P}{6}-\frac{5P}{6^2}=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+\frac{3}{6^4}+\dots \\ \Rightarrow {\left(\frac{5}{6}\right)}^{2}P=1+3\left(\frac{1}{6}+\frac{1}{6^2}+\frac{1}{6^3}+\dots \right) \\ \Rightarrow {\left(\frac{5}{6}\right)}^{2}P=1+3\left(\frac{\frac{1}{6}}{1-\frac{1}{6}}\right) \end{gathered}$$

$$\begin{gathered} \Rightarrow \mathrm{P}=\frac{8}{5}\times \frac{36}{25} \\ P=\frac{288}{125} \end{gathered}$$

$\begin{array}{l}{S}_{10}=390\\ \frac{10}{2}\left[2a\left(10-1\right)d\right]=390\end{array}$

$\Rightarrow$$2a+9d=78$ ........(1)

$\frac{t_{10}}{t_5}=\frac{15}{7}$

$\Rightarrow$$\frac{a+9d}{a+4d}=\frac{15}{7}$

$\Rightarrow$$8a=3d$ .........(2)

From (1) & (2),

$a=3 \& d=8$

$S_{15}-S_5=\frac{15}{2}\left(6+14\times 8\right)-\frac{5}{2}\left(6+4\times 8\right)$

\(=790\)

$\begin{array}{l}{S}_{10}=390\\ \frac{10}{2}\left[2a\left(10-1\right)d\right]=390\end{array}$

$\Rightarrow$$2a+9d=78$ ........(1)

$\frac{t_{10}}{t_5}=\frac{15}{7}$

$\Rightarrow$$\frac{a+9d}{a+4d}=\frac{15}{7}$

$\Rightarrow$$8a=3d$ .........(2)

From (1) & (2),

$a=3 \& d=8$

$S_{15}-S_5=\frac{15}{2}\left(6+14\times 8\right)-\frac{5}{2}\left(6+4\times 8\right)$

\(=790\)

The sum of the first 12 odd terms is: $\sum_{k=1}^{12} a_{2k-1}$

The $(2k-1)$-th term is:

$a_{2k-1} = a_1 + (2k-2)d = a_1 + 2(k-1)d$

So, the sum of the first 12 odd terms is:

${\sum }_{k=1}^{12}(a_1+2(k-1\left)d\right)$$=12a_1+2d{\sum }_{k=1}^{12}(k-1)$

The sum of the first 11 natural numbers is:

$\sum_{k=1}^{12} (k-1) = \sum_{k=0}^{11} k = \frac{11 \cdot 12}{2} = 66$

So, the sum of the first 12 odd terms is:

$12a_1 + 2d \cdot 66 = 12a_1 + 132d$

$12a_1 + 132d = -\frac{72}{5}a_1$

$d = -\frac{a_1}{5}$

The sum of the first $n$ terms of an A.P. is given by:

$S_n = \frac{n}{2} (2a_1 + (n-1)d)$

We need to find $n$ such that

$S_n = 0$: $\frac{n}{2} (2a_1 + (n-1)d) = 0$

Substitute $d = -\frac{a_1}{5}$ into the second equation:

$2a_1 + (n-1)\left(-\frac{a_1}{5}\right) = 0$

$2a_1 - \frac{(n-1)a_1}{5} = 0$

$n = 11$

The value of $n$ is 11.

Let A.P. be \(3,3+d, 3+2 d, 3+3 d\)

So, \(3+\mathrm{d}=\mathrm{a}, 3+2 \mathrm{~d}=\mathrm{b}, 3+3 \mathrm{~d}=\mathrm{c}\)
Also, Let G.P. be \(3,3 r, 2 r^2, 3 r^3\)
So, \(3 r=a-1,3 r^2=b+1,3 r^3=c+9\)
On comparing,

\(3 r=3+d-1 \Rightarrow d=3 r-2\) \(\ldots (1)\)

\(3 \mathrm{r}^2=3+2 \mathrm{d}+1 \Rightarrow 3 \mathrm{r}^2=2 \mathrm{d}+4\) \(\ldots (2)\)

On solving (1) and (2) we get

\(r=2\)

\(\mathrm{a}=7, \mathrm{~b}=11, \mathrm{c}=15\)

A.M. of \(a, b, c \Rightarrow \frac{7+11+15}{3}=11\)

Since \(2 a_{n+2}=5 a_{n+1}-3 a_n\)

Therefore

\(\begin{aligned} & 2 x^2=5 x-3 \\ & \Rightarrow x=1, \frac{3}{2} \\ & \therefore a_n=P (1)^n+Q\left(\frac{3}{2}\right)^n\end{aligned}\)

for \(\mathrm{n}=0\) we get, \(0=P+Q\)

for \(\mathrm{n}=1\) we get, \(\frac{1}{2}=P+\frac{3}{2} Q\)

On simplifying, we get

\(\begin{aligned} & P=-1, Q=1 \\ & \Rightarrow \mathrm{a}_{\mathrm{n}}=-1+\left(\frac{3}{2}\right)^{\mathrm{n}}\end{aligned}\)

Now

$$\begin{gathered} \sum _{\mathrm{k}=1}^{100}{\mathrm{a}}_{\mathrm{k}}=\sum _{\mathrm{k}=1}^{100}\left(-1\right)+{\left(\frac{3}{2}\right)}^{\mathrm{k}} \\ =-100+\frac{\left(\frac{3}{2}\right)\left({\left(\frac{3}{2}\right)}^{100}-1\right)}{\frac{3}{2}-1} \\ =3\left[{\left(\frac{3}{2}\right)}^{100}-1\right]-100 \\ =3{\mathrm{a}}_{100}-100 \end{gathered}$$

Required = $\frac{4.1}{4+{3.1}^2+1^4}+\frac{4.2}{4+{3.2}^2+2^4}+\frac{4.3}{4+{3.3}^2+3^{4^2}}\dots 20$ terms

$\begin{array}{l}={\sum }_{r=1}^{20}\frac{4r}{4+3r^2+r^4}\\ ={\sum }_{r=1}^{20}\frac{4r}{{(r^2+2)}^2-r^2}=2{\sum }_{r=1}^{20}\frac{2r}{(r^2+2-r)(r^2+2+r)}\\ =2{\sum }_{r=1}^{20}\left[\frac{1}{r^2-r+2}-\frac{1}{r^2+r+2}\right]\\ =2(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{8}+\cdots -\frac{1}{422})\\ \frac{m}{n}=1-\frac{1}{211}=\frac{210}{211}\\ m+n=421\end{array}$

$$\begin{gathered} a_1, a_2, a_3,.........a_{2024} \mathrm{are} \mathrm{in} \mathrm{AP} \\ \mathrm{using} \mathrm{the} \mathrm{property} \mathrm{of} \mathrm{AP}: \\ \text{let} a_1+a_{2024}=a_2+a_{2023}=a_3+a_{2022}=\dots =k \\ {a}_1+\left(a_5+a_{10}+\cdots +a_{2020}\right)+a_{2024}=2233 \\ k+202k = 2233 \\ 203k=2233 \\ k =11 \\ Now, =a_1+a_2+\cdots +a_{2024}= 1012k=11\times 1012=11132 \end{gathered}$$

$$\begin{gathered} a+a+d+a+2d=54 \\ 3a+3d=54 \\ a+d=18............\left(i\right) \\ {S}_{20}=\frac{20}{2}\left[2a+19d\right]=10\left[2\right(18-d)+19d] \\ {S}_{20}=10(36+17d) \\ 1600<S_{20}<1800 \\ 1600<10(36+17d)<1800 \\ 7.29<d<8.47 \\ \Rightarrow d = 8 \\ a+d = 18 \\ a= 10 \\ Now, a_{11}=a+10d \\ = 10+10\left(8\right)=90 \end{gathered}$$