Sequences and Series

The arithmetic mean of $\frac{1}{a}$ and $\frac{1}{b}$ is $\frac{5}{16}$.

$\frac{\frac{1}{a}+\frac{1}{b}}{2}=\frac{5}{16} $

$\Rightarrow \frac{1}{a}+\frac{1}{b}=\frac{5}{8} $

$\Rightarrow $ $ \frac{a+b}{a b}=\frac{5}{8} $

$\Rightarrow 8(a+b)=5 a b \,\,\,\,\,\,\,\,\,\,...(i)$

$a, 4, \alpha, b$ are in A.P. Let the common difference be $d$.

• (A). $4=a+d \Longrightarrow d=4-a$

• (B). $\alpha=a+2 d=a+2(4-a)=8-a$

• (C). $b=a+3 d=a+3(4-a)=12-2 a$

Solve for $a, b$, and $\alpha$

Substitute $b=12-2 a$ into equation (1) :

$ \begin{gathered} 8(a+12-2 a)=5 a(12-2 a) \\ 8(12-a)=60 a-10 a^2 \\ 96-8 a=60 a-10 a^2 \\ 10 a^2-68 a+96=0 \\ 5 a^2-34 a+48=0 \end{gathered} $

Using the quadratic formula: $a=\frac{34 \pm \sqrt{34^2-4(5)(48)}}{2(5)}=\frac{34 \pm \sqrt{1156-960}}{10}=\frac{34 \pm 14}{10}$

$ a=\frac{48}{10}=4.8, \quad a=\frac{20}{10}=2 \text { (Discarded because the problem states } a>2 \text { ) } $

With $a=4.8: \quad \alpha=8-4.8=\mathbf{3 . 2}, \quad b=12-2(4.8)=12-9.6=\mathbf{2 . 4}$

Analyze the Quadratic Equation

The given equation is $\alpha x^2-a x+2(\alpha-2 b)=0$. Substitute the values:

$ \begin{gathered} 3.2 x^2-4.8 x+2(3.2-2(2.4))=0 \\ 3.2 x^2-4.8 x+2(3.2-4.8)=0 \\ 3.2 x^2-4.8 x-3.2=0 \\ 2 x^2-3 x-2=0 \end{gathered} $

Factorise $ (2 x+1)(x-2)=0 $

The roots are $x=2$ and $x=-0.5$.

Looking at the options: Root $x=2$ lies in the interval $(1,4)$. Root $x=-0.5$ lies in the interval $(-2,0)$.

Let $\quad S=\frac{6}{3^{26}}+\frac{10 \cdot 1}{3^{25}}+\frac{10 \cdot 2}{3^{24}}+\frac{10 \cdot 2^2}{3^{23}}+\cdots+\frac{10 \cdot 2^{24}}{3}$

Notice that from the second term onwards, the terms follow a pattern. Let $S^{\prime}$ be the sum of these terms:

$ S^{\prime}=\frac{10 \cdot 2^0}{3^{25}}+\frac{10 \cdot 2^1}{3^{24}}+\frac{10 \cdot 2^2}{3^{23}}+\cdots+\frac{10 \cdot 2^{24}}{3^1} $

We can factor out $\frac{10}{3^{25}}$ :

$ \begin{gathered} S^{\prime}=\frac{10}{3^{25}}\left(1+\frac{2^1}{3^{-1}}+\frac{2^2}{3^{-2}}+\cdots+\frac{2^{24}}{3^{-24}}\right) \\ S^{\prime}=\frac{10}{3^{25}}\left(2^0 \cdot 3^0+2^1 \cdot 3^1+2^2 \cdot 3^2+\cdots+2^{24} \cdot 3^{24}\right) \end{gathered} $

$ S^{\prime}=\frac{10}{3^{25}} \sum\limits_{k=0}^{24}(2 \cdot 3)^k=\frac{10}{3^{25}} \sum\limits_{k=0}^{24} 6^k $

Calculate the sum of the GP

The series $\sum\limits_{k=0}^{24} 6^k$ is a geometric progression with first term $a=1$, common ratio $r=6$, and $n=25$ terms.

The sum formula is $S_n=a \frac{r^n-1}{r-1}$ :

$ \sum\limits_{k=0}^{24} 6^k=\frac{6^{25}-1}{6-1}=\frac{6^{25}-1}{5} $

Substituting this back into $S^{\prime}$ :

$ S^{\prime}=\frac{10}{3^{25}}\left(\frac{6^{25}-1}{5}\right)=\frac{2}{3^{25}}\left(6^{25}-1\right)=\frac{2 \cdot 6^{25}}{3^{25}}-\frac{2}{3^{25}} $

Since $6^{25}=(2 \cdot 3)^{25}=2^{25} \cdot 3^{25}$ :

$ S^{\prime}=\frac{2 \cdot 2^{25} \cdot 3^{25}}{3^{25}}-\frac{2}{3^{25}}=2 \cdot 2^{25}-\frac{2}{3^{25}}=2^{26}-\frac{2}{3^{25}} $

Find the total sum $S$ :

Now, add the first term $\frac{6}{3^{26}}$ back to $S^{\prime}$ :

$S =\frac{6}{3^{26}}+2^{26}-\frac{2}{3^{25}} $

[$ \text { Note that, } \frac{6}{3^{26}}=\frac{2 \cdot 3}{3^{26}}=\frac{2}{3^{25}} $]

$\Rightarrow $ $ S =\frac{2}{3^{25}}+2^{26}-\frac{2}{3^{25}}=2^{26} $

$ S=\sum\limits_{k=1}^{\infty}(-1)^{k+1} \frac{k(k+1)}{k!} $

First, let's simplify the expression inside the summation. We can cancel out the $k$ in the numerator with the $k$ in $k!$ :

$ \frac{k(k+1)}{k!}=\frac{k(k+1)}{k \cdot(k-1)!}=\frac{k+1}{(k-1)!} $

Now, we rewrite the numerator $(k+1)$ in terms of $(k-1)$ to further simplify the fraction:

$ k+1=(k-1)+2 $

Substituting this back into the expression:

$ \frac{k+1}{(k-1)!}=\frac{(k-1)+2}{(k-1)!}=\frac{k-1}{(k-1)!}+\frac{2}{(k-1)!} $

For the first part, $\frac{k-1}{(k-1)!}=\frac{1}{(k-2)!}$ (valid for $k \geq 2$ ). So, the general term $a_k$ becomes:

$ a_k=(-1)^{k+1}\left[\frac{1}{(k-2)!}+\frac{2}{(k-1)!}\right] $

Expand the Series

Let's apply the summation $\sum\limits_{k=1}^{\infty}$ to both parts. Note that for $k=1$, the term $1 /(k-2)$ ! is effectively 0 because the factorial of a negative integer is infinite in the denominator.

$ S=\sum\limits_{k=2}^{\infty} \frac{(-1)^{k+1}}{(k-2)!}+\sum\limits_{k=1}^{\infty} \frac{2(-1)^{k+1}}{(k-1)!} $

Using exponential expansion(Taylor series expansion for $e^x$ :)

$ e^x=\sum\limits_{n=0}^{\infty} \frac{x^n}{n!}=1+\frac{x}{1!}+\frac{x^2}{2!}+\ldots $

for $x=-1$ :

$ e^{-1}=\frac{1}{e}=\sum\limits_{n=0}^{\infty} \frac{(-1)^n}{n!}=1-\frac{1}{1!}+\frac{1}{2!}-\frac{1}{3!}+\ldots $

Evaluate the Summations

$\sum\limits_{k=2}^{\infty} \frac{(-1)^{k+1}}{(k-2)!}$

Let $n=k-2$. When $k=2, n=0$. When $k \rightarrow \infty, n \rightarrow \infty$. The exponent becomes $k+1=(n+2)+1=n+3$.

= $ \sum\limits_{n=0}^{\infty} \frac{(-1)^{n+3}}{n!}=(-1)^3 \sum\limits_{n=0}^{\infty} \frac{(-1)^n}{n!}=-1 \cdot\left(\frac{1}{e}\right)=-\frac{1}{e} $

$\sum\limits_{k=1}^{\infty} \frac{2(-1)^{k+1}}{(k-1)!}$

Let $m=k-1$. When $k=1, m=0$. When $k \rightarrow \infty, m \rightarrow \infty$. The exponent becomes $k+1=(m+1)+1=m+2$.

= $ 2 \sum\limits_{m=0}^{\infty} \frac{(-1)^{m+2}}{m!}=2(-1)^2 \sum\limits_{m=0}^{\infty} \frac{(-1)^m}{m!}=2 \cdot\left(\frac{1}{e}\right)=\frac{2}{e} $

Required S

$ S=-\frac{1}{e}+\frac{2}{e}=\frac{1}{e} $

Find common difference of second A.P.

It is given that $b_{31}=-277$, $b_{43}=-385$

Let $d_b$ is common difference of second A.P.

$ \begin{aligned} & b_{31}=b_1+30 d_b---(1) \\ & b_{43}=b_1+42 d_b---(2) \end{aligned} $

Subtract (1) from (2)

$ \begin{aligned} & b_{43}-b_{31}=(42-30) d_b \\ & -385-(-277)=12 d_b \\ & d_b=\frac{-108}{12}=-9 \end{aligned} $

Find the common difference ( $d_a$ ) of the first A.P.

The problem states that $d_a$ is 13 more than $d_b$ :

$ \begin{gathered} d_a=d_b+13 \\ d_a=-9+13=4 \end{gathered} $

Calculate $a_1$

We are given $a_{78}=327$. Using the formula $a_n=a_1+(n-1) d_a$ :

$ \begin{gathered} 327=a_1+(78-1)(4) \\ 327=a_1+(77)(4) \\ 327=a_1+308 \\ a_1=327-308=19 \end{gathered} $

It is given that $a_1, a_2, a_3, a_4$ are in A.P with common difference $l$.

four terms in A.P are $(a-3 d),(a-d),(a+d),(a+3 d)$ common difference $l=2 d \Rightarrow d=\frac{l}{2}$

$ \begin{aligned} & \text { so } a_1=a-\frac{3 l}{2}, a_2=\left(a-\frac{l}{2}\right) \\ & a_3=a+\frac{l}{2}, a_4=\left(a+\frac{3 l}{2}\right) \end{aligned} $

It is given that $a_1+a_2+a_3+a_4=48$

$ \begin{aligned} & a-\frac{3 l}{2}+a-\frac{l}{2}+a+\frac{l}{2}+a+\frac{3 l}{2}=48 \\ & \Rightarrow 4 a=48 \Rightarrow a=12 \\ & \text { also, } a_1 a_2 a_3 a_4+l^4=361 \\ & \Rightarrow \left(12-\frac{3 l}{2}\right)\left(12-\frac{l}{2}\right)\left(12+\frac{l}{2}\right)\left(12+\frac{3 l}{2}\right)+l^4=361 \\ & \Rightarrow \left(144-\frac{l^2}{4}\right)\left(144-\frac{9 l^2}{4}\right)+l^4=361 \\ & \Rightarrow 20736-324 l^2-36 l^2+\frac{9}{16} l^4+l^4=361 \end{aligned} $

$\Rightarrow $ $ \frac{25}{16} l^4-360 l^2+20375=0 $

multiply both side by $\frac{16}{5}$.

$ 5 l^4-1152 l^2+65200=0 $

using quadratic formula

$ l^2=\frac{1152 \pm \sqrt{(1152)^2-4 \times 5 \times 65200}}{2 \times 5} $

$ \begin{aligned} & \Rightarrow l^2=\frac{1152 \pm \sqrt{23104}}{10}=\frac{1152 \pm 152}{10} \\ & \Rightarrow l^2=\frac{1152+152}{10}, \frac{1152-152}{10} \\ & \Rightarrow l^2=\frac{1304}{10}, \frac{1000}{10} \\ & \Rightarrow l^2=\frac{1304}{10}, 100 \end{aligned} $

Since $l$ is integer so $l= \pm 10$

$\begin{aligned} & \text { If } l=10 \\ & \begin{array}{l}a_1=12-\frac{3}{2} \times 10=-3 \\ \quad a_2=12-\frac{10}{2}=7 \\ \quad a_3=12+\frac{10}{2}=17 \\ a_4=12+\frac{3}{2} \times 10=27\end{array}\end{aligned}$

$\begin{aligned} & \text { If } l=-10 \\ & \begin{array}{c}a_1=12+\frac{3}{2} \times 10=27 \\ a_2=12+\frac{10}{2}=17 \\ a_3=12-\frac{10}{2}=7 \\ a_4=12-\frac{3}{2} \times 10=-3\end{array}\end{aligned}$

So, numbers are same but in reverse order therefore, largest term of A.P is 27.

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

So Question is

$ S_{\infty}=(a+b)+\left(a^2+a b+b^2\right)+\left(a^3+a^2 b+a b^2+b^3\right)+\ldots \infty \text { terms } $

So clearly each bracket is in G.P. having ( $n+1$ ) terms.

So general term is

$ T_n=\left(a^n+a^{n-1} b+a^{n-2} b^2 \cdots+b^n\right) $

$T_n$ is G.P. first term $a^n$, common ratio $=\frac{b}{a},(n+1)$ terms

$ T_n=a^n\left[\frac{\left(\frac{b}{a}\right)^{n+1}-1}{\frac{b}{a}-1}\right] $

Now substitute $a=\frac{1}{3}$ and $b=\frac{4}{7}$

$ \begin{aligned} & T_n=\left(\frac{1}{3}\right)^n\left[\frac{\left(\frac{12}{7}\right)^{n+1}-1}{\frac{12}{7}-1}\right] \\ & T_n=\left(\frac{1}{3}\right)^n \frac{\left[\left(\frac{12}{7}\right)^n \cdot \frac{12}{7}-1\right] \cdot 7}{5} \\ & T_n=\left(\frac{1}{3}\right)^n\left(\frac{12}{7}\right)^n \cdot \frac{12}{7} \times \frac{7}{5}-1 \times \frac{7}{5} \times\left(\frac{1}{3}\right)^n \\ & T_n=\frac{12}{5}\left(\frac{4}{7}\right)^n-\frac{7}{5}\left(\frac{1}{3}\right)^n \end{aligned} $

Now Required sum $\left(S_{\infty}\right)=\sum_{n=1}^{\infty} T_n$

$ \begin{aligned} S_{\infty} & =\sum_{n=1}^{\infty} \frac{12}{5}\left(\frac{4}{7}\right)^n-\frac{7}{5}\left(\frac{1}{3}\right)^n \\ S_{\infty} & =\frac{12}{5} \sum_{n=1}^{\infty}\left(\frac{4}{7}\right)^n-\frac{7}{5} \sum_{n=1}^{\infty}\left(\frac{1}{3}\right)^n \\ S_{\infty} & =\frac{12}{5}\left[\frac{4}{7}+\left(\frac{4}{7}\right)^2+\left(\frac{4}{7}\right)^3 \ldots\right]-\frac{7}{5}\left[\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3 \ldots\right] \end{aligned} $

Both series are infinite G.P.

$ \begin{aligned} & S_{\infty}=\frac{12}{5}\left[\frac{4 / 7}{1-4 / 7}\right]-\frac{7}{5}\left[\frac{1 / 3}{1-1 / 3}\right] \\ & =\frac{12}{5}\left[\frac{4}{3}\right]-\frac{7}{5}\left[\frac{1}{2}\right] \\ & =\frac{16}{5}-\frac{7}{10}=\frac{32-7}{10}=\frac{25}{10}=\frac{5}{2} \end{aligned} $

$729,81,9,1, \ldots$ is a sequence in G.P. Can be written as $9^3, 9^2, 9^1, 9^0, \frac{1}{9},\left(\frac{1}{9}\right)^2, \ldots$

$P_n$ is product of first $n$ term

$ \begin{aligned} & P_n=\left(9^3\right)\left(9^2\right)\left(9^1\right) \ldots \\ & P_n=9^{3+2+1+0+(-1)+(-2) \ldots} \end{aligned} $

Clearly power of 9 are in A.P. with first term $=3$

common difference $=-1$

So, sum of first $n$ terms of this A.P. $=\frac{n}{2}[2 \times 3+(n-1)(-1)]$

$ =\frac{n}{2}[6-n+1]=\frac{n}{2}(7-n) $

So, $P_n=(9)^{\frac{n}{2}(7-n)}$

It is given that

$ \begin{aligned} &2 \sum_{n=1}^{40}\left(P_n\right)^{1 / n}=\frac{3^\alpha-1}{3^\beta}\\ &\text { Solving }\\ &\begin{aligned} & \sum_{n=1}^{40}\left(P_n\right)^{1 / n}=\sum_{n=1}^{40}(9)^{\frac{n}{2}(7-n) \times \frac{1}{n}}=\sum_{n=1}^{40} 9^{\frac{7-n}{2}} \\ & \sum_{n=1}^{40}\left(9^{1 / 2}\right)^{7-n}=\sum_{n=1}^{40}(3)^{7-n}=3^7 \sum_{n=1}^{40}\left(\frac{1}{3}\right)^n \\ & =3^7\left[\frac{1}{3}+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3 \cdots\left(\frac{1}{3}\right)^{40}\right] \\ & =3^7\left[\text { G.P. with first term }=\frac{1}{3} \text { common ratio }=\frac{1}{3}\right] \end{aligned}\\ &=3^7 \times \frac{1}{3}\left[\frac{1-\left(\frac{1}{3}\right)^{40}}{1-\frac{1}{3}}\right] \end{aligned} $

$ \begin{aligned} & \sum_{n=1}^{40}\left(P_n\right)^{1 / n}=3^6\left(\frac{\frac{3^{40}-1}{3^{40}}}{\frac{2}{3}}\right)=\frac{3^{40}-1}{2 \cdot 3^{33}} \\ & \quad 2 \sum_{n=1}^{40}\left(P_n\right)^{1 / n}=\frac{3^{40}-1}{3^{33}}=\frac{3^\alpha-1}{3^\beta} \\ & \text { Comparing } \alpha=40, \beta=33 \\ & \text { g.c.d }(40,33)=1 \\ & \text { So } \alpha+\beta=40+33=73 \end{aligned} $

Given $a_1, a_2, \ldots, a_n$ are in A.P; $a_1>0$

Let $d$ be the common difference of this A.P.

So $d=a_2-a_1=-\frac{3}{4}$ (given)

It is given that $\sum\limits_{i=1}^n a_i=\frac{525}{2}$ and $a_n=\frac{1}{4} a_1$

$\therefore $ $ a_1+a_2+a_3+\ldots a_n=\frac{525}{2} $

Using sum of $n$ terms of A.P $S_n=\frac{n}{2}\left(a_1+a_n\right)$

$\Rightarrow $ $\frac{n}{2}\left(a_1+\frac{1}{4} a_1\right)=\frac{525}{2}$

$\Rightarrow $ $5 a_1 n=525 \times 4$

$ a_1 n=420-(1) $

Also using formula for general term

$ a_n=a_1+(n-1) d $

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

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

$ n=a_1+1-(2) $

Substitute in equation (1)

$ a_1\left(a_1+1\right)=420 $

$\Rightarrow $ $a_1^2+a_1-420=0$

$\Rightarrow $ $a_1^2+21 a_1-20 a_1-420=0$

$\Rightarrow $ $a_1\left(a_1+21\right)-20\left(a_1+21\right)=0$

$\Rightarrow $ $\left(a_1+21\right)\left(a_1-20\right)=0$

Since $a_1>0$ so $a_1=20$

To find $\sum\limits_{i=1}^{17} a_i=a_1+a_2+\ldots a_{17}$

Sum of 17 terms $=\frac{17}{2}\left[2 a_1+(17-1) d\right]$

$\begin{aligned} & =\frac{17}{2}\left[2 \times 20+16 \times-\frac{3}{4}\right] \\ & =\frac{17}{2}[40-12]=\frac{17}{2} \times 28 \\ & =17 \times 14=238\end{aligned}$

Let $S_n=\sum\limits_{k=1}^n a_k=\alpha n^2+\beta n$

$ a_1+a_2+a_3+a_4+\ldots a_n=\alpha n^2+\beta n $

Since the sum $S_n$ is a quadratic expression in $n$ with no constant term then the sequence $a_k$ is in A.P (arithmetic progression).

Then $n$th term of the sequence can be found using the formula

$ a_n=S_n-S_{n-1} $

$\begin{aligned} & a_n=\alpha n^2+\beta n-\left[\alpha(n-1)^2+\beta(n-1)\right] \\ & =\alpha n^2+\beta n-\left[\alpha\left(n^2-2 n+1\right)+\beta n-\beta\right] \\ & =\alpha n^2+\beta n-\alpha n^2+2 \alpha n-\alpha-\beta n+\beta\end{aligned}$

$\Rightarrow $ $ a_n=\beta-\alpha+2 \alpha n-(1) $

It is given that $a_{10}=59$, substitute $n=10$ in equation (1).

$ a_{10}=\beta-\alpha+2 \alpha(10)=59 $

$\Rightarrow $ $19 \alpha+\beta=59-(2)$

also, $a_6=7 a_1$ (given)

substitute $n=6$ and $n=1$ in equation (1).

$ \beta-\alpha+2 \alpha(6)=7(\beta-\alpha+2 \alpha(1)) $

$\Rightarrow $ $11 \alpha+\beta=7 \alpha+7 \beta$

$\Rightarrow $ $4 \alpha=6 \beta$

$\Rightarrow $ $ 2 \alpha=3 \beta \Rightarrow \alpha=\frac{3 \beta}{2} $

Substitute $\alpha=\frac{3 \beta}{2}$ in equation (2)

$ 19\left(\frac{3 \beta}{2}\right)+\beta=59 $

$\Rightarrow $ $57 \beta+2 \beta=118$

$\Rightarrow $ $59 \beta=118 $

$\Rightarrow $$ \beta=\frac{118}{59}=2$

And $\alpha=\frac{3 \beta}{2}=\frac{3 \times 2}{2}=3$

So, value of $\alpha+\beta=3+2=5$.

$ \begin{aligned} & a+(a+d)+(a+2 d)+(a+3 d)=6 \\ & 2 a+3 d=3 \\ & 6 a+15 d=4 \\ & d=-5 / 6 \\ & a=11 / 4 \\ & S_{12}=\frac{12}{2}\left[2\left(\frac{11}{4}\right)+11 \times(-5 / 6)\right]=-22 \end{aligned} $

$ \begin{aligned} & n x^2+(2 x+6 x+10 x+\ldots . .+(4 n-2))+(2.4+4.6+\ldots+(2 n-2) 2 n)=\frac{8 n}{3} \\ & n x^2+2 x(1+3+5+\ldots .+2 n-1)+\sum_{r=1}^n(2 r-2) 2 r=\frac{8 n}{3} \end{aligned} $

$ \begin{aligned} &\begin{aligned} & n x^2+2 n^2 x+4\left(\frac{n(n+1)(2 n-2)}{6}\right)=\frac{8 n}{3} \\ & 3 x^2+6 n x+4\left(n^2-1\right)=8 \end{aligned}\\ &\text { For } \mathrm{n}=3\\ &\begin{aligned} & 3 x^2+18 x+24=0 \\ & x=-2,-4(\text { two consecutive even integers }) \end{aligned} \end{aligned} $

$ \begin{aligned} &\frac{a_2}{2 a_1}=\frac{a_3}{2 a_2}=\frac{a_4}{2 a_3}=\ldots=\frac{a_{10}}{2 a_9}=\frac{1}{\sqrt{2}}\\ &\therefore a_1, a_2, a_2, \ldots . a_{10} \text { are in G.P. With common ratio } \sqrt{2}\\ &\therefore \sum_{i=1}^{10} a_i=\frac{a_1\left((\sqrt{2})^{10}-1\right)}{\sqrt{2}-1}=62 \Rightarrow a_1=2(\sqrt{2}-1) \end{aligned} $

$ \begin{aligned} & a_2 \cdot a_3 \cdot a_4=64 \\ & \frac{a}{r} \cdot a \cdot a r=64 \\ & \Rightarrow \quad a=4 \\ & a_1+a_3+a_5=\frac{813}{7} \\ & =\frac{4}{r^2}+4+4 r^2=\frac{813}{7} \\ & \text { Let } r^2=t \\ & 28 t^2-785 t+28=0 \\ & (28 t-1)(t-28)=0 \\ & \Rightarrow \quad t=28, \frac{1}{28} \end{aligned} $

$ \begin{aligned} & r^2=28(\because \text { G.P. is increasing }) \\ & a_3+a_5+a_7 \\ & =a+a r^2+a r^4=r^2\left(a+\frac{a}{r^2}+r^2\right) \\ & =28 \times \frac{813}{7} \\ & =3252 \end{aligned} $

$\begin{aligned} & \text { If } \frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots . \infty=\frac{\pi^4}{90} \quad\text{....... (i)}\\ & \beta=\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots, \\ & =\frac{1}{16}\left[\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\ldots . .\right], \end{aligned}$

$=\frac{1}{16} \times \frac{\pi^4}{90}$ using (ii) ............ (ii)

$\begin{aligned} & \alpha=\frac{1}{1^4}+\frac{1}{3^4}+\frac{1}{5^4}+\ldots . . \infty \\ & \left(\frac{1}{1^4}+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\ldots . .\right) \\ & -\left(\frac{1}{2^4}+\frac{1}{4^4}+\frac{1}{6^4}+\ldots . .\right) \\ & \alpha=\frac{\pi^4}{90}-\frac{1}{16} \times \frac{\pi^4}{90} \quad \text { [using (i) and (ii)] } \\ & \alpha=\frac{16-1}{16 \times 90} \times \pi^4=\frac{15}{16 \times 90} \pi^4=\frac{\pi^4}{96} \\ & \therefore \frac{\alpha}{\beta}=\frac{\frac{\pi^4}{96}}{\frac{\pi^4}{16 \times 90}}=\frac{16 \times 90}{96}=15 \end{aligned}$

Sum of n terms of an AP ($S_n$): $S_n = \frac{n}{2} [2a + (n-1)d]$ where 'a' is the first term and 'd' is the common difference.

nth term of an AP ($a_n$): $a_n = a + (n-1)d$

Given Information:

$S_n = 700$ (The sum of the first n terms is 700)

$a_6 = 7$ (The 6th term is 7)

$S_7 = 7$ (The sum of the first 7 terms is 7)

Steps to Solve:

Use $S_7$ to find a relationship between 'a' and 'd':

$S_7 = \frac{7}{2} [2a + (7-1)d] = 7$

$\frac{7}{2} [2a + 6d] = 7$

$2a + 6d = 2$

$a + 3d = 1$ (Equation 1)

Use $a_6$ to find another relationship between 'a' and 'd':

$a_6 = a + (6-1)d = 7$

$a + 5d = 7$ (Equation 2)

Solve the system of equations (Equation 1 and Equation 2) to find 'a' and 'd':

Subtract Equation 1 from Equation 2:

$(a + 5d) - (a + 3d) = 7 - 1$

$2d = 6$

$d = 3$

Substitute d = 3 into Equation 1:

$a + 3(3) = 1$

$a + 9 = 1$

$a = -8$

Find 'n' using $S_n = 700$:

$S_n = \frac{n}{2} [2a + (n-1)d] = 700$

$\frac{n}{2} [2(-8) + (n-1)(3)] = 700$

$n[-16 + 3n - 3] = 1400$

$n[3n - 19] = 1400$

$3n^2 - 19n - 1400 = 0$

Solve the quadratic equation for 'n':

We can use the quadratic formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

$n = \frac{19 \pm \sqrt{(-19)^2 - 4(3)(-1400)}}{2(3)}$

$n = \frac{19 \pm \sqrt{361 + 16800}}{6}$

$n = \frac{19 \pm \sqrt{17161}}{6}$

$n = \frac{19 \pm 131}{6}$

We have two possible values for n:

$n = \frac{19 + 131}{6} = \frac{150}{6} = 25$

$n = \frac{19 - 131}{6} = \frac{-112}{6} = -\frac{56}{3}$ (Since 'n' must be a positive integer, we discard this solution)

So, $n = 25$

Find $a_n$ (which is $a_{25}$):

$a_{25} = a + (25-1)d$

$a_{25} = -8 + (24)(3)$

$a_{25} = -8 + 72$

$a_{25} = 64$

Therefore, $a_n = 64$. The correct answer is Option D.

To solve this problem, we have to work with two equations derived from the geometric progression (G.P.):

$\mathrm{ar} + \mathrm{ar}^3 + \mathrm{ar}^5 = 21$

$\mathrm{ar}^7 + \mathrm{ar}^9 + \mathrm{ar}^{11} = 15309$

From these equations, we can extract the terms:

Equation (1): $\mathrm{ar}(1 + \mathrm{r}^2 + \mathrm{r}^4) = 21$

Equation (2): $\mathrm{ar}^7(1 + \mathrm{r}^2 + \mathrm{r}^4) = 15309$

Now, divide equation (2) by equation (1):

$ \frac{\mathrm{ar}^7}{\mathrm{ar}} = \frac{15309}{21} $

From this division, we get:

$ \mathrm{r}^6 = 729 $

Which implies:

$ \mathrm{r} = 3 \quad \text{(since both terms and ratios are positive)} $

Using this value of $\mathrm{r}$, calculate the sum of the first nine terms of the G.P. using the formula for the sum of a G.P.:

$ S_n = \frac{\mathrm{a}(\mathrm{r}^9 - 1)}{\mathrm{r} - 1} $

Substitute known values:

$ \Rightarrow \frac{\mathrm{a}(3^9 - 1)}{3 - 1} = \frac{\mathrm{a} \cdot (19683 - 1)}{2} $

Given that $\frac{7}{91} \times (19683 - 1)/2 = \frac{7 \times 19682}{91 \times 2}$:

$ = \frac{7 \times 19682}{91 \times 2} = \frac{9841}{13} = 757 $

Therefore, the sum of the first nine terms of the G.P. is 757.

Given the sequence $x_1, x_2, x_3, x_4$ in geometric progression:

$x_1 = a$

$x_2 = ar$

$x_3 = ar^2$

$x_4 = ar^3$

When you subtract 2, 7, 9, and 5 from $x_1, x_2, x_3, x_4$ respectively, the sequence becomes an arithmetic progression. Thus, the new sequence is:

$a - 2$

$ar - 7$

$ar^2 - 9$

$ar^3 - 5$

For these to form an arithmetic progression, the common differences must be equal, so:

$ (ar - 7) - (a - 2) = (ar^2 - 9) - (ar - 7) $

Simplifying gives:

$ a(r - 1) - 5 = ar(r - 1) - 2 $

$ a(r - 1)(r - 1) = -3 \quad (i) $

$ (ar - 7) - (a - 2) = (ar^3 - 5) - (ar^2 - 9) $

Simplifying gives:

$ a(r - 1) - 5 = ar^2(r - 1) + 4 $

$ a(r - 1)(r^2 - 1) = -9 \quad (ii) $

Using the ratio of equations (ii) and (i):

$ \frac{a(r - 1)(r^2 - 1)}{a(r - 1)(r - 1)} = \frac{-9}{-3} $

$ r + 1 = 3 \implies r = 2 $

Plugging back into equation (i):

$ a(1)(1) = -3 \implies a = -3 $

So, the sequence $x_1, x_2, x_3, x_4$ is:

$x_1 = -3$

$x_2 = -6$

$x_3 = -12$

$x_4 = -24$

The expression for $\frac{1}{24}(x_1 \cdot x_2 \cdot x_3 \cdot x_4)$ is:

$ \frac{1}{24}((-3) \cdot (-6) \cdot (-12) \cdot (-24)) = 216 $

$\begin{aligned} & S_n=\sum_{r=1}^n \frac{4 r}{4+3 r^2+r^4} \\ & =2 \sum_{r=1}^n \frac{2 r}{\left(r^2+2\right)^2-r^2}=2 \sum_{r=1}^n \frac{\left(r^2+2+r\right)-\left(r^2+2-r\right)}{\left(r^2+2+r\right)\left(r^2+2-r\right)} \\ & =2 \sum_{r=1}^n\left(\frac{1}{r^2+2-r}-\frac{1}{r^2+2+r}\right) \\ & S_{20}=2\left[\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{4}-\frac{1}{8}\right)+\ldots\right] \\ & \quad=2\left(\frac{1}{2}-\frac{1}{20^2+2+20}\right) \\ & \quad=2\left(\frac{1}{2}-\frac{1}{422}\right) \\ & \quad=2\left(\frac{422-2}{422 \times 2}\right)=\frac{420}{422}=\frac{210}{211}=\frac{m}{n} \\ & m+n=421 \end{aligned}$

Let the terms in $A$ be $a_1-d, a_1, a_1+d$

and in $B$ be $a_2-D, a_2, a_2+D$

Now $3 a_1=36$

$\Rightarrow \quad a_1=12$

and $3 a_2=36$

$\Rightarrow \quad a_2=12$

Now $(12-d)(12)(12+d)=p$

and $(12-D)(12)(12+D)=q$

Also $\frac{p+q}{p-q}=\frac{19}{5}$

$\begin{aligned} & \Rightarrow 12 q=7 p \\ & \Rightarrow 12(12-D)(12)(12+D)=7(12-d)(12)(12+\mathrm{d}) \\ & \Rightarrow 12(9-d)(12)(15-d)=7(12-d)(12)(12+\mathrm{d}) \\ & \Rightarrow 12\left(135-d^2-6 d\right)=7\left(144-d^2\right) \\ & \Rightarrow d=6, D=9 \\ & p=6 \times 12 \times 18=1296 \\ & q=756 \\ & p-q=540 \end{aligned}$

$\begin{array}{ll} 1^{\text {st }} \text { A.P. }: 1,6,11 \ldots & \Rightarrow T_n=S_n-4 \\ 2^{\text {nd }} \text { A.P.: } 9,16,23 \ldots & \Rightarrow T_m=2+7 m \end{array}$

Let's find when they are equal for the first time:

$\begin{aligned} & 5 n-4=2+7 m \\ & \Rightarrow 5 n-7 m=6 \\ & \Rightarrow n=4, m=2 \end{aligned}$

$\Rightarrow 16$ is the first term, common difference will be

$\operatorname{LCM}\left(d_1, d_2\right)=\operatorname{LCM}(5,7)=35$

$\Rightarrow$ Common terms will be 16, 51, $86 \ldots$

The last term of $1^{\text {st }}$ A.P.

$=T_{2025}=5 \times 2025-4=10121$

$\begin{aligned} &\Rightarrow \text { Common term must be less than that }\\ &\begin{aligned} & \Rightarrow 35 n-19 \\ & \Rightarrow 35 n-19 \leq 10121 \Rightarrow 35 n \leq 10140 \\ & \quad \Rightarrow n \leq 289.7 \\ & \quad \Rightarrow n=289 \\ & \Rightarrow \operatorname{in} n(A \cup B)=n(A)+n(B)-n(A \cap B) \\ & \quad=2025+2025-289 \\ & \quad=3761 \end{aligned} \end{aligned}$

Step 1: Breaking the sequence into two parts

The given sequence is $1+3+5^2+7+9^2+\ldots$ up to 40 terms. Let's group the terms as follows:

The terms are arranged like this: $1$, $3$, $5^2$, $7$, $9^2$, $11$, $13^2$, $15$, $17^2$, ... You can see that every third term is being squared, while others are just numbers. So, split the sequence into: all squared terms ($1^2, 5^2, 9^2, ...$) and all other numbers ($3, 7, 11, ...$).

Step 2: Counting the terms in each group

Out of 40 terms, half will be squared terms and half will be non-squared terms. So we have 20 squared terms and 20 ordinary numbers.

Step 3: Sums for each group

The squared numbers follow this pattern: $1^2, 5^2, 9^2, ...$, which can be written as $(4k-3)^2$ for $k$ from 1 to 20.

The other numbers are $3, 7, 11, ..., (4k-1)$ for $k$ from 1 to 20.

Step 4: Write these as formulas

Total sum = Sum of squares part $+ $ Sum of ordinary numbers part.

The sum of the squared numbers is $\sum_{k=1}^{20} (4k-3)^2$.

The sum of the other numbers (arithmetic series): $3 + 7 + 11 + ...$ for 20 terms.

The first term is 3, the last term is $3 + (20-1)\times4 = 3 + 76 = 79$. So, the sum is $\frac{20}{2} [3 + 79] = 10 \times 82 = 820$.

Step 5: Expanding and simplifying

Expand $(4k-3)^2$: $(4k-3)^2 = 16k^2 - 24k + 9$

So, sum up these for $k=1$ to $20$: $16 \sum k^2 - 24 \sum k + 9 \times 20$

We know formulas: $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$
$\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$

Plug in $n=20$: $\sum_{k=1}^{20} k^2 = \frac{20 \times 21 \times 41}{6}$
$\sum_{k=1}^{20} k = \frac{20 \times 21}{2}$

Now, sum up:

$= 16 \left(\frac{20 \times 21 \times 41}{6}\right) - 24\left(\frac{20 \times 21}{2}\right) + 180 + 820$

Step 6: Calculating

$16 \left(\frac{20 \times 21 \times 41}{6}\right) = 16 \times 2870 = 45920$
$24 \left(\frac{20 \times 21}{2}\right) = 24 \times 210 = 5040$
$9 \times 20 = 180$
Sum of the other numbers = $820$

So, total = $45920 - 5040 + 180 + 820$
$= 45920 - 5040 = 40880$
$40880 + 180 = 41060$
$41060 + 820 = 41880$

Final Answer

The sum of the 40 terms is $41880$.

The series given is:

$ 1 + \frac{1+3}{2!} + \frac{1+3+5}{3!} + \frac{1+3+5+7}{4!} + \ldots $

In general, the $ r $-th term of the series, denoted as $ T_r $, takes the form:

$ T_r = \frac{r^2}{r!} $

To simplify $ T_r $, we write it in terms of factorials:

$ T_r = \frac{r}{(r-1)!} = \frac{(r-1)+1}{(r-1)!} = \frac{1}{(r-2)!} + \frac{1}{(r-1)!} $

Thus, the series can be expressed as:

$ \sum\limits_{r=1}^{\infty} T_r = \sum\limits_{r=1}^{\infty} \left( \frac{1}{(r-2)!} + \frac{1}{(r-1)!} \right) $

This breaks into two parts:

$ = \sum\limits_{r=1}^{\infty} \frac{1}{(r-2)!} + \sum\limits_{r=1}^{\infty} \frac{1}{(r-1)!} $

Each part is a well-known series. Specifically:

$\sum\limits_{r=1}^{\infty} \frac{1}{(r-2)!} = e$, starting from the term where $ r-2 = 0$, which aligns with the expansion of $ e $.

$\sum\limits_{r=1}^{\infty} \frac{1}{(r-1)!} = e$, for the terms starting where $ r-1 = 0$.

Consequently, the sum of the entire series is:

$ e + e = 2e $

We start with the given sequence $a_1, a_2, a_3, \ldots$ of an increasing geometric progression (G.P.). Two key conditions are provided:

$a_3 \cdot a_5 = 729$

$a_2 + a_4 = \frac{111}{4}$

Calculating using given conditions:

Using the sequence terms:

$ a_3 = a \cdot r^2, \quad a_5 = a \cdot r^4, $

Then:

$ a_3 \cdot a_5 = (a \cdot r^2)(a \cdot r^4) = a^2 \cdot r^6 = 729 = 27^2 $

It follows:

$ a \cdot r^3 = 27 \quad \Rightarrow \quad a_4 = a \cdot r^3 = 27 \quad \text{(i)} $

Using the second condition:

$ a_2 + a_4 = \frac{111}{4} $

Substituting $a_4 = 27$:

$ a_2 = \frac{111}{4} - 27 $

$ a_2 = a \cdot r = \frac{3}{4} \quad \text{(ii)} $

Solving for $r$ and $a$:

From equation (i) and (ii), we derive $r^2$:

$ r^2 = \frac{4 \cdot 27}{3} = 4 \times 9 = 36 $

Since the G.P. is increasing, $r = 6$.

Solve for $a$:

Substituting $r = 6$ into $a \cdot r = \frac{3}{4}$:

$ a \cdot 6 = \frac{3}{4} \quad \Rightarrow \quad a = \frac{3}{24} = \frac{1}{8} $

Calculating the requested sum:

The task is to find:

$ 24(a_1 + a_2 + a_3) = 24(a + ar + ar^2) $

Substitute the values:

$ = 24 \times \frac{1}{8}(1 + 6 + 6^2) = 3 \times (1 + 6 + 36) = 3 \times 43 = 129 $

Therefore, the value is 129.

$\begin{aligned} &T_r=3 r^2-7 r+5 \text { using second order difference }\\ &\begin{aligned} & \sum_{r=1}^{20}\left(3 r^2-7 r+5\right)=3 \Sigma r^2-7 \Sigma r+5 \Sigma(1) \\ & =\frac{3(n)(n+1)(2 n+1)}{6}-\frac{7 n(n+1)}{2}-5 n, n=20 \\ & =7240 \end{aligned} \end{aligned}$

Let the number of terms be $2 n$

$\begin{aligned} & n d=6 \\ & (a+(2 n+1) d)-a=\frac{21}{2} \\ & \Rightarrow 2 n d-d=\frac{21}{2} \\ & \Rightarrow 12-\frac{21}{2}=d \end{aligned}$

$\begin{aligned} & \Rightarrow d=\frac{3}{2} \\ & \therefore n=4 \\ & \therefore \text { Total terms }=8 \end{aligned}$

Given:

$ \sum\limits_{k=1}^{12} a_{2k-1} = -\frac{72}{5} a_1 $

This means:

$ a_1 + a_3 + \cdots + a_{23} = -\frac{72}{5} a_1 $

Express each term in A.P.:

The general term of the A.P. is given by $ a_k = a + (k-1)d $. Thus, for the odd indices:

$ a_1 + (a + 2d) + (a + 4d) + \cdots + (a + 22d) $

Simplify the equation:

$ 12a + 2d(1 + 2 + \cdots + 11) = -\frac{72}{5} a $

Use the formula for sum of an arithmetic series (first 11 terms):

$ 1 + 2 + \cdots + 11 = \frac{11 \times 12}{2} = 66 $

So, the equation becomes:

$ 12a + 2d(66) = -\frac{72}{5} a $

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

Solve for the relationship between $a$ and $d$:

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

$ 132d = -\frac{132}{5} a $

So,

$ a = -5d \quad \text{(i)} $

Second condition:

$ \sum_{k=1}^{n} a_k = 0 \quad \Rightarrow \quad S_n = 0 $

$ \frac{n}{2} [2a + (n-1)d] = 0 $

$ 2a = -(n-1)d \quad \text{(ii)} $

Combine equation (i) and (ii):

Substitute $ a = -5d $ into equation (ii):

$ 2(-5d) = -(n-1)d $

$ -10d = -(n-1)d $

Therefore:

$ n-1 = 10 $

$ n = 11 $

Thus, $ n = 11 $.

$\begin{aligned} &\begin{aligned} & \mathrm{S}_3=3 \mathrm{a}+3 \mathrm{~d}=54 \\ & \Rightarrow \mathrm{a}+\mathrm{d}=18 \\ & \mathrm{~S}_{20}=10(2 \mathrm{a}+19 \mathrm{~d}) \\ & \Rightarrow 10(36+17 \mathrm{~d}) \\ & \Rightarrow 1600<10(36+17 \mathrm{~d})<1800 \\ & \Rightarrow 160<36+17 \mathrm{~d}<180 \\ & \Rightarrow 124<17 \mathrm{~d}<144 \\ & \Rightarrow 7 \frac{5}{17}<\mathrm{d}<8 \frac{8}{17} \end{aligned}\\ &\text { Common difference will be natural number }\\ &\begin{aligned} & \Rightarrow d=8 \Rightarrow a=10 \\ & \Rightarrow a_{11}=10+10 \times 8=90 \end{aligned} \end{aligned}$

$\begin{aligned} \mathrm{a}_{\mathrm{n}} & =\frac{\mathrm{n}^2+5 \mathrm{n}+6}{4} \\ \mathrm{~S}_{\mathrm{n}} & =\mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{a}_{\mathrm{k}}}=\sum_1^{\mathrm{n}} \frac{4}{\mathrm{k}^2+5 \mathrm{k}+6} \\ & =4 \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{(\mathrm{k}+2)(\mathrm{k}+3)} \\ & =4 \sum_{\mathrm{k}=1}^{\mathrm{n}} \frac{1}{\mathrm{k}+2}-\frac{1}{\mathrm{k}+3} \\ & =4\left(\frac{1}{3}-\frac{1}{4}\right)+4\left(\frac{1}{4}-\frac{1}{5}\right)+\ldots \ldots . . \end{aligned}$

$\begin{array}{l} 4\left(\frac{1}{\mathrm{n}+2}-\frac{1}{\mathrm{n}+3}\right) \\ =4\left(\frac{1}{3}-\frac{1}{\mathrm{n}+3}\right) \\ =\frac{4 \mathrm{n}}{3(\mathrm{n}+3)} \\ { }^{507} \mathrm{~S}_{2025}=\frac{(507)(4)(2025)}{3(2028)} \\ =675 \end{array}$

$\begin{aligned} & a_0=0, a_1=\frac{1}{2} \\ & 2 a_{n+2}=5 a_{n+1}-3 a_n \\ & 2 x^2-5 x+3=0 \Rightarrow x=1,3 / 2 \\ & \therefore a_n=A(1)^n+B\left(\frac{3}{2}\right)^n \\ &\left.\begin{array}{cc} \mathrm{n}=0 & 0=\mathrm{A}+\mathrm{B} \\ \mathrm{n}=1 & \frac{1}{2}=\mathrm{A}+\frac{3}{2} \mathrm{~B} \end{array}\right] \begin{aligned} & \mathrm{A}=-1 \\ & \mathrm{~B}=1 \end{aligned}\\ & \Rightarrow a_n=-1+\left(\frac{3}{2}\right)^n \\ & \sum_{k=1}^{100} a_k=\sum_{k=1}^{100}(-1)+\left(\frac{3}{2}\right)^k \end{aligned}$

$\begin{aligned} & =-100+\frac{\left(\frac{3}{2}\right)\left(\left(\frac{3}{2}\right)^{100}-1\right)}{\frac{3}{2}-1} \\ & =-100+3\left(\left(\frac{3}{2}\right)^{100}-1\right) \\ & =3 \cdot\left(\mathrm{a}_{100}\right)-100 \end{aligned}$

To solve this problem, we start by analyzing the terms of an arithmetic progression (A.P.) where:

$ T_m = \frac{1}{25} $

$ T_{25} = \frac{1}{20} $

$ 20 \sum\limits_{r=1}^{25} T_r = 13 $

The formula for the $ r^{\text{th}} $ term of an A.P. is:

$ T_r = a + (r-1)d $

Given:

$ T_m = a + (m-1)d = \frac{1}{25} \quad \text{(Equation 1)} $

$ T_{25} = a + 24d = \frac{1}{20} $

The sum of the first 25 terms ($ S_{25} $) is given by:

$ S_{25} = \frac{25}{2} \times (2a + 24d) $

Substituting into the equation for the sum:

$ 20 \times \frac{25}{2} \times (2a + 24d) = 13 $

This simplifies to:

$ a = \frac{1}{500} $

Substituting $ a = \frac{1}{500} $ into $ T_{25} = a + 24d = \frac{1}{20} $, we find:

$ \frac{1}{500} + 24d = \frac{1}{20} $

$ 24d = \frac{1}{20} - \frac{1}{500} $

$ d = \frac{1}{500} $

Using Equation 1 again:

$ \frac{1}{500} + \frac{m-1}{500} = \frac{1}{25} $

$ \frac{m}{500} = \frac{1}{25} $

$ m = 20 $

Now to find $ 5m \sum\limits_{r=m}^{2m} T_r $:

$ 5m = 5 \times 20 = 100 $

$ \sum\limits_{r=20}^{40} T_r = \frac{21}{2} \times (T_{20} + T_{40}) $

Since $ m = 20 $, the summation covers terms from $ T_{20} $ to $ T_{40} $, and:

$ 100 \times \sum\limits_{r=20}^{40} T_r = 126 $

Therefore, $ 5m \sum\limits_{r=m}^{2m} T_r = 126 $.

Let a & d are first term and common diff of an AP.

$\begin{aligned} & \mathrm{S}_{40}=\frac{40}{2}[2 \mathrm{a}+39 \mathrm{~d}]=1030 \quad\text{..... (1)}\\ & \mathrm{~S}_{12}=\frac{12}{2}[2 \mathrm{a}+11 \mathrm{~d}]=57 \quad\text{..... (2)} \end{aligned}$

by (1) & (2)

$\begin{aligned} & \mathrm{a}=-\frac{7}{2} \quad \mathrm{~d}=\frac{3}{2} \\ & \therefore \mathrm{~S}_{30}-\mathrm{S}_{10}=\frac{30}{2}[2 \mathrm{a}+29 \mathrm{~d}]-\frac{10}{2}[2 \mathrm{a}+9 \mathrm{~d}] \\ & =20 \mathrm{a}-390 \mathrm{~d} \\ & =515 \end{aligned}$

$\begin{aligned} & \text { Let } S=5+\frac{1}{7}(5+\alpha)+\frac{1}{7^2}(5+2 \alpha)+\ldots \\ & \frac{1}{7} S=\frac{1}{7}(5)+\frac{1}{7^2}(5+\alpha)+\ldots \infty \end{aligned}$

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

To find the value of $ S_{2025} $, calculate the partial sum

$ S_n = \sum\limits_{n=1}^{2025} \frac{1}{n(n+1)} = \sum\limits_{n=1}^{2025} \left( \frac{1}{n} - \frac{1}{n+1} \right) $

This series is telescopic, meaning most terms cancel out with each other:

$ = \left( \frac{1}{1} - \frac{1}{2} \right) + \left( \frac{1}{2} - \frac{1}{3} \right) + \cdots + \left( \frac{1}{2025} - \frac{1}{2026} \right) $

Simplifying the expression, we find:

$ = \frac{1}{1} - \frac{1}{2026} = 1 - \frac{1}{2026} = \frac{2025}{2026} $

Now, evaluate $\sqrt{2026 \cdot S_{2025}}$:

$ \sqrt{2026 \cdot \frac{2025}{2026}} = \sqrt{2025} = 45 $

For the arithmetic progression with the first term $-p$ and common difference $p$, the sum of the first six terms is given by:

$ \frac{6}{2} [-2p + (6-1)p] = 3(5p) = 15p $

Given that:

$ 15p = 45 $

We find:

$ p = 3 $

To determine the absolute difference between the $20^{\text{th}}$ and $15^{\text{th}}$ terms of the A.P., compute:

$ |A_{20} - A_{15}| = |(-p + 19p) - (-p + 14p)| $

Substituting the value of $p$:

$ = |18p - 13p| = |5p| = 5 \times 5 = 25 $

Thus, the absolute difference is $ \boxed{25} $.

The first term, $ a = 3 $

Common difference, $ d $

The formula for the sum of the first $ n $ terms of an A.P. is:

$ S_n = \frac{n}{2} [2a + (n-1)d] $

Given:

$ S_4 = \frac{1}{5}(S_8 - S_4) $

This implies:

$ 5S_4 = S_8 - S_4 \quad \Rightarrow \quad 6S_4 = S_8 $

Substituting the sum formulas:

$ 6 \cdot \frac{4}{2}[2 \times 3 + (4-1)d] = \frac{8}{2}[2 \times 3 + (8-1)d] $

Simplifying:

$ 6 \times 2 [6 + 3d] = 4 [6 + 7d] $

$ 12(6 + 3d) = 4(6 + 7d) $

$ 72 + 36d = 24 + 28d $

$ 36d - 28d = 24 - 72 $

$ 8d = -48 $

$ d = -6 $

Now, to find $ S_{20} $:

$ S_{20} = \frac{20}{2} [2 \times 3 + (20-1)(-6)] $

$ S_{20} = 10 [6 + 19 \times (-6)] $

$ S_{20} = 10 [6 - 114] $

$ S_{20} = 10 \times (-108) $

$ S_{20} = -1080 $

Thus, the sum of the first 20 terms is $-1080$.

$\begin{aligned} & a_1, a_2, a_3, \ldots \ldots, a_{2 k} \quad \rightarrow \mathrm{A.P.}\\ & \sum_{\mathrm{r}=1}^{\mathrm{k}} \mathrm{a}_{2 \mathrm{r}-1}=40, \sum_{\mathrm{r}=1}^{\mathrm{k}} \mathrm{a}_{2 \mathrm{r}}=55, \mathrm{a}_{2 \mathrm{k}}-\mathrm{a}_1=27 \\ & \frac{\mathrm{k}}{2}\left[2 \mathrm{a}_1+(\mathrm{k}-1) 2 \mathrm{~d}\right]=40, \frac{\mathrm{k}}{2}\left[2 \mathrm{a}_2+(\mathrm{k}-1) 2 \mathrm{~d}\right]=55, \\ & \mathrm{~d}=\frac{27}{2 \mathrm{k}-1} \\ & \mathrm{a}_1=\frac{40}{\mathrm{k}}-(\mathrm{k}-1) \mathrm{d}=\frac{55}{\mathrm{k}}-\mathrm{kd} \\ & \mathrm{~d}=\frac{15}{\mathrm{k}} \Rightarrow \frac{27}{2 \mathrm{k}-1}=\frac{15}{\mathrm{k}} \Rightarrow 9 \mathrm{k}=10 \mathrm{k}-5 \\ & \therefore \mathrm{k}=5 \end{aligned}$

First, let us denote the first term of the G.P. by $a_1 = A$ and the common ratio (which is $> 1$, since the G.P. is increasing) by $r$. Then the terms are:

$ a_1 = A,\quad a_2 = Ar,\quad a_3 = Ar^2,\quad a_4 = Ar^3,\quad a_5 = Ar^4,\quad a_6 = Ar^5,\;\dots $

We are given:

$a_1 \cdot a_5 = 28$, i.e.

$ A \cdot (A r^4) \;=\; A^2 r^4 \;=\; 28. \quad (1) $

$a_2 + a_4 = 29$, i.e.

$ Ar \;+\; A r^3 \;=\; A(r + r^3) \;=\; 29. \quad (2) $

We want to find $a_6 = A r^5$.

1. Solve for $r$

From $\text{(1)}$:

$ A^2 r^4 = 28 \quad\Longrightarrow\quad A^2 = \frac{28}{r^4}. $

From $\text{(2)}$:

$ A\,\bigl(r + r^3\bigr) = 29 \quad\Longrightarrow\quad A = \frac{29}{r + r^3}. $

Plug $A$ from $\text{(2)}$ into $\text{(1)}$. After some algebra (or by a systematic approach), one finds that

$ r^2 = 28 \quad\Longrightarrow\quad r = \sqrt{28} \;=\; 2\sqrt{7} $

(since $r>1$, we take the positive root).

2. Solve for $A$

From (2), using $r = 2\sqrt{7}$:

$ r + r^3 = 2\sqrt{7} + (2\sqrt{7})^3 = 2\sqrt{7} + 56\sqrt{7} = 58\sqrt{7}. $

Hence,

$ A = \frac{29}{\,58\sqrt{7}\,} = \frac{1}{2\sqrt{7}}. $

3. Find $a_6$

$ a_6 = A\,r^5 = \frac{1}{2\sqrt{7}}\;\times\;(2\sqrt{7})^5. $

Compute $(2\sqrt{7})^5$. First, $(2\sqrt{7})^2 = 28$; thus

$ (2\sqrt{7})^5 = (2\sqrt{7})^4 \cdot (2\sqrt{7}) = 784 \times (2\sqrt{7}) = 1568\sqrt{7}. $

Therefore,

$ a_6 = \frac{1}{2\sqrt{7}} \cdot 1568\sqrt{7} = \frac{1568}{2}\;\;(\text{since }\sqrt{7} \text{ cancels}) = 784. $

Answer: $a_6 = 784$.

Hence, the correct option is Option B.

Given,

$ \begin{aligned} & S_n=t_1+t_2+t_3+\ldots+t_n \\ & S_1=1^2 \Rightarrow t_1=1 \\ & S_2=9 \Rightarrow t_1+t_2=9 \Rightarrow t_2=8 \\ & S_3=36 \Rightarrow t_1+t_2+t_3=36 \\ \Rightarrow \quad & t_3=27 \\ \Rightarrow \quad & S_{10}=1^3+2^3+\ldots+10^3 \\ \Rightarrow \quad & k^2=\left(\frac{10 \times 11}{2}\right)^2 \\ \Rightarrow \quad & k=55 \end{aligned} $

$ \begin{aligned} K=-8 & +9+\frac{4}{3}-\frac{5}{4}+\frac{4}{9}+\frac{5}{16}+\ldots+\infty \\ = & 1+4\left(\frac{1}{3}+\frac{1}{3^2}+\ldots+\infty\right) \\ & -5\left(\frac{1}{4}-\frac{1}{4^2}+\ldots+\infty\right) \\ = & 1+\frac{4 \times \frac{1}{3}}{\frac{2}{3}}-5\left(\frac{\frac{1}{4}}{\frac{4}{4}}\right) \\ = & 1+2-1=2 \end{aligned} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { } \begin{aligned} 2^{n+4} & +12 \geq k(n+4) \\ k & \leq \frac{2^{n+4}+12}{n+4} \\ \text { Let } n+4 & =t \\ \Rightarrow \quad k & \leq \frac{2^t+12}{t}, t \geq 5 \\ t & =5, k \leq \frac{2^5+12}{5}=8.81 \\ t & =6, k \leq \frac{2^6+12}{6}=1267 \\ t & =7, k \leq \frac{2^7+12}{7}=20 \\ \because \quad t & =5 \Rightarrow k \text { will be minimum }=8.8 \\ \therefore \quad[k] & =8 \end{aligned} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { Given, }\\ &\begin{aligned} & \frac{1}{2 \cdot 7}+\frac{1}{7 \cdot 12}+\frac{1}{12 \cdot 17}+\ldots+10 \text { terms. } \\ & =\frac{1}{5}\left[\frac{5}{2 \cdot 7}+\frac{5}{7 \cdot 12}+\frac{5}{12 \cdot 17}+\ldots+10 \text { terms }\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{1}{5}\left[\frac{7-2}{2 \cdot 7}+\frac{12-7}{7 \cdot 12}+\frac{17-12}{12 \cdot 17}\right.+\ldots+10 \text { terms] } \\ & =\frac{1}{5}\left[\begin{array}{l} \frac{1}{2}-\frac{1}{7}+\frac{1}{7}-\frac{1}{12}+\left(\frac{1}{12}-\frac{1}{17}\right)+\ldots \\ +\left(\frac{1}{2+9 \times 5}-\frac{1}{7+9 \times 5}\right) \end{array}\right] \\ & =\frac{1}{5}\left(\frac{1}{2}-\frac{1}{52}\right)=\frac{1}{5}\left(\frac{25}{52}\right)=\frac{5}{52} \end{aligned} $

We have, $49^k+1$ is a factor of

$ \begin{aligned} & 48\left(49^{125}+49^{124}+\ldots+49^2+49+1\right) \\ & \text { Let } S=49^{125}+49^{124}+\ldots+49+1 \\ & \Rightarrow \quad S=\frac{49^{126}-1}{49-1}=\frac{49^{126}-1}{48} \Rightarrow 48 S=49^{126}-1 \end{aligned} $

Since, $49^k+1$ divides $49^{126}-1$

$\therefore k$ divides 126 .

$\Rightarrow \quad 2 k$ divides 126

$\Rightarrow \quad$ Greatest possible integer $k=63$

Given series

$ 1+(1+3)+(1+3+5)+(1+3+5+7)+\ldots $

to 10 terms.

The $n$th term of the series is the sum of first $n$ odd numbers.

$ \text { } \begin{aligned}So,\,\, T_n & =1+3+5+\ldots+(2 n-1)=n^2 \\ \therefore \quad T_1 & =1^2=1 \\ T_2 & =2^2=4 \\ T_3 & =3^2=9 \\ \vdots & \\ T_{10} & =10^2=100 \end{aligned} $

Thus, $S_{10}=\sum_{n=1}^{10} n^2$

We know that the sum of the first $k$ squares is

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

For $k=10$

$ \begin{aligned} S_{10} & =\frac{10(10+1)(2 \times 10+1)}{6} \\ & =\frac{10 \times 11 \times 21}{6}=\frac{2310}{6}=385 \\ \therefore S_{10} & =385 \end{aligned} $

As we know,

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

and $T_n=1+2+3+\ldots+n$

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

Hence, $S_n=T_n^2$

$ \begin{aligned} &\\ &\begin{aligned} & \text { } S_n=\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\frac{1}{7 \cdot 9}+\ldots \\ & S_{24}=\sum_{n=1}^{24} \frac{1}{(2 n+1)(2 n+3)} \\ & =\frac{1}{2} \sum_{n=1}^{24} \frac{1}{2 n+1}-\frac{1}{2 n+3} \\ & =\frac{1}{2}\left[\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)\right. \left.+\ldots+\left(\frac{1}{49}-\frac{1}{51}\right)\right] \\ & =\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}\left(\frac{16}{51}\right)=\frac{8}{51} . \end{aligned} \end{aligned} $

$1+\frac{4}{15}+\frac{4 \cdot 10}{15 \cdot 30}+\frac{4 \cdot 10 \cdot 16}{15 \cdot 30 \cdot 45}+\ldots+\infty$

We know that

$ \begin{aligned} (1+x)^n=1+n x & +\frac{n(n-1)}{2!} x^2 \\ & +\frac{n(n-1)(n-2)}{3!} x^3+\ldots \end{aligned} $

On comparing with given expansion, we get

$ n x=\frac{4}{15} \Rightarrow x=\frac{4}{15 n} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\frac{n(n-1)}{2!} x^2=\frac{4 \cdot 10}{15 \cdot 30}=\frac{40}{450}=\frac{4}{45}$

Using Eq. (i), $\frac{n(n-1)}{2!}\left(\frac{4}{15 n}\right)^2=\frac{4}{45}$

$ \begin{array}{lrl} \Rightarrow & \frac{n(n-1)}{2} \times \frac{4 \times 4}{15 n \times 15 n} & =\frac{4}{45} \\ \Rightarrow & \frac{n(n-1)}{2} \times \frac{16}{225 n^2} & =\frac{4}{45} \\ \Rightarrow & & n=\frac{-2}{3} \end{array} $

$ \begin{aligned} & \text { Also, } x=\frac{4}{15}\left(\frac{-3}{2}\right)=\frac{-2}{5} \\ & \begin{aligned} \therefore \quad(1+x)^n & =\left(1-\frac{2}{5}\right)^{-2 / 3} \\ & =\left(\frac{3}{5}\right)^{-2 / 3}=\left(\frac{5}{3}\right)^{2 / 3} \end{aligned} \end{aligned} $

Given, $t_n=\frac{1}{4}(n+2)(n+3), n \in N \,\,\,\,\,\,\,\,\,\,\,\,\,\ldots$ (i)

Also, $\frac{1}{t_1}+\frac{1}{t_2}+\ldots+\frac{1}{t_{2003}}=\frac{2003}{3009}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i), we get

$ \begin{aligned} \frac{1}{t_n} & =\frac{4}{(n+2)(n+3)} \\ & =\frac{4}{n+2}-\frac{4}{n+3} \end{aligned} $

(By partial fraction)

$ \begin{aligned} \Rightarrow \quad & \sum_{n=1}^{2003} \frac{1}{t_n}=\sum_{n=1}^{2003}\left(\frac{4}{n+2}-\frac{4}{n+3}\right) \\ & =4\left(\frac{1}{3}-\frac{1}{2006}\right)=4\left(\frac{2006-3}{3 \times 2006}\right) \\ & =4 \times \frac{2003}{3 \times 2006} \\ & =\frac{8012}{6018}=\frac{4006}{3009} \end{aligned} $

So, Assertion is false.

Also, by putting $n=2003$ in Reason then we observe that it is wrong.

The integers divisible by 5 from 1 to 100 are $5,10,15,20,25, \ldots, 100$. Which are in A.P. with first term (a) $=5$ and common difference $(d)=5$ ) last term $(l)=100$

So, total number of terms,

$ \begin{aligned} (n) & =\frac{l-a}{d}+1=\frac{100-5}{5}+1=\frac{95}{5}+1 \\ & =19+1=20 \\ \therefore S_n & =\frac{n}{2}(a+l) \\ & =10(5+100)=1050 \end{aligned} $

Similarly, the integers divisible by 13 from 1 to 100 are $13,26,39,52, \ldots, 91$ which are in AP with to $a=13, d=13$, $l=91$.

$ \begin{aligned} & \text { } \begin{array}{l}So,\,\, n=\frac{l-a}{d}+1=\frac{91-13}{13}+1=\frac{78}{13}+1 \\ =6+1=7 \end{array} \\ & \therefore \quad S_n=\frac{7}{2}(13+91)=7 \times 52=364 \end{aligned} $

Now, the integer divisible by both 5 and 13 from 1 to 100 is 65 .

Hence, the sum of integer from 1 to 100 that are divisible by 5 or 13 is $=1050+364-65=1414-65=1349$

Given, $x>\sqrt{3} \Rightarrow x^2>3 \Rightarrow 0<\frac{1}{x^2}<\frac{1}{3}$ Also, $\frac{x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{-1}{x^2+2}+\frac{2}{x^2+3} =(-1)\left(x^2+2\right)^{-1}+2\left(x^2+3\right)^{-1}$

$ \begin{aligned} &\begin{aligned} & =(-1)\left(x^2\right)^{-1}\left(1+\frac{2}{x^2}\right)^{-1}+2\left(x^2\right)^{-1}\left(1+\frac{3}{x^2}\right)^{-1} \\ & =\frac{-1}{x^2}\left(1-\frac{2}{x^2}+\frac{4}{x^4}-\frac{8}{x^6}+\frac{16}{x^8}-\ldots \infty\right) \\ & \quad+\frac{2}{x^2}\left(1-\frac{3}{x^2}+\frac{9}{x^4}-\frac{27}{x^6}+\ldots \infty\right) \end{aligned}\\ &\text { So, the coefficient of }\\ &\begin{aligned} x^{-8} & =(-1)(-8)+(2)(-27) \\ & =8-54=-46 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { From the given, we have }\\ &\begin{aligned} \sum_{k=1}^n k & (k+1)(k+2 \ldots(k+r-1) \\ & =\sum_{K=1}^n \frac{(k+r-1)!}{(k-1)!} \\ & =r!\sum_{k=1}^n{ }^{k+r-1} C_r \\ & =r!{ }^{n+r} C_{r+1} \\ & =\frac{n(n+1)(n+2) \ldots(n+r)}{(r+1)} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } \mathrm{AM} \geq \mathrm{GM}\\ &\begin{array}{lc} \therefore & \frac{1+3+3^2+3^3+3^{n-1}}{n} \\ & \geq\left(1 \cdot 3 \cdot 3^2 \cdot 3^3 \cdot \ldots \cdot 3^{n-1}\right)^{\frac{1}{n}} \\ \Rightarrow & \frac{3^n-1}{2 n} \geq\left(3^{1+2+3+\ldots n-1}\right)^{\frac{1}{n}} \\ \Rightarrow & \frac{3^n-1}{2} \geq n\left(3^n \frac{(n-1)}{2}\right)^{\frac{1}{n}} \\ \Rightarrow & \frac{3^n-1}{2} \geq n 3^{\frac{n-1}{2}} \end{array} \end{aligned} $