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.

$\begin{array}{ll} \sum_{n=0}^{\infty} a r^n=57 & \Rightarrow \frac{a}{1-r}=57 \quad \text{.... (i)}\\ \sum_{n=0}^{\infty} a^3 r^{3 n}=9747 & \Rightarrow \frac{a^3}{1-r^3}=9747 \quad \text{.... (ii)} \end{array}$

$\begin{aligned} & \frac{\left(1-r^3\right)}{(1-r)^3}=\frac{(57)^3}{9747}=19 \\ & \Rightarrow \quad \frac{(1-r)\left(1+r+r^2\right)}{(1-r)^3}=19 \\ & \Rightarrow \quad 18 r^2-39 r+18=0 \\ & \Rightarrow \quad r=\frac{2}{3}, \frac{3}{2} \text { (rejected) } \\ & \therefore \quad a=19 \\ & \quad a+18 r \\ & \quad=19+12=31 \end{aligned}$

$\frac{1}{1 \cdot(1+d)}+\frac{1}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}=5$

Multiply and divide by $d$

$\begin{aligned} & \frac{1}{d}\left[\frac{d}{1 \times(1+d)}+\frac{d}{(1+d)(1+2 d)}+\ldots+\frac{1}{(1+9 d)(1+10 d)}\right]=5 \\ & \frac{1}{d}\left[\left(1-\frac{1}{1+d}\right)+\left(\frac{1}{1+d}-\frac{1}{1+2 d}\right)+\ldots+\left(\frac{1}{1+9 d}-\frac{1}{1+10 d}\right)\right]=5 \\ & \frac{1}{d}\left[1-\frac{1}{1+10 d}\right]=5 \\ & \frac{1}{d}\left[\frac{1+10 d-1}{1+10 d}\right]=5 \end{aligned}$

$\begin{aligned} & \frac{10}{1+10 d}=5 \\ & 1+10 d=2 \\ & d=\frac{1}{10} \\ & 50 d=50 \times \frac{1}{10}=5 \end{aligned}$

Let's denote the first term of the geometric progression by $a$ and the common ratio by $r$. The terms of the geometric progression can be written as follows:

First term: $a$

Second term: $ar$

Third term: $(ar^2)$

Fourth term: $(ar^3)$

Fifth term: $(ar^4)$

Sixth term: $(ar^5)$

Eighth term: $(ar^7)$

We are given two key pieces of information:

1. The sum of the second and sixth terms is $\frac{70}{3}$:

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

2. The product of the third and fifth terms is 49:

$(ar^2) \cdot (ar^4) = 49$

$a^2 r^6 = 49$

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

$a = \frac{7}{r^3}$

Substituting $a = \frac{7}{r^3}$ into the first equation:

$\frac{7}{r^3} \cdot r + \frac{7}{r^3} \cdot r^5 = \frac{70}{3}$

$\frac{7r}{r^3} + \frac{7r^5}{r^3} = \frac{70}{3}$

$\frac{7}{r^2} + \frac{7r^2}{1} = \frac{70}{3}$

Let $x = r^2$. Then:

$\frac{7}{x} + 7x = \frac{70}{3}$

Multiply through by 3x to clear the denominator:

$21 + 21x^2 = 70x$

Rearrange into a standard quadratic equation:

$21x^2 - 70x + 21 = 0$

Divide by 7 to simplify:

$3x^2 - 10x + 3 = 0$

Solve this quadratic equation using the quadratic formula:

$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where $a = 3$, $b = -10$, and $c = 3$. Thus:

$x = \frac{10 \pm \sqrt{100 - 36}}{6}$

$x = \frac{10 \pm \sqrt{64}}{6}$

$x = \frac{10 \pm 8}{6}$

$x = 3$ or $x = \frac{1}{3}$

Since $x = r^2$ and $r$ is positive, we get $r = \sqrt{3}$ or $r = \frac{1}{\sqrt{3}}$. We need to choose the value that results in positive, increasing terms:

If $r = \sqrt{3}$:

$a = \frac{7}{r^3} = \frac{7}{(\sqrt{3})^3} = \frac{7}{3\sqrt{3}} = \frac{7}{3} \cdot \frac{1}{\sqrt{3}} = \frac{7\sqrt{3}}{9}$

Now we can determine the sum of the 4th, 6th, and 8th terms:

The 4th term is: $ar^3 = \frac{7\sqrt{3}}{9} \cdot 3\sqrt{3} = 7$

The 6th term is: $ar^5 = \frac{7\sqrt{3}}{9} \cdot 9\sqrt{3} = 21$

The 8th term is: $ar^7 = \frac{7\sqrt{3}}{9} \cdot 27(\sqrt{3}) = 49$

Adding these together:

$(4th + 6th + 8th terms) = 7 + 21 + 63 = 91$

Therefore, the sum of the 4th, 6th, and 8th terms is 91.

Correct answer:

Option C: 91

$\triangle A B C$ is an equilateral triangle having side $=a$ unit

Now, perimeter $=$ sum of all sides $=3 a$

Area $=\frac{\sqrt{3}}{4} a^2$

Now, $\ln \triangle D E F, D E=\frac{a}{2}=E F=D F$

$\begin{aligned} & \text { Perimeter }=3 \times \frac{a}{2}=\frac{3 a}{2} \\ & \text { Area }=\frac{\sqrt{3}}{4} \times\left(\frac{a}{2}\right)^2=\frac{\sqrt{3} a^2}{16} \end{aligned}$

Now, $P=3 a+\frac{3 a}{2}+\frac{3 a}{4}+\cdots$

$Q=\frac{\sqrt{3}}{4} a^2+\frac{\sqrt{3}}{16} a^2+\frac{\sqrt{3}}{64} a^2+\cdots$

$P=\frac{3 a}{1-\frac{1}{2}}=3 a \times 2 \Rightarrow P=6 a \quad \text{.... (i)}$

$Q=\frac{\frac{\sqrt{3}}{4} a^2}{1-\frac{1}{4}}=\frac{4}{3} \times \frac{\sqrt{3}}{4} a^2 \quad Q=\frac{\sqrt{3}}{3} a^2 \quad \text{.... (ii)}$

From equation (i) & (ii)

$\begin{aligned} & P=6 a \\ & Q=\frac{\sqrt{3}}{3} a^2 \\ & Q=\frac{\sqrt{3}}{3} \times \frac{P^2}{36} \\ & P^2=36 \sqrt{3} Q \end{aligned}$

To determine the value of $\mathrm{m}$, we need to formulate the problem using some basic concepts of arithmetic progression and work. Let's first understand the nature of the problem:

Initially, there are $\mathrm{m}$ computers, and it is estimated that with these $\mathrm{m}$ computers, the assignment can be completed in 17 days.

However, due to the crash of 4 computers every day starting from the second day onward, the total time taken extends by 8 days, making it 25 days in total.

To begin with, let's define the total work (W) in terms of the number of computers and days:

The total work (W) is given by:

The amount of work completed each day with $\mathrm{m}$ computers for 17 days:

$W = 17m$

When computers crash, the number of working computers each day forms an arithmetic sequence. On the first day, there are $\mathrm{m}$ computers. On the second day, there are $\mathrm{m} - 4$ computers, on the third day, there are $\mathrm{m} - 8$ computers, and so on. We need to sum this series until 25 days are completed.

This can be formulated as:

Total work done over 25 days with decrement in the number of computers:

$W = m + (m - 4) + (m - 8) + \ldots + \left[m - 4 \times (n - 1)\right]$

where $n$ is the number of days. Here, $n = 25$.

Notice that we form an arithmetic series where the first term (a) is $\mathrm{m}$ and the common difference (d) is -4. The sum of the first n terms of an arithmetic series is:

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

Plugging in the values:

$S_{25} = \frac{25}{2} \left[ 2m + (25 - 1)(-4) \right]$

$S_{25} = \frac{25}{2} \left[ 2m - 96 \right]$

$S_{25} = \frac{25}{2} \left[ 2m - 96 \right] = 25(m - 48)$

This work should be equivalent to the work calculated earlier, so:

$17m = 25(m - 48)$

Solving for $\mathrm{m}$:

$17m = 25m - 1200$

$8m = 1200$

$m = 150$

Thus, the value of $\mathrm{m}$ is equal to:

Option C: 150

To determine the least value of $\mathrm{K}$ for which the terms $4^{1+x} + 4^{1-x}, \frac{\mathrm{K}}{2}, 16^x + 16^{-x}$ form an arithmetic progression (A.P.), we need to establish the relationship among these terms in an A.P.

For three numbers to be in an arithmetic progression, the middle term must be the average of the other two terms. Therefore, we can write:

$ \frac{4^{1+x} + 4^{1-x} + 16^x + 16^{-x}}{2} = \frac{K}{2} $

First, simplify each term individually:

1. Consider $4^{1+x} + 4^{1-x}$:

$4^{1+x} = 4 \cdot 4^x = 4 \cdot (2^2)^x = 4 \cdot 2^{2x} = 4 \cdot 2^{2x}$

and

$4^{1-x} = 4 \cdot 4^{-x} = 4 \cdot (2^2)^{-x} = 4 \cdot 2^{-2x} = 4 \cdot 2^{-2x}$

Thus,

$4^{1+x} + 4^{1-x} = 4 \cdot 2^{2x} + 4 \cdot 2^{-2x} = 4(2^{2x} + 2^{-2x})$

2. Consider $16^x + 16^{-x}$:

$16^x = (2^4)^x = 2^{4x}$

and

$16^{-x} = (2^4)^{-x} = 2^{-4x}$

Thus,

$16^x + 16^{-x} = 2^{4x} + 2^{-4x}$

3. Combine the terms and set up the equation:

$ \frac{4(2^{2x} + 2^{-2x}) + 2^{4x} + 2^{-4x}}{2} = \frac{K}{2} $

Multiply both sides by 2:

$ 4(2^{2x} + 2^{-2x}) + 2^{4x} + 2^{-4x} = K $

To find the least value of $\mathrm{K}$, let's assume $x = 0$ (since $x$ can range over non-negative values):

For $x = 0$:

$4(2^{2 \cdot 0} + 2^{-2 \cdot 0}) + 2^{4 \cdot 0} + 2^{-4 \cdot 0}$

This simplifies to:

$4(2^0 + 2^0) + 2^0 + 2^0$

$ = 4(1 + 1) + 1 + 1 $

$ = 4 \cdot 2 + 1 + 1 $

$ = 8 + 1 + 1 $

$ = 10 $

Therefore, the least value of $\mathrm{K}$ that ensures the values form an arithmetic progression is $ 10 $. Hence, the correct option is:

Option A: 10