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} $

1. 1. A series sum is defined : $\boldsymbol{\alpha}=3+4+8+9+13+14+\ldots . . \boldsymbol{t}_{40}$

2. 2. $(\tan \boldsymbol{\beta})^{\frac{\alpha}{1020}}$ is a root of the quadratic equation $\boldsymbol{x}^2+\boldsymbol{x}-2=0$ where $\boldsymbol{\beta} \in\left(0, \frac{\boldsymbol{\pi}}{2}\right)$

3. 3. To evaluate the expression $\sin ^2 \beta+3 \cos ^2 \beta$

$\therefore $ $\boldsymbol{\alpha}=3+4+8+9+13+14+\ldots . . \boldsymbol{t}_{40}$

$=7+17+27+\ldots . .+\boldsymbol{T}_{20}$

$=\frac{20}{2}\{(2) \cdot(7)+(20-1) \cdot(10)\}$

$=(10) \cdot(14+190)=2,040 \ldots \ldots(\boldsymbol{i})$

Equation (i) gives the value of $\alpha=2,040$ :

$ (\tan \beta)^{\frac{\alpha}{1020}}=(\tan \beta)^{\frac{2040}{1020}}=\tan ^2 \beta \ldots \ldots(i i) $

Now let us solve the quadratic equation $x^2+x-2=0$ :

$x^2+x-2=0 \Rightarrow x^2+2 x-x-2=0 \Rightarrow(x+2)(x-1)=0$

$ \Rightarrow \begin{array}{|l|l|} \hline p & q \\ \hline-2 & 1 \\ \hline \end{array} $

where $\boldsymbol{p}$ and $\boldsymbol{q}$ are the roots

We have derived in equation $(i i)(\tan \beta)^{\frac{\alpha}{1020}}=\tan ^2 \beta$ which is also a root of the equation $x^2+x-2=0$ :

From the pair of roots we get $\tan ^2 \beta$ =1$ \Rightarrow \tan \beta= \pm 1 $ As $ \tan ^2 \beta \geqslant 0$ always

$\boldsymbol{\beta} \in\left(0, \frac{\pi}{2}\right) \Rightarrow \tan \boldsymbol{\beta}=1 \Rightarrow \boldsymbol{\beta}=\frac{\pi}{4}$

$\Rightarrow \sin ^2 \beta+3 \cos ^2 \beta=\sin ^2 \frac{\pi}{4}+3 \cos ^2 \frac{\pi}{4}=\left(\frac{1}{2}\right)+\left(\frac{3}{2}\right)=2$

For the quadratic equation

$ \left(n^2-2n+2\right)x^2-3x+\left(n^2-2n+2\right)^2=0, $

let us first compare it with the standard quadratic form

$ ax^2+bx+c=0. $

So,

$ a=n^2-2n+2,\qquad b=-3,\qquad c=\left(n^2-2n+2\right)^2. $

Now rewrite $a$ in a simpler form:

$ n^2-2n+2=(n-1)^2+1. $

Hence,

$ a=(n-1)^2+1. $

Since $(n-1)^2 \ge 0$, we get

$ a \ge 1. $

Now, for a quadratic equation, by Vieta’s formulas:

• Sum of roots $= -\dfrac{b}{a}$

• Product of roots $= \dfrac{c}{a}$

So here,

$ \text{Sum of roots}=\frac{3}{a} $

and

$ \text{Product of roots}=\frac{a^2}{a}=a. $

Therefore,

$ \text{Product of roots}=(n-1)^2+1. $

So the minimum value of the product is clearly

$ \alpha=1, $

attained when $n=1$.

Now the sum of roots is

$ \frac{3}{(n-1)^2+1}. $

Since the denominator is minimum when $(n-1)^2=0$, the maximum value of the sum is

$ \beta=\frac{3}{1}=3. $

So we have

$ \alpha=1,\qquad \beta=3. $

Now the G.P. has

• first term $a=\alpha=1$

• common ratio $r=\dfrac{\alpha}{\beta}=\dfrac{1}{3}$

We need the sum of first six terms:

$ S_6=a\cdot \frac{1-r^6}{1-r}. $

Substitute the values:

$ S_6=1\cdot \frac{1-\left(\frac13\right)^6}{1-\frac13}. $

Now,

$ \left(\frac13\right)^6=\frac{1}{729} $

and

$ 1-\frac13=\frac23. $

Hence,

$ S_6=\frac{1-\frac{1}{729}}{\frac23} =\frac{\frac{728}{729}}{\frac23} =\frac{728}{729}\cdot \frac32. $

So,

$ S_6=\frac{1092}{729} =\frac{364}{243}. $

Therefore, the required sum is

$ \boxed{\frac{364}{243}}. $

So the correct option is

$ \boxed{\text{Option C}}. $

$\begin{aligned} & \mathrm{T}_{\mathrm{n}}=\frac{1}{\mathrm{n}}\left(1^2+2^2+\ldots+\mathrm{n}^2\right) \\ & \mathrm{T}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6 \mathrm{n}} \\ & \mathrm{T}_{\mathrm{n}}=\frac{1}{6}\left[2 \mathrm{n}^2+3 \mathrm{n}+1\right] \\ & \mathrm{S}_{\mathrm{n}}=\Sigma \mathrm{T}_{\mathrm{n}}=\frac{1}{6}\left[2 \Sigma \mathrm{n}^2+3 \Sigma \mathrm{n}+\Sigma 1\right] \\ & \Sigma \mathrm{T}_{\mathrm{n}}=\frac{1}{6}\left[2 \cdot \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{3 \mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n}\right] \\ & \mathrm{S}_{10}=\frac{1}{6}\left[\frac{2 \times 10 \times 11 \times 21}{6}+\frac{3}{2} \times 10 \times 11+10\right] \\ & =\frac{10}{6}\left[77+\frac{33}{2}+1\right]=\frac{315}{2}\end{aligned}$

We need to find

$ 1^3-2^3+3^3-4^3+\cdots+15^3 $

This is an alternating sum of cubes.

Let us group the terms in pairs:

$ (1^3-2^3)+(3^3-4^3)+(5^3-6^3)+\cdots+(13^3-14^3)+15^3 $

Now calculate each pair:

$ 1^3-2^3=1-8=-7 $

$ 3^3-4^3=27-64=-37 $

$ 5^3-6^3=125-216=-91 $

$ 7^3-8^3=343-512=-169 $

$ 9^3-10^3=729-1000=-271 $

$ 11^3-12^3=1331-1728=-397 $

$ 13^3-14^3=2197-2744=-547 $

Also,

$ 15^3=3375 $

Now add all these:

$ -7-37-91-169-271-397-547+3375 $

First add the negative terms:

$ 7+37+91+169+271+397+547=1519 $

So,

$ 3375-1519=1856 $

Hence, the value is

$ \boxed{1856} $

So, the correct option is B.

Let the first term of the A.P. be $a$ and common difference be $d$.

Then, from the question:

• first term of the G.P. $= d$

• common ratio of the G.P. $= a$

Now use the given conditions one by one.

1. Sum of first 10 terms of the A.P.

We know,

$ S_{10}=\frac{10}{2}\left[2a+(10-1)d\right] $

So,

$ 160=5(2a+9d) $

$ 2a+9d=32 \qquad ...(1) $

2. Sum of first 2 terms of the G.P.

The G.P. has first term $d$ and common ratio $a$.

So its first two terms are:

$ d,\ da $

Their sum is $8$:

$ d+da=8 $

$ d(1+a)=8 \qquad ...(2) $

3. Express $d$ from equation (1)

From

$ 2a+9d=32 $

we get

$ 9d=32-2a $

$ d=\frac{32-2a}{9} $

Substitute this in equation (2):

$ \frac{32-2a}{9}(1+a)=8 $

Multiply by $9$:

$ (32-2a)(1+a)=72 $

Expand:

$ 32+32a-2a-2a^2=72 $

$ 32+30a-2a^2=72 $

$ -2a^2+30a-40=0 $

Divide by $-2$:

$ a^2-15a+20=0 $

4. Find values of $a$

Using quadratic formula:

$ a=\frac{15\pm\sqrt{225-80}}{2} $

$ a=\frac{15\pm\sqrt{145}}{2} $

Now we need the possible values of the first term of the G.P., which is $d$.

5. Find corresponding values of $d$

From equation (1):

$ d=\frac{32-2a}{9} $

For the two values of $a$, the two values of $d$ are obtained. But the question asks for the sum of all possible values of the first term of the G.P., i.e. sum of both possible values of $d$.

Let the two roots of the quadratic in $a$ be $a_1$ and $a_2$.

Then corresponding values of $d$ are

$ d_1=\frac{32-2a_1}{9}, \qquad d_2=\frac{32-2a_2}{9} $

So,

$ d_1+d_2=\frac{32-2a_1}{9}+\frac{32-2a_2}{9} $

$ d_1+d_2=\frac{64-2(a_1+a_2)}{9} $

For the quadratic

$ a^2-15a+20=0 $

sum of roots is

$ a_1+a_2=15 $

Hence,

$ d_1+d_2=\frac{64-2(15)}{9} $

$ d_1+d_2=\frac{64-30}{9}=\frac{34}{9} $

Therefore, the required sum is

$ \boxed{\frac{34}{9}} $

So the correct option is A.

Let the four roots $\alpha, \beta, \gamma, \delta$, which are in G.P., be represented as $a, ar, ar^2$, and $ar^3$ respectively. Here, $a$ is the first term and $r$ is the common ratio.

For the first quadratic equation $x^2-x+\mathrm{p}=0$, the roots are $\alpha$ and $\beta$.

Using the sum and product of the roots, we get:

1. Sum of roots: $\alpha + \beta = -\frac{\text{Coefficient of } x}{\text{Coefficient of } x^2}$

$a + ar = 1 \implies a(1+r) = 1 \quad \text{--- (Equation 1)}$

2. Product of roots: $\alpha\beta = \frac{\text{Constant term}}{\text{Coefficient of } x^2}$

$a(ar) = p \implies a^2r = p \quad \text{--- (Equation 2)}$

For the second quadratic equation $x^2-4x+\mathrm{q}=0$, the roots are $\gamma$ and $\delta$.

Similarly:

3. Sum of roots: $\gamma + \delta = -\frac{-4}{1}$

$ar^2 + ar^3 = 4 \implies ar^2(1+r) = 4 \quad \text{--- (Equation 3)}$

4. Product of roots: $\gamma\delta = \frac{q}{1}$

$(ar^2)(ar^3) = q \implies a^2r^5 = q \quad \text{--- (Equation 4)}$

Now, we can solve for $a$ and $r$ by dividing Equation 3 by Equation 1:

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

$r^2 = 4$

$r = 2 \quad \text{or} \quad r = -2$

Let us check both cases using the condition given in the problem that $p, q \in \mathbf{Z}$ (they are integers).

Case 1: When $r = 2$

Substitute $r = 2$ into Equation 1:

$a(1 + 2) = 1 \implies 3a = 1 \implies a = \frac{1}{3}$

Now, finding the value of $p$ using Equation 2:

$p = a^2r = \left(\frac{1}{3}\right)^2(2) = \frac{2}{9}$

Since $p$ must be an integer ($p \in \mathbf{Z}$), $\frac{2}{9}$ is not valid. Thus, we reject $r = 2$.

Case 2: When $r = -2$

Substitute $r = -2$ into Equation 1:

$a(1 - 2) = 1 \implies -a = 1 \implies a = -1$

Now, finding the value of $p$ using Equation 2:

$p = a^2r = (-1)^2(-2) = 1 \times (-2) = -2$

Since $-2$ is an integer, this is a valid value for $p$.

Next, we find the value of $q$ using Equation 4:

$q = a^2r^5$

$q = (-1)^2(-2)^5$

$q = 1 \times (-32) = -32$

Both $p$ and $q$ are integers, which satisfies the given condition.

Finally, we need to calculate the value of $|p + q|$:

$|p + q| = |-2 + (-32)|$

$|p + q| = |-34| = 34$

Correct Answer:

Option C

The given series is

$ \frac{1}{1+4\cdot 1^4}+\frac{2}{1+4\cdot 2^4}+\frac{3}{1+4\cdot 3^4}+\cdots $

So the general term is

$ \frac{r}{1+4r^4} $

We need the sum of the first $10$ terms:

$ S_{10}=\sum_{r=1}^{10}\frac{r}{1+4r^4} $

Now factor the denominator:

$ 1+4r^4=(2r^2-2r+1)(2r^2+2r+1) $

So,

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

We now try to split it:

$ \frac{r}{(2r^2-2r+1)(2r^2+2r+1)} =\frac{1}{4}\left(\frac{1}{2r^2-2r+1}-\frac{1}{2r^2+2r+1}\right) $

Check:

$ \frac{1}{4}\left(\frac{1}{2r^2-2r+1}-\frac{1}{2r^2+2r+1}\right) $

Taking LCM,

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

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

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

Correct.

Also note that

$ 2r^2+2r+1=2(r+1)^2-2(r+1)+1 $

So if we define

$ a_r=2r^2-2r+1 $

then

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

Hence,

$ \frac{r}{1+4r^4} =\frac{1}{4}\left(\frac{1}{a_r}-\frac{1}{a_{r+1}}\right) $

Therefore,

$ S_{10}=\frac{1}{4}\sum_{r=1}^{10}\left(\frac{1}{a_r}-\frac{1}{a_{r+1}}\right) $

This is a telescoping series:

$ S_{10}=\frac{1}{4}\left(\frac{1}{a_1}-\frac{1}{a_{11}}\right) $

Now compute:

$ a_1=2(1)^2-2(1)+1=1 $

and

$ a_{11}=2(11)^2-2(11)+1=242-22+1=221 $

So,

$ S_{10}=\frac{1}{4}\left(1-\frac{1}{221}\right) $

$ =\frac{1}{4}\cdot \frac{220}{221} $

$ =\frac{55}{221} $

Thus,

$ m=55,\quad n=221 $

So,

$ m+n=55+221=276 $

Hence, the correct answer is:

$ \boxed{276} $

So, Option C is correct.

Since there are $39$ arithmetic means between $59$ and $159$, the full arithmetic progression has

$ 59, A_1, A_2, A_3, \ldots, A_{39}, 159 $

So total number of terms is

$ 39+2=41 $

Hence, the number of intervals is

$ 41-1=40 $

Therefore, the common difference is

$ d=\frac{159-59}{40}=\frac{100}{40}=2.5 $

Now the terms are:

$ A_1=59+2.5=61.5 $

In general,

$ A_n=59+n(2.5) $

Now find the required terms:

$ A_{25}=59+25(2.5)=59+62.5=121.5 $

$ A_{28}=59+28(2.5)=59+70=129 $

$ A_{31}=59+31(2.5)=59+77.5=136.5 $

$ A_{36}=59+36(2.5)=59+90=149 $

Now their mean is

$ \frac{A_{25}+A_{28}+A_{31}+A_{36}}{4} = \frac{121.5+129+136.5+149}{4} $

$ = \frac{536}{4}=134 $

So the correct answer is:

$ \boxed{134} $

Hence, Option D is correct.

Given,

$ S_n = 3n^2 + 5n $

where $S_n$ is the sum of the first $n$ terms of the A.P.

To get the $n$th term, we use

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

So,

$ S_{n-1} = 3(n-1)^2 + 5(n-1) $

$ = 3(n^2 - 2n + 1) + 5n - 5 $

$ = 3n^2 - 6n + 3 + 5n - 5 $

$ = 3n^2 - n - 2 $

Now,

$ a_n = (3n^2 + 5n) - (3n^2 - n - 2) $

$ a_n = 6n + 2 $

So the terms of the A.P. are:

$ 8,\ 14,\ 20,\dots $

Now we need the sum of squares of the first $10$ terms:

$ \sum_{n=1}^{10} (6n+2)^2 $

Expand:

$ (6n+2)^2 = 36n^2 + 24n + 4 $

So,

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

Now use the standard sums:

$ \sum_{n=1}^{10} n^2 = \frac{10 \cdot 11 \cdot 21}{6} = 385 $

$ \sum_{n=1}^{10} n = \frac{10 \cdot 11}{2} = 55 $

$ \sum_{n=1}^{10} 1 = 10 $

Substitute:

$ 36(385) + 24(55) + 4(10) $

$ = 13860 + 1320 + 40 $

$ = 15220 $

So the correct answer is

$ \boxed{15220} $

Option C

$ S=\sum_{n=1}^{10}\frac{528}{n(n+1)(n+2)} $

We will simplify the general term using partial fractions.

Step 1: Use the standard identity

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

So,

$ \frac{528}{n(n+1)(n+2)} =528\cdot \frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right) $

$ =264\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right) $

Hence,

$ S=264\sum_{n=1}^{10}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right) $

Step 2: Write a few terms

$ S=264\Bigg[\left(\frac{1}{1\cdot 2}-\frac{1}{2\cdot 3}\right) +\left(\frac{1}{2\cdot 3}-\frac{1}{3\cdot 4}\right) +\left(\frac{1}{3\cdot 4}-\frac{1}{4\cdot 5}\right) +\cdots +\left(\frac{1}{10\cdot 11}-\frac{1}{11\cdot 12}\right)\Bigg] $

Now the middle terms cancel out telescopically.

So we get

$ S=264\left(\frac{1}{1\cdot 2}-\frac{1}{11\cdot 12}\right) $

$ S=264\left(\frac{1}{2}-\frac{1}{132}\right) $

Take LCM:

$ \frac{1}{2}=\frac{66}{132} $

So,

$ \frac{1}{2}-\frac{1}{132}=\frac{66-1}{132}=\frac{65}{132} $

Therefore,

$ S=264\cdot \frac{65}{132} $

$ S=2\cdot 65=130 $

$ \boxed{130} $

So, the correct option is Option B.

To solve this problem, we will use the standard formulas for an Arithmetic Progression (A.P.) as taught in the NCERT syllabus. Let us break down the information given in the question step by step.

Given:

1. The number of terms in the A.P. is $n = 30$.

2. The first term is $a = \frac{10}{3}$.

3. All the terms in the A.P. are non-negative.

4. The sum of the A.P. is equal to the cube of its last term.

Let the common difference of the A.P. be $d$ and the last term (which is the 30th term) be $l$.

The formula for the $n$-th term of an A.P. is:

$l = a + (n - 1)d$

$l = \frac{10}{3} + (30 - 1)d$

$l = \frac{10}{3} + 29d \quad \text{--- (Equation 1)}$

The formula for the sum of $n$ terms of an A.P. when the first and last terms are known is:

$S_n = \frac{n}{2}(a + l)$

For $n = 30$, we substitute our known values:

$S_{30} = \frac{30}{2} \left( \frac{10}{3} + l \right)$

$S_{30} = 15 \left( \frac{10}{3} + l \right)$

$S_{30} = 50 + 15l$

According to the problem, the sum of this A.P. is equal to the cube of its last term. So, we can write:

$S_{30} = l^3$

Now, substitute the value of $S_{30}$ into this equation:

$50 + 15l = l^3$

Rearranging the terms to form a standard cubic equation, we get:

$l^3 - 15l - 50 = 0$

To find the roots of this cubic equation, we can use the trial and error method for integer values.

Let us check $l = 5$:

$(5)^3 - 15(5) - 50 = 125 - 75 - 50 = 0$

Since $l = 5$ satisfies the equation, $(l - 5)$ is a factor.

Now, we can factorise the cubic polynomial completely:

$l^3 - 5l^2 + 5l^2 - 25l + 10l - 50 = 0$

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

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

This gives us two parts:

1. $l - 5 = 0 \implies l = 5$

2. $l^2 + 5l + 10 = 0$

For the quadratic equation $l^2 + 5l + 10 = 0$, the discriminant is:

$D = (5)^2 - 4(1)(10) = 25 - 40 = -15$

Since the discriminant is negative ($D < 0$), this part has no real roots. Therefore, the only real solution for the last term is:

$l = 5$

Now, substitute $l = 5$ back into Equation 1 to find the common difference $d$:

$5 = \frac{10}{3} + 29d$

Subtract $\frac{10}{3}$ from both sides:

$29d = 5 - \frac{10}{3}$

$29d = \frac{15 - 10}{3}$

$29d = \frac{5}{3}$

Now, divide both sides by 29:

$d = \frac{5}{3 \times 29}$

$d = \frac{5}{87}$

Thus, the common difference of the given A.P. is $\frac{5}{87}$. Since the first term and the common difference are positive, all terms are indeed non-negative, satisfying all conditions.

Option A is the correct answer.

Let the common difference of the given A.P. be $d$ and the common ratio of the given G.P. be $r$.

From the general formulas for the $n$-th term of an A.P. and a G.P., we know:

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

$g_n = g_1 r^{n - 1}$

We are given that $g_1 = a_1$ and $a_1 = 1$.

Thus, the first term of both the A.P. and the G.P. is $1$. So, $g_1 = 1$.

Next, we are given the first relationship:

$a_2 + g_2 = 1$

Using our formulas, we can write the second terms as $a_2 = 1 + d$ and $g_2 = 1 \cdot r = r$. Let us substitute these values into the given equation:

$(1 + d) + r = 1$

Subtracting $1$ from both sides, we establish a relationship between $d$ and $r$:

$d + r = 0 \implies d = -r$

We are also given a second relationship:

$a_3 + g_3 = 4$

Again, applying the $n$-th term formulas, we have the third terms as $a_3 = 1 + 2d$ and $g_3 = 1 \cdot r^2 = r^2$. Substituting these in the equation gives:

$(1 + 2d) + r^2 = 4$

Now, let us substitute $d = -r$ into this equation to form a quadratic equation in terms of $r$:

$1 + 2(-r) + r^2 = 4$

$r^2 - 2r + 1 - 4 = 0$

$r^2 - 2r - 3 = 0$

Let us factorize this quadratic equation by splitting the middle term:

$r^2 - 3r + r - 3 = 0$

$r(r - 3) + 1(r - 3) = 0$

$(r - 3)(r + 1) = 0$

This gives us two possible values for $r$: $r = 3$ or $r = -1$.

The problem clearly states that the G.P. is an increasing sequence. Since the first term $g_1 = 1$ is positive, an increasing sequence requires the common ratio to be strictly greater than $1$. Therefore, we reject $r = -1$ and take the valid value:

$r = 3$

Since $d = -r$, we find the common difference of the A.P.:

$d = -3$

Now, we are asked to find the value of $a_{10} + g_5$. Let us find $a_{10}$ and $g_5$ individually.

For the 10th term of the A.P.:

$a_{10} = a_1 + (10 - 1)d$

$a_{10} = a_1 + 9d$

$a_{10} = 1 + 9(-3)$

$a_{10} = 1 - 27 = -26$

For the 5th term of the G.P.:

$g_5 = g_1 r^{5 - 1}$

$g_5 = g_1 r^4$

$g_5 = 1 \cdot (3)^4$

$g_5 = 81$

Finally, adding these two values together:

$a_{10} + g_5 = -26 + 81 = 55$

Therefore, the value of $a_{10} + g_5$ is $55$.

Correct Answer:

Option D: $55$

$\begin{aligned} & \mathrm{T}_{\mathrm{r}}=\frac{1^3+2^3+3^3+\ldots \ldots \mathrm{r}^3}{1+3+5+\ldots . .+(2 \mathrm{r}-1)}=\frac{\left(\frac{\mathrm{r}(\mathrm{r}+1)}{2}\right)^2}{\mathrm{r}^2} \\ & =\frac{\mathrm{r}^2+2 \mathrm{r}+1}{4}\end{aligned}$

$\begin{aligned} & \mathrm{S}_{\mathrm{n}}=\sum_{\mathrm{r}=1}^{\mathrm{n}} \mathrm{T}_{\mathrm{r}} \\ & \mathrm{S}_{\mathrm{n}}=\frac{1}{4} \sum\left(\mathrm{r}^2+2 \mathrm{r}+1\right) \\ & =\frac{1}{4}\left[\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+2 \frac{\mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n}\right] \\ & \mathrm{S}_8=\frac{1}{4}\left[\frac{8 \times 9 \times 17}{6}+8 \times 9+8\right] \\ & =\frac{1}{4}[204+72+8]=71\end{aligned}$

Understand Set A

Set $A$ consists of the first 101 terms of an A.P. with the first term $a = 1$ and common difference $d_1 = 5$.

The $n$-th term of this A.P. is given by the formula:

$a_n = a + (n - 1)d_1$

Therefore, the last term (101st term) of Set A is:

$a_{101} = 1 + (101 - 1) \times 5 = 1 + 100 \times 5 = 501$

So, $A = \{1, 6, 11, 16, 21, \dots, 501 \}$.

Understand Set B

Set $B$ consists of the first 71 terms of an A.P. with the first term $b = 9$ and common difference $d_2 = 7$.

The $m$-th term of this A.P. is:

$b_m = b + (m - 1)d_2$

Therefore, the last term (71st term) of Set B is:

$b_{71} = 9 + (71 - 1) \times 7 = 9 + 70 \times 7 = 499$

So, $B = \{9, 16, 23, 30, \dots, 499 \}$.

Find the elements in $A \cap B$

The elements common to both sets $A$ and $B$ will themselves form an A.P.

From the initial terms listed above, we can easily see that the first common term is $16$.

The common difference of this new A.P. will be the Least Common Multiple (LCM) of the common differences of the two original sets:

$\text{Common difference of } A \cap B = \text{LCM}(5, 7) = 35$

The $k$-th term of the common sequence is given by:

$C_k = 16 + (k - 1)35$

Since a common term must exist in both sets, it cannot be greater than the smallest of their last terms. Thus, $C_k \le \text{min}(501, 499)$.

$16 + (k - 1)35 \le 499$

$35(k - 1) \le 483$

$k - 1 \le \frac{483}{35}$

$k - 1 \le 13.8$

Since $k$ must be a positive integer, the maximum possible value for $(k - 1)$ is $13$. Thus, $k \le 14$.

There are altogether 14 common terms in $A \cap B$.

Find elements in $A \cap B$ divisible by 3

We want to find terms $C_k$ that are divisible by $3$, which means:

$C_k \equiv 0 \pmod 3$

$16 + 35(k - 1) \equiv 0 \pmod 3$

Since $16 = 3 \times 5 + 1 \implies 16 \equiv 1 \pmod 3$ and $35 = 3 \times 11 + 2 \implies 35 \equiv 2 \pmod 3$, we can simplify:

$1 + 2(k - 1) \equiv 0 \pmod 3$

Checking the first few integer values for $k$:

• If $k = 1$: $1 + 2(0) = 1$ (Not divisible by 3)

• If $k = 2$: $1 + 2(1) = 3 \equiv 0 \pmod 3$ (Divisible by 3)

• If $k = 3$: $1 + 2(2) = 5 \equiv 2 \pmod 3$ (Not divisible by 3)

The condition holds true when $k = 2$. Because we are operating in modulo $3$, the values of $k$ that satisfy this condition will form an A.P. with a common difference of $3$.

Thus, the valid values of $k$ within the range $1 \le k \le 14$ are:

$k \in \{2, 5, 8, 11, 14\}$

Counting these, there are exactly $5$ valid values of $k$.

For confirmation, the terms are:

• For $k = 2 \rightarrow 51$

• For $k = 5 \rightarrow 156$

• For $k = 8 \rightarrow 261$

• For $k = 11 \rightarrow 366$

• For $k = 14 \rightarrow 471$

All 5 of these terms are common elements between the sets and are completely divisible by 3.

Correct Answer:

Option B

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

$=\frac{1}{16} \times \frac{\pi^4}{90}$ 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 $.