Sequences and Series

Given,

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

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

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

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

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

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

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

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

$\therefore k$ divides 126 .

$\Rightarrow \quad 2 k$ divides 126

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

Given series

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

to 10 terms.

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

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

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

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

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

For $k=10$

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

$ \begin{aligned} & \text t_n=(2 n-1)(2 n+1)(2 n+3) \\\\ & t_n=8 n^3+12 n^2-2 n-3 \\\\ & S_n=\Sigma S_n=8 \Sigma n^3+12 \Sigma n^2-2 \Sigma n-\Sigma 3 \\\\ & =8\left(\frac{n(n+1)}{2}\right)^2+12 \frac{n(n+1)(2 n+1)}{6} -2 \frac{n(n+1)}{2}-3 n \\\\ & =n\left[\frac{8 \cdot n(n+1)^2}{4}+2(n+1)(2 n+1)-n-1-3\right] \\\\ & =n\left[2 n\left(n^2+1+2 n\right)+2\left(2 n^2+3 n+1\right)-n-4\right] \\\\ & =n\left[2 n^3+4 n^2+2 n+4 n^2+6 n+2-n-4\right] \\\\ & =n\left[2 n^3+8 n^2+7 n-2\right] \\\\ & =n(n+1) f(n) \\\\ & \therefore f(n)=\frac{2 n^3+8 n^2+7 n-2}{n+1} \\\\ & \quad f(2)=\frac{16+32+14-2}{3}=\frac{60}{3}=20 \end{aligned} $

The given series is related to the binomial expansion for any index. The binomial expansion for $(1-x)^{-n}$ when $|x| < 1$ is given by :

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

To relate the series with the binomial expansion, consider setting $x = \frac{1}{2}$ and $n = \frac{2}{3}$. The substitution into the binomial formula results in :

$ \left(\frac{1}{2}\right)^{-\frac{2}{3}} = 1 + \frac{2 \cdot 1}{3 \cdot 2} + \frac{2 \cdot 5 \cdot 1}{3 \cdot 6 \cdot 4} + \ldots $

This series equates to $\sqrt[3]{4}$ :

$ \sqrt[3]{4} = 1 + \frac{2 \cdot 1}{3 \cdot 2} + \frac{2 \cdot 5 \cdot 1}{3 \cdot 6 \cdot 4} + \ldots $

Thus, the assertion that this infinite series sums to $\sqrt[3]{4}$ is correct, and it is derived from the application of the binomial series expansion with the appropriate substitutions for $x$ and $n$.

We have to check each option, From option (a), we have,

$ \begin{array}{l} (2 n+7) < (n+3)^{2} \\\\ \text { Let } P(n) \equiv(2 n+7) < (n+3)^{2} \\\\ \qquad P(1)=(2+7)=9 < (1+3)^{2}=4^{2}=16 \end{array} $

Let us assume that $ P(K) $ is true.

We have to prove that $ P(K+1) $ is true

$ \begin{array}{l} (2 K+7) < (K+3)^{2} \\\\ 2 K+7 < K^{2}+9+6 K \end{array} $

Add 2 on LHS and $ 7+2 K $ on RHS

as $ 2 < 7+2 K $

$ \begin{array}{l} \Rightarrow 2 K+9 < K^{2}+16+8 K \\\\ 2(K+1)+7 < ((K+1)+3)^{2} \end{array} $

$ \therefore P(K+1) $ is true.

Hence, $ P(x) $ is true, $ \forall n \in N $.

From option b,

$ \begin{array}{r} \text { Let } Q(n) \equiv 1^{2}+2^{2}+\ldots+n^{2} > \frac{n^{3}}{3} \\\\ \qquad Q(1)=1^{2} > \frac{1}{3} \end{array} $

Let $ Q(K) $ be true.

$ \begin{array}{c} 1^{2}+2^{2}+3^{2}+\ldots+K^{2} > \frac{K^{3}}{3} \\\\ 1^{2}+2^{2}+3^{2}+\ldots .+K^{2} \\\\ +(K+1)^{2} > \frac{K^{3}}{3}+(K+1)^{2} \\\\ 1^{2}+2^{2}+3^{2}+\ldots .+K^{2}+(K+1)^{2} \\\\ > \frac{K^{3}+3 K^{2}+6 K+3}{3} \\\\ 1^{2}+2^{2}+3^{2}+\ldots+K^{2}+(K+1)^{2} > \\\\ \frac{K^{3}+3 K^{2}+3 K+1+3 K+2}{3} \\\\ =\frac{(K+1)^{3}+3 K+2}{3} > \frac{(K+1)^{3}}{3} \end{array} $

$ \therefore Q(K+1)^{2} $ is true.

Hence, $ Q(n) $ is true, $ \forall n \in N $

From option (c),

$ \begin{array}{l} R(n)=35^{2 n+1}+2^{3 n+1} \\\\ R(1)=3 \times 125+2^{4}=375+16=391 \end{array} $

which is divisible by 23

Let us assume it is true for $ n=k, k \in N $

$ \therefore S(k)=3 \cdot 5^{2 k+1}+2^{3 k+1} $ is divisible by

$ 23 \forall k \in N $

Let $ 3 \cdot 5^{2 k+1}+2^{3 k+1}=23 t, t \in N $

$ \therefore 3 \cdot 5 k^{2 k+1}=23 t-2^{3 k+1} $

Now, we have to prove that for $ n=k+1 $

$ \begin{aligned} \therefore S(k+1) & =3 \cdot 5^{2(k+1)+1}+2^{3(k+1)+1} \\\\ & =3 \cdot 5^{2 k+1} \cdot 5^{2}+2^{3 k+1} \cdot 2^{3} \\\\ & =\left(23 t-2^{3 k+1}\right) 25+8 \cdot 2^{3 k+1} \\\\ & =23 t \times 25-25 \times 2^{3 k+1}+8 \cdot 2^{3 k+1} \\\\ \therefore S(k+1) & =23 t \times 25-172^{3 k+1} \end{aligned} $

Hence, $ 3 \cdot 5^{2 n+1}+2^{3 n+1} $ is not divisible by 23.

Hence, $ R(n) $ is not divisible by $ 23, \forall n \in N $. From option (d),

$ 2+7+12+\ldots+(5 n-3) $ is an AP with common difference as 5 .

$ \begin{aligned} S_{n} & =\frac{n}{2}[2 \times 2+(n-1) 5] \\\\ & =\frac{n}{2}(5 n-1) \end{aligned} $

$\begin{aligned} \frac{1}{3 \cdot 6} & +\frac{1}{6 \cdot 9}+\frac{1}{9 \cdot 12}+\ldots \text { to } 9 \text { terms } \\ = & \frac{1}{9}\left(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{9 \cdot 10}\right) \\ = & \frac{1}{9}\left(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)\right. \left.+\ldots+\left(\frac{1}{9}-\frac{1}{10}\right)\right) \\ & = \left(1-\frac{1}{10}\right)=\frac{1}{9}\left(\frac{9}{10}\right)=\frac{1}{10}\end{aligned}$

We have,

$ \frac{x}{(x-2)(x-3)}=\frac{A}{(x-2)}+\frac{B}{(x-3)} $

On comparing, we get

$ \begin{aligned} & A=-2 B=3 \\\\ & \Rightarrow \quad \frac{x}{(x-2)(x-3)}=\frac{-2}{x-2}+\frac{3}{x-3} \\\\ &=\frac{1}{\left(1-\frac{x}{2}\right)}+\frac{-1}{\left(1-\frac{x}{3}\right)} \end{aligned} $

$\begin{aligned} & =\left(1-\frac{x}{2}\right)^{-1}-\left(1-\frac{x}{3}\right)^{-1} \\\\ & =\left[1+\frac{x}{2}+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^3+\ldots .+\left(\frac{x}{2}\right)^n\right]- \\\\ & \quad\left[1+\frac{x}{3}+\left(\frac{x}{3}\right)^2+\ldots .+\left(\frac{x}{3}\right)^n\right]\end{aligned}$

$\Rightarrow$ Coefficient of $x^2=\frac{1}{4}-\frac{1}{9}=\frac{5}{36}$

$k x^3-18 x^2-36 x+8=0 x \rightarrow \frac{1}{x}$, then roots will be in AP

$ \begin{aligned} & \Rightarrow \quad \frac{k}{x^3}-\frac{18}{x^2}-\frac{36}{x}+8=0 \\ & \Rightarrow \quad 8 x^3-36 x^2-18 x+k=0 \end{aligned} $

Let roots are $a-d, a$ and $a+d$ Sum of roots, $3 a=\frac{36}{8} \Rightarrow a=\frac{3}{2}$

Root $a=3 / 2$ will satisfy equation

$ \begin{array}{rlrl} \Rightarrow 8 \times\left(\frac{3}{2}\right)^3-36\left(\frac{3}{2}\right)^2-18\left(\frac{3}{2}\right)+k & =0 \\ \Rightarrow & k & =81 \end{array} $

$\because f(x+y)=f(x)+f(y)$

and $\quad f(1)=7$

Put $\quad x=y=1$

Then, $\quad f(2)=\mid 2 f(1)$

$\Rightarrow f(2)=2 \times 7=14$

On putting $x=2, y=1$ in Eq. (i), we get $f(3)=f(2)+f(1)=14+7=21$

On putting $x=3, y=1$ in Eq. (i), we get $f(4)=f(3)+f(1)=21+7=28$

i.e., $f(1), f(2), f(3), \ldots$, are in AP .

First term of $\mathrm{AP}=f(\mathrm{l})=7$

and common difference $=f(2)-f(1)=7$

$ \begin{aligned}Therefore,\,\, & \sum_{r=1}^n f(r)=f(1)+f(2)+\ldots+f(n) \\ & \quad=7+14+21+\ldots+7 n=S_n \end{aligned} $

$So,\,\, S_n=\frac{n}{2}(7+7 n)=\frac{7 n}{2}(n+1) $

Given, $i=\sqrt{-1}$

Then,

$ \begin{array}{r} \sum_{n=0}^{\infty}\left(\frac{i}{3}\right)^n=\left(\frac{i}{3}\right)^0+\left(\frac{i}{3}\right)^1+\left(\frac{i}{3}\right)^2+\left(\frac{i}{3}\right)^3 \\ +\ldots+\left(\frac{i}{3}\right)^{\infty} \end{array} $

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

Clearly, it forms a GP with infinite terms and first term as 1 and common ratio as $\left(\frac{i}{3}\right)$.

$ \begin{aligned} \Rightarrow S_{\infty} \frac{1}{1-\frac{i}{3}} & =\frac{3}{3-i}\left[\begin{array}{l} \text { sum of a GP with } \\ \text { infinite term }=\frac{a}{1-r} \end{array}\right] \\ & =\frac{3(3+i)}{(3-i)(3+i)} \\ & =\frac{9+3 i}{9-(-1)}=\frac{9+3 i}{10} \end{aligned} $

$ \begin{aligned} & \text { } 3 x=1+\frac{5}{8}+\frac{5}{8} \cdot \frac{9}{13}+\frac{5}{16}+\ldots \\ & \qquad \begin{array}{l} 3 x=1+\frac{5}{8}+\frac{5 \cdot 9}{8 \cdot 13}+\frac{5 \cdot 9 \cdot 13}{8 \cdot 13 \cdot 18}+\ldots \\ \Rightarrow 3 x=1+\int_0^{\infty} \frac{1+4 x}{3+5 x} d x \\ \Rightarrow 3 x=1+\int_0^{\infty} \frac{1+4 \frac{(k-3)}{5}}{3+5 x}\left[\begin{array}{l} \text { Put } 3+5 x=k \\ \Rightarrow 5 d x=d k \end{array}\right] \end{array} \end{aligned} $

$ \begin{aligned} & \Rightarrow 3 x=1+\int_3^{\infty}\left(\frac{1}{k}+\frac{4}{5}-\frac{12}{5 k}\right) d x \\ & 3 x=1+0 \Rightarrow-x=\frac{1}{3} \\ & \begin{aligned} \therefore x^4+ & 4 x^3+6 x^2+4 x \\ \left(\frac{1}{3}\right)^4 & +4 \times\left(\frac{1}{3}\right)^3+6 \times\left(\frac{1}{3}\right)^2+\frac{4}{3} \\ & =\frac{1}{81}+\frac{6}{9}+\frac{4}{3}\left[\frac{1}{9}+1\right] \\ & =\frac{1}{81}+\frac{6}{9}+\frac{4}{3} \times \frac{10}{9} \\ & =\frac{1+54+120}{81}=\frac{175}{81} \end{aligned} \end{aligned} $

$x^3-14 x^2+56 x-64=0$

Using Hit and Trial, we can say $x=2$ is one of the root

$ \begin{array}{ll} & (x-2)\left(x^2-12 x+32\right)=0 \\ & x=2, x^2-12 x+32=0 \\ \Rightarrow & (x-4)(x-8)=0 \\ \Rightarrow & x=4 \text { or } x=8 \\ \Rightarrow & \text { 2,4,8 are roots. } \\ \Rightarrow \quad & \text { Roots are in GP. } \end{array} $

Let $\alpha, \beta, \gamma$ be the roots of the equation

$ \begin{gathered} 3 x^3-26 x^2+52 x-24=0 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\\ \alpha+\beta+\gamma=\frac{26}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \alpha \beta+\beta \gamma+\gamma \alpha=\frac{52}{3} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ \alpha \beta \gamma=8 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{gathered} $

$\alpha, \beta, \gamma$ are in GP $\Rightarrow \beta^2=\alpha \gamma$

From Eq. (iii), $\beta^3=8 \Rightarrow \beta=2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

∴ From Eqs. (i) and (iv),

$ \begin{aligned} & \alpha+\gamma=\frac{20}{3} \quad \alpha \gamma=4 \Rightarrow \gamma=\frac{4}{\alpha} \\ & \therefore \alpha+\frac{4}{\alpha}=\frac{20}{3} \Rightarrow 3 \alpha^2-20 \alpha+12=0 \\ & \Rightarrow(\alpha-6)(3 \alpha-2)=0 \quad \alpha=6, \alpha=\frac{2}{3} \end{aligned} $

For $\alpha=6, \gamma=\frac{2}{3}$ and for $\alpha=\frac{2}{3}, \gamma=6$

$ \begin{aligned} & \therefore \quad \alpha=6, \beta=2 \text { and } \gamma=\frac{2}{3} \\ & \text { or } \quad \alpha=\frac{2}{3}, \beta=2 \text { and } \gamma=6 \end{aligned} $

$ \text { } \begin{aligned}Now, 3 \alpha+2 \beta+\gamma & =3 \times \frac{2}{3}+2 \times 2+6 \\ & =2+4+6=12 \end{aligned} $

We have, $f(n)=A(-2)^n+B(-3)^n$

$ \begin{aligned} f(n)+a f(n-1)+b f(n-2) & =0 \\ \therefore A(-2)^n+B(-3)^n+a\left(A(-2)^{n-1}\right. & \\ \left.+B(-3)^{n-1}\right)+b\left(A(-2)^{n-2}+B(-3)^{n-2}\right. & =0 \\ \Rightarrow A(-2)^{n-2}[4-2 a+b]+(-3)^{n-2} & \\ B[9-3 a+b] & =0 \end{aligned} $

∴ It is possible only

$ 4-2 a+b=0 \text { and } 9-3 a+b=0 $

Solving, we get $a=5, b=6$

$ \therefore \quad(a+b)(b-a)=(5+6)(6-5)=11 $

We have,

$ \begin{array}{r} 1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots . \\ =\frac{a-2 \cos \theta}{5+b \cos \theta} \end{array} $

Put, $\theta=0$

$ \begin{align*} & \therefore & 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8} \ldots . & =\frac{a-2}{5+b} \\ & \Rightarrow & \frac{1}{1-\frac{1}{2}} & =\frac{a-2}{5+b} \\ & \Rightarrow & 2 & =\frac{a-2}{5+b} \\ & \Rightarrow & 10+2 b & =a-2 \\ & \Rightarrow & a-2 b & =12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

Put, $\theta=\pi$

$ 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots=\frac{a+2}{5-b} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & \frac{1}{1+\frac{1}{2}}=\frac{a+2}{5-b} \\ \Rightarrow & \frac{2}{3}=\frac{a+2}{5-b} \\ \Rightarrow & 10-2 b=3 a+6 \\ \Rightarrow & 3 a+2 b=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array}\\ &\text { From Eq. (i) and (ii), we get }\\ &\begin{gathered} a=4, b=-4 \\ \therefore \quad(a-b)^2=(4+4)^2=8^2=64 \end{gathered} \end{aligned} $

We have,

$ \begin{aligned} & \begin{aligned} S_n= & 1^2+2 \times 2^2+3^2+2 \times 4^2+5^2 \\ & +2 \times 6^2+\ldots \infty \end{aligned} \\ \therefore \quad & T_n=(2 n-1)^2+2 \times(2 n)^2 \\ & =4 n^2+1-4 n+8 n^2 \\ = & 12 n^2-4 n+1 \\ & \Sigma T_n=12 \Sigma n^2-4 \Sigma n+\Sigma 1 \\ = & 12 \frac{n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n \\ = & 2 n(n+1)(2 n+1)-2 n(n+1)+n \\ = & 2 n(n+1)(2 n+1-1)+n \\ = & 2 n(n+1)(2 n)+n \\ = & 4 n^2(n+1)+n \\ = & n[4 n(n+1)+1] \\ = & n\left(4 n^2+4 n+1\right)=n(2 n+1)^2 \end{aligned} $

Here, $S_n$ is even

$ \begin{aligned} & S_m=\frac{m}{2}\left(2 \frac{m}{2}+1\right)^2 n=\frac{m}{2} \\ \Rightarrow & S_m=\frac{m}{2}(m+1)^2 \\ \therefore & S_n=\frac{n}{2}(n+1)^2 \text { when } n \text { is even } \end{aligned} $

$ \begin{aligned} &\text { Let roots of the equation }\\ &\begin{aligned} & 8 x^3+6 p x^2+3 q x-27=0 \text { are } \frac{a}{r}, a, \text { ar } . \\ & \therefore \quad \frac{a}{r}+a+a r=\frac{-6 p}{8}=\frac{-3 p}{4} \\ & \Rightarrow \quad a\left(\frac{1}{r}+1+r\right)=\frac{-3 p}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \left(\frac{a}{r} \cdot a\right)+(a \cdot a r)+\left(\frac{a}{r} \cdot a r\right)=\frac{3 q}{8} \\ & \Rightarrow \quad a^2\left(\frac{1}{r}+r+1\right)=\frac{3 q}{8} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \text { and } \quad \frac{a}{r} \times a \times a r=\frac{27}{8} \\ & \Rightarrow \quad a^3=\frac{27}{8} \Rightarrow a=\frac{3}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} \end{aligned} $

From Eqs. (i) and (ii), we get

$ a=\frac{3 q / 8}{-3 p / 4}=-\frac{q}{2 p} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

From Eqs. (iii)and (iv), we have

$ \frac{-q}{2 p}=\frac{3}{2} \Rightarrow-q=3 p $

Now,

$ q^2+9 p^2+6 p q+\frac{q}{p} $

$ \begin{array}{ll} \Rightarrow & 9 p^2+9 p^2+6 p(-3 p)+\frac{(-3 p)}{p} \\ \Rightarrow & 18 p^2-18 p^2-3=-3 \end{array} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } \frac{m}{n}=\frac{1}{1 \cdot 7}+\frac{1}{7 \cdot 13}+\frac{1}{13 \cdot 19} \\ & +\ldots+\frac{1}{[1+(20-1) 6][7+(20-1) 6]} \\ & =\frac{1}{1 \cdot 7}+\frac{1}{7 \cdot 13}+\frac{1}{13 \cdot 19}+\ldots+\frac{1}{115 \times 121} \\ & =\frac{1}{6}\left[\frac{6}{1 \cdot 7}+\frac{6}{7 \cdot 13}+\frac{6}{13 \cdot 19}+\ldots+\frac{6}{115 \times 121}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{1}{6}\left[\frac{7-1}{1 \times 7}+\frac{13-7}{7 \times 13}+\frac{19-13}{13 \times 19}+\ldots+\frac{121-115}{115 \times 121}\right] \\ & =\frac{1}{6}\left[\left(\frac{1}{1}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{19}\right)\right. \\ & \left.+\ldots+\left(\frac{1}{115}-\frac{1}{121}\right)\right] \\ & =\frac{1}{6}\left[1-\frac{1}{121}\right]=\frac{1}{6} \times \frac{120}{121}=\frac{20}{121} \\ & \begin{array}{ll} \therefore & m=20, n=121 \\ \therefore & 5 m+2 n=5 \times 20+2 \times 121 \\ & =100+242=342 \end{array} \end{aligned} $