Sequences and Series

  $?$  Roots of 2ax²  - 8 ax + 1 = 0 are

$\frac{1}{p}and\frac{1}{r}$ and roots of 6bx² + 12bx + 1 = 0 are

  $\frac{1}{q}and\frac{1}{8}$  

Let  $\frac{1}{p},\frac{1}{q},\frac{1}{r},\frac{1}{8}$

as    $\alpha -3\beta ,\alpha -\beta ,\alpha +\beta ,\alpha +3\beta$

$So,\frac{1}{a}-\frac{1}{b}=38$

x² – x – 4 = 0

$P_n={\alpha }^n-{\beta }^n$

$?=\frac{P_{15}{P}_{16}-P_{14}{P}_{16}-P_{15}^2+P_{14}{P}_{15}}{P_{13}{P}_{14}}$

$=\frac{\left(P_{15}-P_{14}\right)\left(P_{16}-P_{15}\right)}{P_{13}{P}_{14}}$

$=\frac{4P_{13}\cdot 4P_{14}}{P_{13}\cdot P_{14}}=16$

P_n =  αⁿ - βⁿ

$={\alpha }^{n-1}\cdot \alpha -{\beta }^{n-1}\cdot \beta$

$={\alpha }^{n-1}\left({\alpha }^2-4\right)-{\beta }^{n-1}\left(\beta -4\right)$

$P_n={\alpha }^{n+1}-{\beta }^{n+1}-4\left({\alpha }^{n-1}-{\beta }^{n-1}\right)$

$P_n=P_{n+1}-4P_{n-1}\Rightarrow P_{n+1}-P_n=4P_{n-1}$

${\displaystyle \sum _{}^{}{x}_0^1=\frac{3{\left(1-\left(\frac{1}{2}\right)\right)}^{20}}{1-\frac{1}{2}}=6\left(1-\frac{1}{2^{20}}\right)}$

$={\displaystyle \sum _{i=1}^{20}{\left(x_i\right)}^2+{\left(i\right)}^2-2x_{i}i}$

Now, ${\displaystyle \sum _{i=1}^{20}{\left(x_i\right)}^2=\frac{9{\left(1-\left(\frac{1}{4}\right)\right)}^{20}}{\left(1-\frac{1}{4}\right)}=12\left(1-\frac{1}{2^{40}}\right)}$

$\overline{x}=\frac{2858}{20}+\left(\frac{-12}{2^{40}}+\frac{22}{2^{20}}\right)*\frac{1}{20}$

$\left[\overline{x}\right]=142$

$t_n=\frac{3^n-2^n}{3^n}=1-{\left(\frac{2}{3}\right)}^n$

$S_{100}=100-\frac{\frac{2}{3}\left({\left(\frac{2}{3}\right)}^{100}-1\right)}{\frac{2}{3}-1}=100+2.\frac{\left(2^{100}-3^{100}\right)}{3^{100}}$

$=100-2+\frac{2^{101}}{3^{100}}=98+{\left(\frac{2}{3}\right)}^{100}.2<98+1$

  ${\displaystyle \sum _{i=1}^n{a}_i}=192\Rightarrow a_1+a_2+a_3+......+a_n=192$

$\therefore n\left(a_1+a_n\right)=384..............(i)$

and ${\displaystyle \sum _{i=1}^{n/2}{a}_{2i}}=120\Rightarrow a_2+a_4+a_6+......+a_n=120$  

 n (a₂ + a_n) = 480 ……………… (ii)

n (a₂ – a₁) = 96

$\therefore n=96$

$\left\{a\in \left(1,2,3,......,100\right):HCF\left(a,24\right)=1\right\}$

HCF of (a, 24) = 1 $\therefore$ a = 1, 5, 7, 11, 13, 17, 19, 23 sum of these numbers = 96

$\therefore$ There are four such blocks and a number 97 is there upto 100.

$\therefore$ complete sum = 96 + (24 * 8 + 96) + (48 * 8 + 96) + (72 * 8 + 96) + 97 = 1633

  a + a₁……a_n, 100               n arithmetic mean 100

a + n = 33 ……. (i)

$\left(\frac{a_1}{a_n}=\frac{1}{7}\right)$

$\frac{\left(a+d\right)}{\left(100-d\right)}=\frac{7}{7}$

7a + 8d = 100…………… (ii)

a + (n + 1)d = 100………………. (iiI)

Solving these equations (i), (ii) & (iii), we get

n = 23 & d =    $\frac{15}{4}$

a = 10

First common term to both AP's is 9

t₇₈ of $\left(3,6,9,......\right)=78*3=234$

t₅₉ of $\left(5,9,13,........\right)=5+\left(5-1\right)4=237$

n^th common term $\le 234$

9 + (n – 1) 12 $\le$ 234

n < $\frac{237}{12}\Rightarrow n=19$

Now sum of 19 terms with a = 9, d = 12

$=\frac{19}{2}\left(2.9+\left(19-1\right)\cdot 12\right)=2223$

Let S = 1 $+\frac{5}{6}+\frac{12}{6^2}+\frac{22}{6^3}+....$

$\underset{\_}{-\frac{S}{6}=\frac{1}{6}+\frac{5}{6^2}+....}$

$\frac{5S}{6}=1+\frac{4}{6}+\frac{7}{6^2}+\frac{10}{6^3}+....$

$\underset{\_}{-\frac{5S}{36}=\frac{1}{6}+\frac{4}{6^2}+\frac{7}{6^3}+.....}$

$\frac{25S}{36}=1+\frac{3}{6}+\frac{3}{6^2}+\frac{3}{6^3}+......$

$\frac{25S}{36}=1+\frac{3/6}{1-1/6}$

$\frac{25S}{36}=1+\frac{3/5}{1}$

$S=\frac{288}{125}$

  $f\left(x\right)=min\left\{1,1+xsinx\right\},0\le x\le 2\pi$

$f\left(x\right)=\{\begin{array}{l}1,0\le x<\pi \\ 1+xsinx,\pi \le x\le 2\pi \end{array}$

Now at x = π,

$\therefore f\left(x\right)$ is not differentiable at x = π

$\therefore$ (m, n) = (1, 0)