Sequences and Series

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

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

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)

$2021\equiv -2$ mod (7)

$\Rightarrow {\left(2021\right)}^{2023}\equiv {\left(-2\right)}^{2023}mod\left(7\right)$ …… (i)

Now,  ${\left(-2\right)}^3\equiv -1mod\left(7\right)$

$\Rightarrow {\left(-2\right)}^{2023}\equiv \left(-2\right)mod\left(7\right)\equiv 5mod\left(7\right)$ ……. (ii)

(i) & (ii)

$\Rightarrow {\left(2021\right)}^{2023}\equiv 5mod\left(7\right)$

$\therefore$ Remainder = 5

Given ${\left(2x^2+3x+4\right)}^{10}={\sum }_{r=0}^{20}?a_r{x}^r$

replace $x$ by $\frac{2}{x}$ in above identity :-

$\begin{array}{rr}& \frac{2^{10}{\left(2x^2+3x+4\right)}^{10}}{x^{20}}=\sum _{r=0}^{20}??\frac{a_r{2}^r}{x^r}\\ & \Rightarrow 2^{10}\sum _{r=0}^{20}??a_r{x}^r=\sum _{r=0}^{20}??a_r{2}^r{x}^{(20-r)}\left(\text{from}\right(i\left)\right)\end{array}$

now, comparing coefficient of $x^7$ from both sides

(take $r=7$ in L.H.S. and $r=13$ in R.H.S.)

$2^{10}{a}_7=a_{13}{2}^{13}\Rightarrow \frac{a_7}{a_{13}}=2^3=8$