What is the weightage of Sequence and Series chapter in class 11 annual exams?

$l=\underset{n\to \infty }{lim}{\left(u_n\right)}^{\frac{-4}{n^2}}=\underset{n\to \infty }{lim}{\left\{\left(1+\frac{1}{n^2}\right){\left(1+\frac{2^2}{n^2}\right)}^2{\left(1+\frac{3^2}{n^2}\right)}^3......{\left(1+\frac{n^2}{n^2}\right)}^n\right\}}^{\frac{-4}{n^2}},$ Taking log on both the sides

$\underset{n\to \infty }{lim}\frac{-4}{n^2}{\displaystyle \sum _{r=1}^{n}rlog\left(1+\frac{r^2}{n^2}\right)}=\underset{n\to \infty }{lim}\frac{-4}{n}{\displaystyle \sum _{r=1}^n\frac{r}{n}log\left(1+{\left(\frac{r}{n}\right)}^2\right)}$

$\therefore logl=-4{\displaystyle \underset{0}{\overset{1}{\int }}xlog\left(1+x^2\right)dx}$

$logl=-2\left[log4-1\right]=-2log\frac{4}{e}=log\frac{e^2}{16}$       

$l=\frac{e^2}{16}$

$S_n=\left\{z\in C:\left|z-3+2i\right|=\frac{n}{4}\right\}$

represents a circle with centre C1 (3, 2) and radius

$r_1=\frac{n}{4}$

Similarly Tn represents circle with centre C2 (2, 3) and radius

$r_1=\frac{1}{n}$

$\sqrt[]{2}<\left|\frac{n}{4}-\frac{1}{n}\right|$

n take infinite values.

${\alpha }_n=19^n-12^n$

$\frac{31{\alpha }_9-{\alpha }_{10}}{57.{\alpha }_2}=\frac{31\left(19^9-12^9\right)-\left(19^{10}-12^{10}\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9\left(31-19\right)-12^9\left(31-12\right)}{57\left(19^8-12^8\right)}$

$=\frac{19^9.12-12^9.19}{57\left(19^8-12^8\right)}=4$

$a_{n+2}=2\cdot a_{n+1}-a_{n+1}$

$a_2=2a_1-a_0+1$

a₂ = 1

a₃ = 3

a₄ = 6

So for n $\ge 2,a_n=\frac{n\left(n-1\right)}{2}$

${\displaystyle \sum _{n=2}^{}\frac{n\left(n-1\right)}{2\cdot 7^n}=\frac{1}{7^2}+\frac{3}{7^3}+\frac{6}{7^4}+......}$

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

$\frac{\frac{S}{7}=\frac{1}{7^3}+\frac{3}{7^4}+....}{S-\frac{S}{7}=\frac{1}{7^2}+\frac{2}{7^3}+\frac{3}{7^4}+....}$

$\frac{\frac{6}{7}S\cdot \frac{1}{7}=\frac{1}{7^3}+\frac{2}{7^3}+....}{\frac{6}{7}S-\frac{6}{49}S=\frac{1}{7^2}+\frac{1}{7^3}+....}$

$\frac{36}{49}S=\frac{\frac{1}{7^2}}{1-\frac{1}{7}}$

$S=\frac{1}{7^2}\times \frac{7}{6}\times \frac{49}{36}$

$S=\frac{7}{216}$

Given series $\left\{3\times 1\right\},\left\{3\times 2,3\times 3,3\times 4\right\},\left\{3\times 5,3\times 6,3\times 7,3\times 8,3\times 9\right\}.........$

$\therefore$ 11th set will have 1 + (10)2 = 21 terms

Also up to 10th set total 3 × k type terms will be 1 + 3 + 5 + ……… +19 = 100 terms

$\therefore Set11=\left\{3\times 101,3\times 102,......3\times 121\right\}$ $\therefore$ Sum of elements = 3 × (101 + 102 + ….+121)

$=\frac{3\times 222\times 21}{2}=6993.$