Sequences and Series

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

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

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

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

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

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

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

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

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

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

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

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

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

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

$\frac{25S}{36}=1+\frac{3/5}{1}$  

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

Let $\frac{A_2}{A_1}=\frac{A_3}{A_2}=...=r$  

$\therefore A_1{A}_3{A}_5{A}_7=\frac{1}{1296}$

$\Rightarrow A_1{r}^3=\frac{1}{6}.......\left(i\right)$  

Again, A₂ + A₄ = $\frac{7}{36}$  

$A_{1}r=\frac{7}{36}-\frac{1}{6}=\frac{1}{36}........\left(ii\right)$

$\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$  

  $\therefore A_6+A_8+A_{10}=A_1{r}^5\left(1+r^2+r^4\right)=43$  

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   234

n <  $\frac{237}{12}\Rightarrow n=19$  

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

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

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

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

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

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

$=\frac{3*222*21}{2}=6993.$   

Given G.P's 2, 2², 2³, …60 term and 4, 4², 4³, … of 60

Now G.M. =  ${\left(2\right)}^{\frac{225}{8}}\Rightarrow {\left(2,2^2,2^3,....\right)}^{\frac{1}{60+n}}={\left(2\right)}^{\frac{225}{8}}\Rightarrow n=\frac{57}{8},20son=20$

$\therefore {\displaystyle \sum _{k=1}^{n}k(n-k)}20*\frac{20*21}{2}-\frac{20*21*41}{6}=1330$  

$P=1+\frac{2}{3}+\frac{6}{3^2}+\frac{10}{3^3}+\frac{14}{3^4}+...+\infty$

$\frac{P}{3}=\frac{1}{3}=\frac{2}{3^2}+\frac{6}{3^3}+\frac{10}{3^4}+...+\infty$

$\frac{2P}{3}=1+\frac{1}{3}+\left[\frac{4}{3^2}+\frac{4}{3^3}+\frac{4}{3^4}+...+\infty \right]$

$\frac{2P}{3}=\frac{\frac{4}{3}}{1-\frac{1}{3}}\Rightarrow P=3$

By property

If first and Last term of Andhra Pradesh and HP are equal (of nn  terms) then $a_r{H}_S=$  first *  x last term

If $r+s=n+1$ r+s=n+1

$S_1=1+2+3\dots \dots \dots .n$

$S_2=1+3+5\dots \dots \dots .n$

$S_3=1+4+7\dots \dots \dots .$

$\begin{array}{ll}\left(S_1+S_3\right)=\lambda S_2& S_1=\frac{n}{2}[2a+(n-1\left)1\right]\\ \frac{n}{2}[2a+(n-1\left)1\right]+\frac{n}{2}[2a+(n-1\left)3\right]& S_2=\frac{n}{2}[2a+(n-1\left)2\right]\\ =\lambda \frac{n}{2}[2a+(n-1\left)2\right]& S_3=\frac{n}{2}[2a+(n-1\left)3\right]\end{array}$

$\lambda =2$