$\operatorname{GM}(a, b)=\sqrt{a b} \text {. }$
$ \operatorname{AM}(1,2)=\frac{1+2}{2}=1.5 $
$\operatorname{GM}(1,2)=\sqrt{2} $
$ \operatorname{AM}(1,4)=2.5 $
$ \operatorname{GM}(1,4)=\sqrt{4}=2 $
$ \operatorname{AM}(2,8)=5 $
$ \operatorname{GM}(2,8)=4$
Let $a, b$ be two positive reals.
$ \operatorname{AM}(a, b)=\frac{a+b}{2} $
$ \operatorname{GM}(a, b)=\sqrt{a b.}$ $\quad A M \geqslant G M \ \ \rightarrow ?$.
$\quad \frac{a+b}{2} \geqslant \sqrt{a b} \ \ \rightarrow ?$
consider
$\frac{a+b}{2}-\sqrt{a b}$ $ =\frac{a+b-2 \sqrt{a b}}{2} $
$ =\frac{(\sqrt{a}-\sqrt{b})^2}{2} $
$ \geqslant 0$
$ A M \geqslant G M $
$\frac{a+b}{2} \geqslant \sqrt{a b}$
Equality holds iff $a=b$.
consider a rectangle with Side lengths " $a$ " and " $b$ "
Perimeter $=2 a+2 b$
Area $=a b$
Consider square of Side lengths
$\sqrt{a b}$.
$\text { Area } =a b $
$\text { Perimeter } =4 \sqrt{a b} $
AM - GM inequality gives
$\frac{a+b}{2} \geqslant \sqrt{a b} .$
$2 a+2 b \geqslant 4 \sqrt{a b}$
Given
“$n$” real numbers $a_1, a_2 \ldots a_n$. The $AM$ of these numbers is defined as $A M\left(a_1, a_2, a_n\right)=\frac{a_1+a_2+ ~+a_n}{n}$
“$n$” positive reals $a_1, a_2, . . a_n$. the Geometric mean of these numbers. $\operatorname{GM}\left(a_1, a_2 \quad a_n\right) = \left(a_1 a_2 \ldots a_n\right)^{\frac{1}{n}}$
i.e.
Given “$n$” positive reals $a_1, a_2, \ldots a_n$
Given an $ A P $
$a, a+d, a+2 d, . a+(n-1) d,….$
$\sum_{n=1}^{\infty}[a+(n-1) d]$
$Let \sum_{n=1}^{\infty} a_n \text{~be convergent.}$
i.e., sequence of partial sum $S_n=a_1+a_2+\ldots+a_n$ is convergent.
$\underset{n \rightarrow \infty}{L t} \quad S_n= S$
As " $n$ " becomes large both $S_n$ and $S_{n-1}$ are close to $S.$
$ a_n=S_n-S_{n-1} $
$ \underset{n \rightarrow \infty}{L t} a_n=0$
$\sum_{n=1}^{\infty} a_n \text{~is convergent}$
$\Rightarrow \quad a_n \rightarrow 0$ as $n \rightarrow \infty$.
$P \Longrightarrow \ Q$ is logically
equivalent to
Not $Q \Longrightarrow$ Not $P$.
If $a_n \not\longrightarrow 0$ as $n \rightarrow \infty$ then $\sum a_n$ is NOT convergent.
Given an AP
$a, a+d, a+2 d, \cdots a+(n-1) d$;
$ n^{\text {th }} \text { term } =a+(n-1) d $
$ d \neq 0 $
$\underset{n \rightarrow \infty}{L t} a+(n-1) d = \pm \infty $
$\therefore\sum_{n=1}^{\infty} a+(n-1) d$ is NOT convergent.
$\underline{d = 0}$
$ {n^{\text {th }} \text { term }}=a $
$ \underset{n \rightarrow \infty}{L t} \quad a \neq 0 $
$ \quad(\text { if } a \neq 0)$
$\sum_{n=1}^{\infty} a_n \text { is convergent } $
$ \rArr a_n \text{ is close to zero as }$
“$n$” $\text{becomes large}$
$ \underset{n \rightarrow \infty}{L t} a_n = 0 .$
$\underline{\text{Remark}}$
If $a_n \longrightarrow 0$ as $n \longrightarrow \infty$, then we cannot conclude that $\sum a_n$ is convergent.
$ 1, \frac{1}{2}, \frac{1}{3}, \cdots \frac{1}{n}$,…..
$ \sum \frac{1}{n} = 1+\frac{1}{2}+\frac{1}{3}+$……
$ S_{2^n} \geqslant 1+\frac{n}{2} $
$\sum \frac{1}{n}$ is NOT Convergent
But $\frac{1}{n} \longrightarrow 0$
as $n \rightarrow \infty$.
$ 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$
$\sum_{n=1}^{\infty} \frac{1}{2^{n-1}}\text{ is convergent}$
$\frac{1}{2^{n-1}} \longrightarrow 0$
$\text{as} ~n \rightarrow \infty$
Sum of $n$ terms of two $AP$ are in the ratio $\frac{3n+8}{7n + 15}.$
Find the ratio of their $12^{\text{th}}$ terms.
Let us assume that first $AP$ has first term $a_1$ and common difference $d_1$
$a_1, a_1 + d_1, a_1 + 2d_1,……$
Let second $AP$ has first term $a_2$ and common difference $d_2$
$a_2, a_2 + d_2, a_2 + 2d_2,…..$
$n^{\text{th}}$ term $I$ $AP$
$n^{\text{th}}$ term $II$ $AP$
$ \text{sum of}$ “$n$” terms of $I$ $AP$
$ \text{sum of}$ “$n$” terms of $II$ $AP$
$\frac{\frac{n}{2}(2a_1 + (n-1) ~d_1)}{\frac{n}{2}(2a_2 + (n-1) ~d_2)}=\frac{3_n+8}{7_n+15}$
$\frac{2a_1 + (n-1) ~d_1}{2a_2 + (n-1) ~d_2}=\frac{3_n+8}{7_n+15}$
$12^{\text{th}}$ term of $I$ $AP$
$12^{\text{th}}$ term of $II$ $AP$
$\frac{a_1 + 11 ~d_1}{a_2 + 11 ~d_2} = ?$
$\frac{2a_1 + 22 ~d_1}{2a_2 + 22 ~d_2} = ? …(*)$
$\text{Putting} ~n= 23 \ \text{in} ~(*)$
$\frac{2a_1 + 22 ~d_1}{2a_2 + 22 ~d_2} = \frac{3.23 + 8}{7.23+15}$
$ = \frac{7}{16}$
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