consider a G.P with first term " $a$ " and common ratio " $r$ "
$a, a r, a r^2, \ldots, a r^{n-1}, \ldots$
$ S_n=a+a r+a r^2+\cdots+a^{n-1}$
$\text { Let, }\quad r=1$
$S_n = \underbrace{a+a+ \cdots +a}_{n \ \text{times}}$
i.e $S_n=n a$.
Let $r \neq 1$
(1) $S_n=a+a r+\cdots+ar^{n-1}$
(2) $S_n=a r+a r^2+\ldots+a r^{n-1}+a r^n$
(1) - (2)
$S_n-r S_n $
$ =a-a r^n $
$ (1-r) S_n=a\left(1-r^n\right)$
$ S_n=\frac{a\left(1-r^n\right)}{1-r }$
Sum of $n$ terms of $a$ G.P
$= \biggl \{ n a \quad \text { if } \quad r=1 \\ \quad \frac{a\left(1-r^n\right)}{1-r}\quad \text { if } r\neq 1$
$r \neq 1$
$ S_n= \frac{a\left(r^n-1\right)}{r-1}$
Given a sequence of real numbers $\{a_n\}_{n=1}^{\infty}$ the expression
$ a_1+a_2+\ldots = \sum_{n=1}^{\infty} a_n \ \ \text {is called a series. }$
Sequence of Partial sum
$S_n=a_1+a_2+\cdots+a_n$
$\left(S_n\right)_{n=1}^{\infty}$
$\operatorname{Lt}_{n \rightarrow \infty}\quad S_n=S$
$S=\sum_{n=1}^{\infty} a_n$
$\sum_{n=1}^{\infty} a_n \ $ is summable (convergent)
Series of the form
$a+a r+\cdots+a r^{n-1}+\cdots \cdot =\sum_{n=1}^{\infty} a r^{n-1}$
Note that
$S_n=n a \quad \text { if }\quad r=1$
$\frac{a\left(r^n-1\right)}{r-1} \quad \text { if } \quad r \neq 1$
$ \text { Let } r=1$
$ S_n=n a $
$\underset{n \rightarrow \infty}{\operatorname{Lt}} S_n= \pm \infty \text {, depend on sign of} \ a $
Thus, $\sum_{n=1}^{\infty} a r^{n-1}$
is NoT convergent for $r=1$.
$\text { Let } r=-1$ $ a,-a, a, \cdots $
$ S_1=a $ $S_2=a+-a=0 $ $ S_3=a+-a+a=a$ $\vdots$ $\left(S_n\right)$ alternate between " $a$ " and “$0$”
$Lt_{n \rightarrow \infty} S _n$ does Not exist.
$\sum_{n=1}^{\infty} a r^{n-1}$ is NOT Convergent.
$r \neq 1, -1$
$S_n =\frac{a\left(1-r^n\right)}{1-r}$
$ =\frac{a}{1-r}-\frac{a r^n}{1-r}$
$|r|<1$
$\underset{n \rightarrow \infty}{\operatorname{Lt}} r^n=0 .$
$ \underset{n \rightarrow \infty} {Lt} \quad S_n=\frac{a}{1-r} .$
$ \sum_{n=1}^{\infty} a r^{n-1}=\frac{a}{1-r}$
${\text { Let }|r|>1}$
$\underset{n \rightarrow \infty}{\operatorname{Lt}}{|r|^n}=\infty$
$\therefore \ \underset {{n \rightarrow \infty} } {\operatorname{Lt}} \quad S_n \text { does not exist }$
i.e. $\sum_{n=1}^{\infty} a r^{n-1}$ is NOT convergent
$(|r|>1)$
A geometric series
$\sum_{n=1}^{\infty} a r^{n-1}$ is convergent if $|r| < 1.$
Further, for $|r| <1$
$\sum_{n=1}^{\infty} a r^{n-1}=\frac{a}{1-r}$
For $|r| \geqslant 1$
$\sum_{n=1}^{\infty} a r^{n-1}$
is not convergent .
Given two real numbers " $a$ " and “b”, there exists a number
A such that
$a, A, b$ forms A.P.
$A=\frac{a+b}{2}$
Arithmetic Mean
$A M(a, b)=\frac{a+b}{2}$
Given two numbers " $a$ " and " $b$ “, does there exist $G$ Such that
$a, G, b$ forms consecutive terms of a G.P. ?
$\frac{G}{a}=\frac{b}{G} \ \text{if} \ a, G, b$ are in $G . P$.
$ G^2=a b .$ $ G=\sqrt{a b} .$
In this case $a, G, b$ forms $a$ GP.
Given two positive numbers “$a$” and " $b$ “, the Geometric Mean (GM) of
" $a$ " and " $b$ " is defined as
$G M(a, b)=\sqrt{a b} $
Given two positive numbers $a$ and $b$ can we insert finality many numbers $G_1, G_2, \ldots G_n$
Such that
$a, G_1, G_2, \ldots G_n, b$ forms a G.P.?
$b$ is $(n+2)$ - th term
Let " $r$ " be common ratio.
$a r^{n+2-1}=b $
$ r^{n+1}=\frac{b}{a} $ $\Rightarrow r=\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$
$G_1 =a \cdot r =a \cdot\left(\frac{b}{a}\right)^{\frac{1}{n+1}}$
$G_2 =a \cdot r^2 =a\left(\frac{b}{a}\right)^{\frac{2}{n+1}}$