A sequence $(a_n)^{\infty}_{n=1}$ is called an AP, if

$a_{n+1}=a_n+d \ \ \forall n \geqslant 1$

i.e. If $\{a_ n\}_ {n=1}^{\infty}$ is an AP with $a_ 1 =a$, called first term and common difference d, then AP can be written in standard form as

$a, a+d, a+2 d, \cdots$

1) If $\left(a_n\right) _{n=1}^{\infty}$ is an AP, then sequence

$\left(b_n\right)_{n=1}^{\infty}$ obtained by $b_n=a_n+k \ \ \forall n$ is an AP.

3) If $\left(a_n\right)_{n=1}^{\infty}$ is an AP, then sequence obtained by multiplying each term of $\{a_n\}$ with a constant is again an AP.

$\left(a_n\right)_{n=1}^{\infty} \text { is an } A P$

$a_{n+1}-a_n=d$

$\text { Consider }\left(b_n\right)_{n=1}^{\infty} \ \ \forall n \in \mathbb{N}$

$b_{n+1}-b_n$

$\ =c a_{n+1}-c a_n$

$\ \ =c\left(a_{n+1}-a_n\right)$

$\quad =c d \ \ \forall n \in \mathbb{N}$

Let $a, b$ be given numbers. Can we "insert" a number $A$ such that $a, A, b$ are terms of an AP ?

Solution:

Given two numbers $a, b$ the number $\frac{a+b}{2}$ is called

Arithmetic Mean (AM) of $a$ and $b$.

$A M(a, b)=\frac{a+b}{2}$

Let $a$ and $b$ be two numbers given.

Can we insert numbers

$A_1, A_2\cdot\cdot\cdot A_n$,

So that

$a, A_1, A_2$.. $A_n, b$ are successive terms of a sequence which is AP?

$a, A_1, A_2 \ldots A_n, b$ are in AP.

The $(n+2)$ th term $=a+(n+2-1) d$

$b=a+(n+1) d$

$\Rightarrow d=\frac{b-a}{n+1}$

Then

$A_1=a+d=a+\frac{b-a}{n+1}$

$A_2=a+2 d=a+2\frac{(b-a)}{n+1}$

AP, $a_{n+1}-a_n=d$ $\forall n \in \mathbb{N}$

Std form $a, a+d, a+2 d +\cdot\cdot\cdot\cdot\cdot$

$n^{\text {th }}$ term $a_n=a+(n-1) d$

Given two numbers $a$ and $b$,

$\operatorname{AM}(a, b)=\frac{a+b}{2}$

Solution:

$a+(a+d)+\cdots+(a+(n-1) d)$

$\begin{array}{ccccc} 1 & 2 & 3 & 4 & 5 \\ \cdots & & 100 \\ 1 & 2 & 3 & 4 & 5 \end{array}$

100, 99, 98........

Total sum $=50 \times 101$

What is sum of $n$ terms of the AP $a, a+d, a+2 d+\cdot\cdot ?$

Solution:

$S_n =a+(a+d) +\ldots+a+(n-1) d$

$n^{\text {th }} \text { term } a+(n-1) d =l$

$S_n=a+a+d+\ldots+l$

$\begin{aligned} & S_n=l+l-d+l-2 d \cdot\cdot\cdot\cdot + a \\ & 2 S_n=(a+l)+(a+l) \cdot\cdot\cdot +(a+l) \\ & 2 S_n=n(a+l) \end{aligned}$

$\begin{aligned} & S_n=\frac{n}{2}(a+l) \\ & S_n=\frac{n}{2}[a+a+(n-1) d] \\ & S_n=\frac{n}{2}[2 a+(n-1) d]\end{aligned}$

A sequence $\left(a_n\right)_{n=1}^{\infty}$

is Said to be a Geometric progression $(G P)$ if no terms are zero and

$\begin{aligned} & \frac{a_{n+1}}{a_n}=r \ & \forall n \in \mathbb{N}\end{aligned}$

$r$ is called common ratio.

If first term is " $a$ " and common ratio is " r " then we can write GP (in std form) as

$a, r a, r^2 a, \ldots . .$

$\begin{gathered}n^{\text {th }} \text { term } \ =a r^{n-1}\end{gathered}$