# Arithmatic progression

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## 05 ARITHMATIC PROGRESSION

Arithmetic Progression:

An arithmetic progression is list of numbers in which the difference between any two consecutive terms is constant.

e.g   4,710,13,……   is A. P with    a=4      and   d=3

Each member of the sequence is called term.

• Common difference d can be zero, positive or negative.
• General form of AP is: a,a+d,a+2d
• A.P. can be finite or infinite.

General Term of A.P. THEOREM Let ‘a ’ be the first term and ‘d ’ be the common difference

PROOF

$\to {a}_{1}=a+\left(1–1\right)d......\left(1\right)$

Since each term of an A.P. is obtained by adding common difference to the preceding term. Therefore,

$\to {a}_{2}=a+\left(2–1\right)d$

………(2)

Similarly, we have

${a}_{3}={a}_{2}+d$
$\to {a}_{3}=a+\left(3–1\right)d$
$\to {a}_{4}=\left(a+2d\right)+d......\left(3\right)$
$\to {a}_{4}=a+3d$
${a}_{4}=a+\left(4–1\right)d....\left(4\right)$

Observing the pattern in equation (1), 2), (3) and (4), we find that

${a}_{n}=a+\left(n–1\right)d\phantom{\rule{0ex}{0ex}}$

Let there be an A.P. with first term ‘a ’ and common difference ‘d ’. If there are ‘m ’ terms in

$=a+\left(m–n+1–1\right)d$

Also, if I is the last term of the A.P., then  term from the end is the nth term of an A.P. whose first term is I and common difference is –d.

MIDDLE TERM(S) OF A FINITE A.P.

Let there be a finite A.P. with ‘ ’n terms whose first term is ‘a ’ and common difference is ‘d’.

If n is odd, then

${\left(\frac{n+1}{2}\right)}^{th}$

term is the middle term and is given by

$a+\left(\frac{n+1}{2}–1\right)d$

If n is even, then

are middle term given by

$a+\left(\frac{n}{2}–1\right)d$

respectively

Sum of n terms of AP:

Sum of first n terms of A.P. is given by

THEOREM The sum of n terms of an A.P. with first term ‘ a’ and common difference ‘ d’ is

${s}_{n}=\frac{n}{2}\left\{2a+\left(n–1\right)d\right\}$

Writing the above series in a reverse order, we get

Adding the corresponding terms of equations (1) and (2), we get

$2{s}_{n}=\left\{2a+\left(n–1\right)d\right\}+\left\{2a+\left(n–1\right)d\right\}+......+\left\{2a+\left(n–1\right)d\right\}$

Now,