Arithmetic Progression: A.P is a sequence if the difference of a term and it’s predecessor is always constant.

Eg:

5, 8, 11, _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ C.D = 8 – 5 = 3

4, 0, -4, _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _C.D = 0 -4 = -4

3, 5/2, 2, _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _C.D = 5/2 – 3 = – 1/2

5.0, 5.5, 6.0, 6.5 _ _ _ _ _ _ _ _ _ _ _ _ _ C.D = 5.5 – 5 = 0.5

For an A.P a1, a1 + d, a1 + 2d, a1 + 3d _ _ _ _ _ _ _ _ _ _ _

General Term: an = [a1 + (n – 1)d]

Here knowing each term is very important

a = value of the nth term

n = position of the nth term

a1 = first term

d = common difference

Beware of confusion between an & n. In question, 5th is the position whereas 5 is the value.

1 A. Important application of knowledge of an = a1 + (n – 1)d.

<1> Questions involving two unknown quantities

Tech → Use concept of Linear equations in two variables

E.g.: The 7th term of an AP is 20 and its 13th term

Q.

Sol a1 + 6d = 20 – ①

a1 + 12d = 32 – ②

Solve above equations: by equalising the co-efficient.

a1 + 6d = 20

– a1 ±12d = 32

– 6d = – 12

d = 2

Hence: a = 8

1 – B

*nth – term from end Concept*

Algorithm: (1) Make last term as first term

(2) Change sign of common difference

(3) Just use an = a1 + (n – 1)d

Example:

Find 10th term from end 4, 9, 14, …………………… 254

If we reverse the series.

a1 = 254 c.d = – (9 – 4) = -5

∴ a10 = a1 + 9d

= 254 + 9 (- 5)

= 254 – 45

a10 = 209

∴ Tenth term from end is 209

Practice Exercise :

AP-2 Find 10th term from end in following

4, 12, 20, 28, …………………. 996 Ans: 924

5, 15/2, 10 …………………………. 135/2 Ans: 45

Find the 8th term from the end of the A.P. 7, 10, 13, ……184 Ans: 163

Find the 10th term from the end of the A.P. 8, 10, 12,…, 126. Ans: 108

1 – C Tracing the A.P

Example

Q.1 an = 2 – 3n Find the A.P

Sol. a1 = 2 – 3(1) By putting different values of n.

a1 = -1 By putting different values of n you will get different tems.

a2 = 2 – 3 (2)

a2 = -4

Hence A.P is -1, -4, …..

1-D Ist negative term Type:

Concept: Inequations follow all rules of equations except one, that is if -ve term goes to other side for multiplication & division, then sign of inequation changes.

eg: 4 – 3x < 76

– 3x < 76 – 4

– 3x < 72 X > 72/(- 3)

eg:

Q.1 Which term of the progression 19, 181/5, 172/5, ………… is the first negative term.

Sol: Here a1 = 19 c.d = (- 4)/5,

Let an < 0

∴ a1 + (n – 1)d < 0

19 + (n – 1)(- 4/5) < 0

(n – 1)(- 4/5 ) < – 19 n – 1 > -19 x (- 5/4)

n – 1 > 95/4 + 1

n > 99/4

n > 243/4 ∴ 25th term is 243/4

② Sum of ‘n’ terms of an A.P

Sn = n/2 [ 2a + (n – 1)d]

Sn = n/2 [ a + l ] [ a = first – term, l = last term]

Here Sn = sum of n terms

n = no. Of terms

a = first term

d = common difference

Regarding Sn must note that

Questions could be of following types:

Ist type: This relation(Sn = n/2 [ 2a + (n – 1)d ) involves four quantities out of which they will give three and ask to find fourth one.

IInd type : Two quantities are not known, but they will give 2 conditions, hence make 2 equations and solve

IIIrd type : Sn = 4 n2 + 5n. Find the AP

Put n = 1

S1 = a1 = 4 +5 = 9 => a1 = 9

put n = 2

S2 = a1 + a2 = 26 => 9 + a2 = 26

∴ a2 = 26 – 9 => 17

∴ A.P is 9, 17, ………………………..

∴ C.D = 17 – 9 = 8

IV type : pth & qth Type Questions

Treat them as equations in two variables of p & q

Concept: 2. Find ‘a’ and ‘d’ in terms of p & q.

Then get the desired results.

Example

Q. The pth term of an A.P is ‘q’ and qth term is P. Find (p + q)th term.

a1 + (p – 1)d = q – ①

a1 + (q – 1)d = p – ②

Find a1 & d by equalising the co-efficient

Sol:-

a1 + (p – 1)d = q

-a1 ± (q – 1)d = p

d(p -1 – q + 1 ) = q – p

d (p – q) = q – pn ∴ d = (q-p)/(p-q) => d = -1

Put this in eq – ①

a1 + (p – 1) = q

a1 – p + 1 = q

a1 = p + q – 1

ap+q = a1 + (p + q)d

= p + q -1 + (p + q – 1) (-1)

= p + q -1 –p –q +1

= 0

Example-②

Q. If pm, qm, rm term of an A.P be x, y, z respectively show that

x(p – r) + y(r – p) + z(p – q) = 0. Let a1 & d be first term and common difference of an A.P

Sol: Given: a1 + (p -1)d = x

a1 + (q -1)d = y

a1 + (r -1)d = z

Now just put values of x, y, z in above expression.

(a1 + (p – 1)d)(p-r) + (a1 + (q-1)d)(r-p) + (a1 + (r-1)d)(p-q) = 0

Let’s multiply by not multiplying

Just follow anticlockwise rotations of p & q & r & s

Open first bracket first

a1p – a1r + p2d – pd –rpd + rd +……………….. = 0

Type III

Explain the double answer

Example:-

Q. How many terms of the A.P. -6, -11/2, -5 are needed to give the sum -25 ………..

Some more Important points in AP

Value of ‘n’ (position) in A.P cannot be a fraction or –ve.

Choosing an A.P

3- terms in AP a-d, a, a+d

4- terms in AP a-3d, a-d, a+d, a+3de

5-terms in AP a-2d, a-d, a, a+d, a+3d

To find Middle term/terms in A.P

Find ‘n’. If n is odd then ((n+1)/2)^th term will be the middle term, Find it.

If n is even then (n/2)^m term & (n/( 2 )+ 1)^mterm are middle terms. Just fine them.