# Maths – Theory & Concepts

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