Coupon Accepted Successfully!


Progression and Series

In this topic we will study Sequence which is a succession of numbers formed according to some definite rule. 


Example : 1, 3, 5, 7,9 ……. is a sequence, here each term of the sequence can be obtained by adding 2 to the preceding term.  (All the odd integers)


There are two types of sequence .i.e Finite sequence and Infinite sequence. A sequence is said to be a finite or infinite depending upon number of terms.  For finite n we call it finite sequence and for infinite n we call it infinite sequence. If X = R then we call the sequence a real sequence and if X = C, we call it a complex sequence.

If {fn} be a sequence then an expression of the form f1 + f2 + …… + fn is called Series. In other word a series is the sum of the terms of the sequence.

Sequence and pattern

Not necessary, if the terms of a sequence follow certain pattern or rule then it is called progression. Here we are discussing following progression:

(i)     Arithmetic Progression (A.P.)

(ii)    Geometric Progression (G.P.)

(iii)    Harmonic Progression (H.P.) 

Arithmetic Progression (A.P.) :- A sequence is said to be in Arithmetic Progression when they increase or decrease by a constant number. This constant number is called the common difference (c.d.) of the arithmetic progression.

         1, 3, 5, 7, ………......             c.d. = 2

        –7, –3, 1, 5, 9 ………             c.d. = 4

        8, 5, 2, –1, –4 ………             c.d. = –3

Since we are adding ‘d’ (common difference) each time (negative value of d accounts here for subtraction) to get next number in the sequence. So by close inspection we can easily say nth term of an A.P. will be given by
tn = a + (n–1)d.

um of the first n term of an A.P.

Our next interest is to find the sum of first ‘n’ terms of an A.P. Let us denote it by Sn

Sn = {a} + {a + d} + {a + 2d} +…+ {a + (n–1)d}                  … (1)

we can write the above series in reverse way also.

 Sn = {a+(n–1)d} + {a+(n–2)d} + {a+(n–3)d} +……+ {a}         … (2)

Adding (1) and (2) 
2Sn = {2a(n–1)d} + {2a(n–1)d} + (2a+(n–1)d}+…+ {2a+(n–1)d}

 or 2 Sn = n {2a+(n–1)d}

 Sn = n × {2a + (n–1)d}  / 2                                                         … (3)

Once can also remember the above formula, in this way

Sn = first term+last term/2 × number of terms = ((a + (a + (n-1)d))/2) x n

Suppose, now you have to find out the sum from j-th term to k-th term of an A.P.

Sj.k = jth term+kth term/2 × (k – j + 1)

(We are also including here j-th term so number of terms = k – j + 1)

or     Sj.k =  a+(j–1)d+a+(k–1)d/2 (k – j + 1)

or     Sj.k = (k–j+1)/2 {2a + (j – k + 2)d}                                     … (4)


Let sum of n terms of a series be n (2n–1). Find its mth terms.


Let Sm and Sm–1 denote the sum of first m and (m – 1) terms respectively.
Sm = T1 + T2 + T3 + ……. + Tm–1 + Tm
Sm = T1 + T2 + T3 + ……. + Tm–1
Sm – Sm–1 = Tm
⇒ Tm = (m(2m–1))–(m–1)(2(m–1)–1))
        = (2m2 – m)–(2m2 – 5m + 3)
        = 4m – 3



The sums of n terms of two A.P.’s are in the ratio 3n + 2: 2n+3. Find the ratio of their 10th terms.


Let a, a + d, a + 2d, a + 3d, ……………
A, A + D, A + 2D, A + 3D, ………………
be two A.P.’s n/2[2a+(n–1)d]/n/2[2A+(n–1)d] = 3n+2/2n+3       (given)
⇒ a+n–1/2d/A+n–1/2D = 3n+2/2n+3
⇒ To get the ratio of 10th terms put n–1/2 = 9 or n = 19
⇒ a+9d/A+9D = 3(19)+2/2(19)+3 = 59/41


Arithmetic mean between two numbers

When three numbers are in A.P., the middle term is called arithmetic mean (AM) of the two others.

 Let a and b are two numbers.

 And A be the arithmetic mean between two numbers.

 So a A b are in A.P.

 or,    A – a = b – A

 or,    2A = a + b

          A =a+b/2

 So, in general

 A.M. = Sum of the numbers/2                                                     …… (5)


Similarly we can find the two arithmetic means between two number. A1 and A2 are two arithmetic mean between two member a and ‘b’ then a, A1, A2 b are in A.P.

 Now using the formula, tn = a+(n–1)d

 or,    d = b–a/3 [ here n = 4]

 so,    A1 = a + d = a +b–a/3 = 2a+b/3

 and   A2 = a + 2d = a + 2(b–a)/3 = a+2b/3


Basic Principle of AP :

 (1)    Addition/Subtraction of constant number to each term of an A.P. also results an A.P.

         e.g. suppose a1, a2, a3, …, an are in A.P.

         then a1 + k, a2, k, a3 + k ……, an + k and

         a1 –k, a2 – k, a3 – k, ……, an–k will also be in A.P.

         where k  R

 (2)    Multiplication/Division by a constant number to each term of an A.P. also results an A.P.

         e.g. suppose a1, a2, a3, …, an are in A.P.

         then ka1, ka2, ka3 ……, kan and

         a1/k, a2/k, a3/k will also be in A.P.

         where k ? R and k ≠ 0

 (3)    Addition/Subtraction of two A.P.’s also results an A.P.

         e.g. suppose a1, a2, a3, …, an and

         b1, b2, b3, …, bn, are in A.P.

         then a1+b1, a2+b2, a3+b3, ……, an+bn. (c.d.) = (a2- a1) + (b2- b1)

         a1–b1, a2–b2, a3–b3, ……, an–bn will also be in A.P. and the common difference will be            (a2- a1) – (b2- b1)

(4)    Reversing the order of an A.P. also results an A.P.

         e.g. Suppose a1, a2, a3, ……, an are in A.P.

         then an, an–1, ……, a3, a2, a1 will also be in A.P.


For simplicity assume numbers in A.P. in the following format. If three numbers are in A.P assume them as: α–ß, α, α+ß (here first term is α–ß and c.d. is ß). For four numbers in A.P.: α–3ß, α–ß, α + ß, α+3ß (here first term is α-3ß and c.d. is 2ß). Five convenient numbers in A.P. are α–2 ß, α–ß, α–ß, α, α + ß, α + 2 ß.(c.d. = ß)
 Basically sum of the above sets of variables eliminate one variable which make easy to solve the problems.


 1.     If nth term of an series is tn = A n + B, then

         tn - tn–1 = A n + B – A(n–1)–B

             = A = constant

         i.e. the series is in A.P.


2.     If tn = A n2 + B n + c, then

         tn – tn–1 = A(n2) + Bn + C – (A(n–1)2 + B(n–1) + c)

                       = A (2n + 1) + B = (n)     (say)  Un – Un–1 = 2A

which is constant, i.e. if the difference of difference of the terms of a series is constant, then the nth term is quadratic in n.


3.     Similarly, if the 3rd degree difference between the terms of a series is constant then the nth term of the series is a cubic in n i.e.

         tn = An3 + Bn2 + Cn + D

         and so on.


 • If a fixed number is added to (or subtracted from) each term of a given A.P., then the resulting sequence is also an A.P. with the same common difference as that of the given A.P.

 • If each term of an A.P. is multiplied by a fixed number (say k) (or divided by a non-zero fixed number), the resulting sequence is also an A.P. with the common difference multiplied by k.

 • If a1, a2, a3, …… and b1, b2, b3, …… are two A.P.’s with common differences d and d’ respectively then a1 + b1, a2 + b2, + a3 + b3, … is also an A.P. with common difference d + d’.

 • If we have to take three terms in an A.P., it is convenient to take them as a – d, a, a + d. In general, we take a – rd, a – (r – 1)d, …., a – d, a, a + rd in case we have to take (2r + 1) terms in an A.P.

 • If we have to take four terms, we take a – 3d, a – d, a + d, a + 3d. In general, we take a –(2r–1)d, a –(2r–3)d,…, a–d, a+d, …, a + (2r–1)d, in case we have to take 2r terms in an A.P.

 • If a1, a2, a3, ……, an are in A.P., then a1 + an = a2 + an–1 = a3 + an–2 = …… and so on.

Arithmetic Mean(s)

• If three terms are in A.P., then the middle term is called the arithmetic mean (A.M.) between the other two i.e. if a, b, c are in A.P. then b =  a+c/2 is the A.M. of a and c.

• If a1, a2, ……, an are n numbers, then the arithmetic mean (A) of these numbers is A = 1/n (a1 + a2 + a3 +…+ an).

• The n numbers A1, A2, ……, An are said to be A.M.’s between the numbers a and b if a, A1, A2, ……, An, b are in A.P. If d is the common difference of this A.P., then b = a + (n + 2 – 1)d d = b–a/n+1.

 A1 = a + b–a/n+1 = na+b A2 = a + 2(b–a)/n+1 ,…, An = a + n(b–a)/n+1 = a+nb/n+1.


If 1st and 2nd terms of an A.P. are 1 and –3 respectively, find the nth term and the sum of the first n terms.


 1st term = a, 2nd term = a + d where a = 1, a + d = –3,
        ⇒ d = –4(common difference of A.P.)
        ⇒ an = a + (n – 1)d = 1 + (n–1)(–4) = 5 – 4n.
       and Sn = n/2{a + an} = n/2{1 + 5 – 4n} = n(3 – 2n).



If 6 arithmetic means are inserted between 1 and 9/2, find the 4th arithmetic mean.


  Let a1, a2, a3, a4, a5, a6 be six arithmetic means
        ⇒ 1, a1, a2, ……, a6, 9/2 will be in A.P.
        Now, 9/2 = 1 + 7d ⇒ 7/2 = 7d ⇒ d = 1/2.
        Hence a4 = 1 + 4 (1/2) = 3.

Test Your Skills Now!
Take a Quiz now
Reviewer Name