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GETTING STARTED

Progressions are of three types – Arithmetic, Geometric and Harmonic

☛ Adjacent terms in Arithmetic Progression(AP) have constant difference between them

eg. 1,4,7,10,13,16

The above series with 6 terms is an example of AP. 

Let us see the difference between the terms.

  4-1   =  3

 7-4    = 3

10-7   = 3

13-10 = 3  

The difference is always 3. Hence the terms are in A.P

Note: The difference can also be negative. If the difference is negative, it will be an A.P where the values of terms will be decreasing

Adjacent terms in Geometric Progression have a  constant ratio between them

eg. 3,6,12,24,48,96

The above series has 6 terms. 

Ratio between terms:

6/3    = 2

12/6 = 2

24/12 =2

48/24 = 2

96/48 = 2

The ratio between the terms is always 2 in this case. Hence terms are in G.P

Note: The ratio can also be a fraction and less than 1. If the ratio is less than 1, it will be a G.P where the values of terms will be decreasing

If reciprocals of terms of a series are in AP, then the terms are said to be in Harmonic progression (H.P)

eg. 1,1/3,1/5,1/7,1/9,1/11

The above series has 6 terms. The reciprocals of the terms are: 1,3,5,7,9,11 and they are in A.P

Hence the series is in H.P

ARITHMETIC PROGRESSION:

☛FINDING n th term of A.P

Let us consider an arithmetic progression where the first term is a and the common difference is d. 

The A.P will be  a, a+d, a+2d….

nth term of the series will be 

[1st Term]+Interval x Difference 

n terms in sequence will have (n-1) intervals. 

 Hence, 

 nth Term will be

 [ a + (n-1) . d] 

 

Specialty of A.P:

The sum of first term and last term is same as sum of second term and second last term which is same as sum of third and third last term and so on, till the central value is reached.

 eg. Let’s consider a series:

1,5,9,13,17,21, 25

In this series there are 7 terms which is odd, let us check how the above property works.

   The sum of end terms

    = 1+ 25 = 26

   The sum of second and second last terms

   = 5 + 21 = 26

   The sum of third and third last terms

   = 9 + 17 = 26

   The sum of fourth and fourth last term (both are same)

  = 13 + 13 = 26

If we find the average of the series, it is 13 and middle term in the series.

 

 

 eg 2. Let us consider another series with even number of terms:

1,4,7,10,13,16

 a = 1 and d= 3

The sum of first and last term = 1+ 16 = 17

   The sum of second and second last term = 4 + 13 = 17

   The sum of third and third last term = 7 + 10 = 17

If we find average of the series, it is 8.5. Though it is not in the series, it is the central value between the central values of the A.P

The central values of AP are 7 and 10. Average of the two terms is 8.5 and hence is average of the series.

 

☛ Finding sum of A.P

 Sum of A. P =  No. of terms of series X Average Value of Series

Average Value of series     =  (First Term + Last Term)/2

=  (a+ [a+(n-1)d]) /2

= [2a + (n-1) d]/2

Therefore, Sum of A.P

= n/2 x [2a+(n-1).d]

 

HACK TO SOLVE A.P Problems:

☛ If the number of terms in A.P is even, say 4

Take the terms as  a-3d, a-d, a+d and a+3d 

The central term of the four terms is a and the common difference is 2d.

The reason for that is that the sum of all terms will be 4a.

☛ Similarly, if  number of terms in A.P is odd, say 5

Take the terms as  a-2d, a-d, a, a+d and a+2d

The central term of the four terms is a and the common difference is d.

The reason for that is that the sum of all terms will be 5a.

GEOMETRIC PROGRESSION

 Let the first term of G.P be a and the common ratio be r.

The terms of G.P, hence, are: a , ar, ar 2 , ar3   

☛ nth term of G.P =  a . r n-1

 

☛ Sum of terms in G.P:

Sum of terms in G.P

= a . (r n – 1)/(r-1) , for r > 1

=  a.(1- r n )/(1 -r), for  -1< r < 1

 

 The sum can also be expressed as 

(r.[Last Term]-[1st Term])/(r-1)

 

☛ Geometric mean of G.P is nth root of the product of all terms in the series

 

HACK TO SOLVE G.P Problems:

☛ If the number of terms in G.P is even, say 4 

Take the terms as  a/d3 , a/d, ad and ad3

The central term of the four terms is a and the common ratio is d2

The reason for that is that the product of all terms will be a4 and geometric mean will be a.

☛ Similarly, if  number of terms in G.P is odd, say 5 

Take the terms as  a/d2, a/d, a, a.d and a.d2

The central term of the five terms is a and the common ratio is d.

The reason for that is that the product of all terms will be a5 and the Geometric mean will be a.

☛ Infinite G.P

Let us consider a G.P
1, 1/2,1/4,1/8…..

The series is an infinite series because it does not terminate at all.

The sum of such a series will not be infinite but will converge at a particular value as the terms will keep on diminishing.

Sum of infinite G.P is given by a/(1-r), where -1< r < 1

 

For common ratio, r > 1, the sum will be infinite.

Eg. 1,2,4,8…

r is 2 in this case and hence the sum is infinite

 

Harmonic Progression

☛ If a, b, c are in harmonic progression, b is said to be the harmonic mean of a and c.

☛ In general, if x1, x2, ……xn  are in harmonic progression,

x2, x3,….xn – 1 are the (n-2) harmonic means between x1 and xn

 

☛ Calculating Harmonic Mean of Two Numbers

HM of two numbers let us say a and b is given as 2a.b/(a+b)

PROPERTY OF NUMBERS :

For any two positive numbers,

                                                                       A.M ≥ G.M ≥ H.M

PRACTICE PROBLEMS

1
Created on By venkateshj

Sequences & Series - Concept Refresher

1 / 20

Which term of the arithmetic progression 3,9,15,21... is 249

2 / 20

Find the sum of the series 1, 3/4, 9/16 ,27/64... ∞

3 / 20

Find the sum of all the two-digit numbers which leave a remainder of 3 when divided by 5.

4 / 20

Find the sum of the terms of the arithmetic progression whose first term, last term and common difference are 5, 104 and 9 respectively

5 / 20

Find the value of (30 x 1) + (29 x 2) + (28 x 3) + ….(1 x 30).

6 / 20

13 1 + 1/2 + 1/4 + 1/8... =

7 / 20

Find the sum of the first 8 terms of the following series 32(1) + 42(2) + 52(3) + 62(4)+.... .

8 / 20

Find the sum of the first 15 terms of the series 1, (1 + 2), (1 + 2 + 3), (1 + 2 + 3 + 4), (1 + 2 + 3 + 4 +5), …

9 / 20

If there are twelve arithmetic means inserted between 12 and 64, then find the seventh arithmetic mean.

10 / 20

Find the sum of the first fifteen terms of S, where S = 3/4 + 5/36 + 7/144 + 9/576 +...

11 / 20

The sum of four numbers in an ascending arithmetic progression is 120 and the product of the two non-extreme terms is 864. Find the smallest of the numbers.

12 / 20

If Sn is the sum of the first n terms of the series 60 + 58 + 56 + …., then find the maximum value of Sn.

13 / 20

If 7/16, m and 16/7 form a G.P., then what is the value of m, if m is a positive integer?

14 / 20

How many three-digit numbers less than 500 are divisible neither by 3 nor by 7?

15 / 20

The number of bacteria in a colony doubles every minute. If there are 1600 bacteria after 6 minutes, find the number of bacteria present initially.

16 / 20

Find the value of - 12 + 22 - 32 + 42 - 52 + 62 + …… - 292 + 302

17 / 20

Find the sum of all the two-digit numbers which leave a remainder of 2 when divided by 7.

18 / 20

The first three terms of an arithmetic progression are 3x+2, 5x -2 and 8x - 11. Find the sum of the first 15 terms.

19 / 20

The sum of ages of 5 children born at the intervals of 4 years each is 70 years. What is the age of the youngest child?

20 / 20

Lakshman is as much younger than Ram as he is older than Bharat. If the sum of the ages of Ram and Bharat is 70 years, what is definitely the difference between Lakshman's and Ram's age?

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