ARITHMETIC PROGRESSION

A sequence of numbers is called arithmetic progression or arithmetic sequence where the difference between any two consecutive terms will be same along with sequence. 

Example : 

4, 8, 12, 16, 20, 24, 28, 32, 36,..................

8 - 4  =  4

12 -  8  =  4

16 - 12  =  4

20 - 24  =  4

In the above sequence of numbers, the difference between any two consecutive terms is 4 along the sequence. So, the above sequence of numbers is arithmetic sequence. 

Key Concepts

General term or n^th term of an arithmetic progression :

a_n  =  a₁ + (n - 1)d

where 'a₁' is the first term and 'd' is the common difference.  

Formula to find the common difference :

d  =  a₂ - a₁

Formula to find number of terms in an arithmetic progression :

n  =  [(l - a₁) / d] + 1

where 'l' is the last term, 'a₁' is the first term and 'd' is the common difference.

General form of an arithmetic progression : 

a₁, (a₁ + d), (a₁ + 2d), (a₁ + 3d),.........

Arithmetic Progression - Practice Questions

Question 1 : 

The first term of an A.P is 6 and the common difference is 5. Find the arithmetic progression its general term.

Solution :

a₁  =  6

d  =  5

Arithmetic Progression :

a₁, (a₁ + d), (a₁ + 2d), (a₁ + 3d),....................

Substitute 6 for a₁ and 5 for d. 

6, (6 + 5), (6 + 2 ⋅ 5), (6 + 3 ⋅ 5),....................

6, 11, 16, 21,....................

General Term : 

a_n  =  a₁ + (n - 1)d

Substitute 6 for a₁ and 5 for d. 

a_n  =  6 + (n - 1)5

a_n  =  6 + 5n - 5

a_n  =  5n + 1

Question 2 : 

Find the common difference and 15^th term of an arithmetic progression : 

125, 120, 115, 110,............... 

Solution :

a₁  =  125

a₂  =  120

Common Difference :

d  =  a₂ - a₁  

d  =  120 - 125

  d  =  -5

15^th Term :

a_n  =  a₁ + (n - 1)d

Substitute 15 for n, 125 for a₁ and -5 for d. 

a₁₅  =  125 + (15 - 1)(-5)

a₁₅  =  125 + (14)(-5)

a₁₅  =  125 - 70

a₁₅  =  55

So, the 15^th term of the given arithmetic progression is 55. 

Question 3 : 

Which term of the arithmetic progression 24, 23¼, 22½, 21¾, ……… is 3? 

Solution :

a₁  =  24

d  =  a₂ - a₁  =  23¼ - 24  =  -3/4

Let 3 be the nth term of the given arithmetic progression. 

Then, 

a_n  =  3

a₁ + (n - 1)d  =  3

Substitute 24 for a₁ and -3/4 for d. 

24 + (n - 1)(-3/4)  =  3

96/4 - 3n/4 + 3/4  =  3

(96 - 3n + 3) / 4  =  3

(-3n + 99) / 4  =  3

Multiply each side by 4. 

-3n + 99  =  12

Subtract 99 from each side. 

-3n  =  -87

Divide each side by -3.

n  =  29

Therefore, 3 is 29^th term in the given arithmetic progression.

Question 4 :

Find the 12th term of the arithmetic progression ;

√2, 3√2, 5√2,…………

Solution :

a₁  =  √2

a₂  =  3√2

Common Difference :

d  =  a₂ - a₁  

d  =  3√2 - √2

  d  =  2√2

12^th Term :

a_n  =  a₁ + (n - 1)d

Substitute 12 for n, √2 for a₁ and 2√2 for d. 

a₁₂  =  √2 + (12 - 1)(2√2)

a₁₂  =  √2 + (11)(2√2)

a₁₂  =  √2 + 22√2

a₁₂  =  23√2

So, the 12^th term of the given arithmetic progression is 23√2. 

Question 5 :

Find the 17th term of the arithmetic progression : 

4, 9, 14,…………

Solution :

a₁  =  4

a₂  =  9

Common Difference :

d  =  a₂ - a₁  

d  =  9 - 4

  d  =  5

17^th Term :

a_n  =  a₁ + (n - 1)d

Substitute 17 for n, 4 for a₁ and 5 for d. 

a₁₇  =  4 + (17 - 1)(5)

a₁₇  =  4 + (16)(5)

a₁₇  =  4 + 80

a₁₇  =  84

So, the 17^th term of the given arithmetic progression is 84. 

Question 6 : 

How many terms are in the following arithmetic sequence ?

7, 11, 15,……………483

Solution :

a₁  =  7

d  =  a₂ - a₁  =  11 - 7  =  4

Formula to find number of terms in an arithmetic progression :

n  =  [(l - a₁) / d] + 1

Substitute 7 for a₁ and 4 for d. 

n  =  [(483 - 7) / 4] + 1

n  =  [476 / 4] + 1

n  =  119 + 1

n  =  120

Therefore, there are 120 terms in the given arithmetic sequence. 

Question 7 : 

How many terms are there in the following Arithmetic progressions?

-16, -12, -8,……………16

Solution :

a₁  =  -16

d  =  a₂ - a₁

  d  =  -12 - (-16)

d  =  -12 + 16

d  =  4

Formula to find number of terms in an arithmetic progression :

n  =  [(l - a₁) / d] + 1

Substitute 16 for l, -16 for a₁ and 4 for d. 

n  =  [{16 - (-16)} / 4] + 1

n  =  [{16 + 16} / 4] + 1

n  =  [32 / 4] + 1

n  =  8 + 1

n  =  9

Therefore, there are 9 terms in the given arithmetic sequence. 

Question 8 : 

The 10^th and 18^th terms of an arithmetic sequence are 41 and 73 respectively. Find the 27th term

Solution :

Solving (1) and (2), 

a₁  =  5

d  =  4

27th Term : 

a₂₇  =  a₁ + (27 - 1)d

a₂₇  =  a₁ + 26d

Substitute 5 for a₁ and 4 for d. 

a₂₇  =  5 + 26(4)

a₂₇  =  5 + 104

a₂₇  =  109

So, the 27^th term of the arithmetic progression is 109. 

Question 9 :

Find n so that the n^th terms of the following two arithmetic progressions are equal. 

1, 7, 13, 19,................

and

100, 95, 90,................

Solution :

First Sequence : 

a₁  =  1

d  =  a₂ - a₁

  d  =  7 - 1 

d  =  6

Formula to find n^th term : 

a_n  =  a₁ + (n - 1)d

Substitute 1 for a₁ and 6 for d. 

a_n  =  1 + (n - 1)(6)

a_n  =  1 + 6n - 6

a_n  =  6n - 5

Second Sequence : 

a₁  =  100

d  =  a₂ - a₁

  d  =  95 - 100 

d  =  -5

Formula to find n^th term : 

a_n  =  a₁ + (n - 1)d

Substitute 100 for a₁ and 5 for d. 

a_n  =  100 + (n - 1)(-5)

a_n  =  100 - 5n + 5

a_n  =  105 - 5n

Given : n^th terms of the two arithmetic progressions are equal. 

Then,

6n - 5  =  105 - 5n

Add 5n and 5 to each side. 

11n  =  110

Divide each side by 11.

n  =  10

Question 10 :

How many two digit numbers are divisible by 13?

Solution :

Tow digit numbers :

10, 11, 12,………… 99

Find the two digit numbers divisible by 13. 

The first two digit number divisible by 13 is

13

The second two digit number divisible by 13 is

13 + 13  =  26

The third two digit number divisible by 13 is

26 + 13  =  39

................

The last two digit number divisible by 13 is 

91

Then, the two digit numbers divisible by 13 are 

13, 26, 39,.......................91

In the above sequence, the difference between any two consecutive term is 13. So, the above sequence is an arithmetic progression. 

Formula to find number of terms in an arithmetic progression :

n  =  [(l - a₁) / d] + 1

Substitute 91 for l, 13 for a₁ and 13 for d. 

n  =  [(91 - 13) / 13] + 1

n  =  [78 / 13] + 1

n  =  6 + 1

n  =  7

Therefore, there are seven two digit numbers divisible by 13.

Apart from the stuff given in this section, if you need any other stuff in math, please use our google custom search here.

You can also visit our following web pages on different stuff in math. 

WORD PROBLEMS

HCF and LCM  word problems

Word problems on simple equations 

Word problems on linear equations 

Word problems on quadratic equations

Algebra word problems

Word problems on trains

Area and perimeter word problems

Word problems on direct variation and inverse variation 

Word problems on unit price

Word problems on unit rate 

Word problems on comparing rates

Converting customary units word problems 

Converting metric units word problems

Word problems on simple interest

Word problems on compound interest

Word problems on types of angles 

Complementary and supplementary angles word problems

Double facts word problems

Trigonometry word problems

Percentage word problems 

Profit and loss word problems 

Markup and markdown word problems 

Decimal word problems

Word problems on fractions

Word problems on mixed fractrions

One step equation word problems

Linear inequalities word problems

Ratio and proportion word problems

Time and work word problems

Word problems on sets and venn diagrams

Word problems on ages

Pythagorean theorem word problems

Percent of a number word problems

Word problems on constant speed

Word problems on average speed 

Word problems on sum of the angles of a triangle is 180 degree

OTHER TOPICS 

Profit and loss shortcuts

Percentage shortcuts

Times table shortcuts

Time, speed and distance shortcuts

Ratio and proportion shortcuts

Domain and range of rational functions

Domain and range of rational functions with holes

Graphing rational functions

Graphing rational functions with holes

Converting repeating decimals in to fractions

Decimal representation of rational numbers

Finding square root using long division

L.C.M method to solve time and work problems

Translating the word problems in to algebraic expressions

Remainder when 2 power 256 is divided by 17

Remainder when 17 power 23 is divided by 16

Sum of all three digit numbers divisible by 6

Sum of all three digit numbers divisible by 7

Sum of all three digit numbers divisible by 8

Sum of all three digit numbers formed using 1, 3, 4

Sum of all three four digit numbers formed with non zero digits

Sum of all three four digit numbers formed using 0, 1, 2, 3

Sum of all three four digit numbers formed using 1, 2, 5, 6