How To Solve Questions on Arithmetic Progression (A.P)

Arithmetic Progression is a popular questions in both objective and theory questions in Mathematics.

In this tutorial, you will learn the formula for finding the nth of an A.P, how to use the formula and how use equations to solve arithmetic progression.

BONUS: A link to real past questions on arithmetic progression is included at the end of this tutorial.

Meaning of Arithmetic Progression

An arithmetic progression is a sequence in which the increase or decrease between the consecutive terms are the same throughout the series.

The increase or decrease in an arithmetic progression is caused by a value called “common difference” which can be either positive or negative.

EXAMPLE 1:

A.P = 2, 4, 6, 8, 10…

The common difference in an arithmetic progression is gotten by subtracting the first number from the second number

And in the example above,

  • 4 – 2 = 2
  • 6 – 4 = 2
  • 8 – 6 = 2
  • 10 – 8 = 2

Therefore, the common difference in the arithmetic progression above is 2

EXAMPLE 2:

In the AP = 24, 15, 6, -3, -12…

The common difference is -9 from subtracting the first number from the second number.

  • 15 – 24 = -9
  • 6 – 15 = -9
  • -3 – 6 = -9
  • -12 – (-3) = -12 + 3 = -9

FORMULA FOR FINDING THE nth TERM OF AN ARITHMETIC PROGRESSION

In any arithmetic progression, each member is called a term. The first term is represented by letter “a” while the common difference is represented by letter “d”

Therefore, a sequence of an A.P can be written as,

  • a = first term
  • a + d = second term
  • a + 2d = third term
  • a + 3d = forth term
  • a + 4d = fifth term
  • a + 5d = sixth term
  • a + 6d = seventh term
  • a + 7d = eighth term
  • a + 8d = ninth term
  • a + 9d = tenth term

You will notice from the above that the number of common difference (d) that must be added to a particular term (a) in order to get the next term is one number less than the term sought.

FORMULA:

Using Tn to represent the nth term:

Tn = a + (n – 1)d

APPLYING THE FORMULA FOR FINDING THE nth TERM OF AN ARITHMETIC PROGRESSION

QUESTION 1:

Find the 20th term of an A.P if the first term is 11 and the common difference is 3

SOLUTION:

:- Tn = a + (n – 1)d

= 11 + (20 – 1)3

= 11 + (19)3

= 11 + 57 = 68

T20 = 68

QUESTION 2:

The 16th term of an A.P is 93. Given that its common difference is 6, find the first and the 28th terms.

SOLUTION:

16th is T16 which is given as 93. And the common difference (d) is given as 6

T16 = a + 15d

Then T16 = a + 15(6)

= a + 90

:- a + 90= 93

a = 93 – 90

a = 3 :- the first term is = 3

To find the 28th term

T28 = a + 27d

= 3 + 27(6)

= 3 + 163 = 165

USING EQUATION TO SOLVE ARITHMETIC PROGRESSION

QUESTION 1:

If 6, p and 14 are consecutive terms in an Arithmetic progression(A.P), find the value of P. [WAEC, 2019]

SOLUTION:

6 = a………..equation 1

P = a + d…….equation 2

14 = a + 2d…….equation 3

Since we already know the first term to be 6, we have to find the common difference (d) next.

And we can find our common difference from equation 3 since it is the only equation that has equal of 14.

Therefore, a + 2d = 14

6 + 2d = 14

Collect like terms

2d = 14 – 6

2d = 8

Divide both sides by 2

d = 4

Now that we know the common difference (d) as 4, we can now find the value of P.

To find the value of P

P = a + d

P = 6 + 4

P = 10

ARITHMETIC PROGRESSION THAT LEADS TO SIMULTANEOUS EQUATION

QUESTION:

The 14th term of an A.P is 96 while the 25th term is 173. Find the 19th term?

SOLUTION

14th term: Tn = a + 13d

25th term: Tn = a + 24d

Therefore,

a + 13d = 95………….(i)

a + 25d = 173………..(ii)

The question has formed simultaneous equation. Go ahead and subtract equation (i) from equations (ii)

a + 13d = 96………(i)

a + 24d = 173…….(ii)

:11d = 77

Divide both side by 11

d = 7

To find the first term “a”

Substitute for d in equation (i)

a + 13d = 96

a + 13(7) = 96

a + 91 = 96

a = 96 – 91

a = 5

Therefore, the first term (a) is = 5, and the common difference (d) = 7

To look for the 19th term

19th term: T19 = a + 18d

= 5 + 18(7)

= 5 + 126

T19 = 131

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