Arithmetic Progression is one of the topics in mathematics that sometime looks difficult to students. However, the challenge faced by the students may result from the fact that they are not properly taught.

The good news is that I am here to unveil the secret behind problems involving Arithmetic Progression. First, do me a favour and get your writing materials and probably your calculator and let’s get started.

The first term of any A.P is represented by “a” while the common difference is represented by “d”. if a set of numbers 2,4,6,8 …… are given, the first term is 2.

The common difference is the new term minus the previous term i.e 4-2, = 2, 6-4 = 2, 8-6 = 2 …

Therefore a Sequence of an A.P can be written thus a, a+d, a+rd, a+3d a+4d…

From the above, it is noted that the number of common difference must be added to a particular term in order to get the next term is less than the term being sought.

Given that an *n*th term is given by T_{n}

Then T_{n} = a+ (n-1) d

This is the formula for finding an *n*th term of an arithmetic progression.

**See: **

Write down the formula in your book for reference purposes. Having known the general formula for solving arithmetic progression, let’s get started by solving problems. Hope you are ready!

**Solution to nth Term of Arithmetic Progression**

Example 1: find the 6th and 15th terms of the A.P whose first term is 6 and common difference is 7

Solution

Get your data ready i.e

The first term a = 6

Common difference d = 7

Recall: the formula for *n*th term of an arithmetic progression

T_{n} = a+(n-1)d

The 6th term means that n=6

Let substitute the terms we have

T_{6} = 6+(6-1)x7

So we get rid of the bracket, remember BODMAS?

T_{6} = 6+5×7

Up next! Multiplication

i.e 5 multiply by 7

T_{6} = 6+35 = 41

Therefore T_{6} = 41

Similarly,

The 15th terms means n = 15

So,

T_{15} = 6+14×7

T_{15} = 6+98 = 104

Therefore T_{15} = 104

Therefore, T_{6} and T_{15} are 41 and 104 respectively

Example 2: if 8, A, B, C, D 38, 44, 50 …. Provides and A.P. find the values of A,B,C and D

Solution

The terms given are 8, A, B, C, D, 38, 44, 50…

The first term is 8

Similarly, the second term is A

The third term is B

The fourth term is C

The fifth term is D

The sixth term is 38

Recall: *n*th term of A.P

T_{n} = a+(n-1)d

i.e T_{6} = a+(6-1)d

a + 5d = 38 —- (1)

Substitute a=8 into equation 1

8+5xd = 58

8+5d = 38

Collect like terms that means the whole numbers to the right hand side

5d = 38-8

You wonder why the sign changed?

N/B: when a member crosses the equal to sign, the sign changes

5d = 30

Divide both sides by

5 (the coefficient of d)

5d/5 = 30/5

d= 6

Second term means n = 2

Using the *n*th term formula we then substitute what we have

T_{n} = a+(n-1)d

T_{2} = a+(2-1)d = A

a+(1)xd = A

but a=8 and d=6

8+1×6=A

A=14

Solving for B, takes the first method

T_{3} = a+(3-1)d = B

a+2d=B

8+2×6=B

8+12=B

B = 20

Again, T_{4} = a+(4-1)d = C

a+3d=C

8+3×6=C

8+18=C

C=26

Lastly, T_{5} = a+(5-1)d = D

a+4d=D

8+4×6=D

8+24=D

D=32

Therefore values of A,B,C, and D are 14,20,26 and 32 respectively

Example

The first term of an A.P is 11 while the last term is 144. If the common difference is 7. Find the number of terms it has.

Solution

From the problem above, the first term is 11, that means our (a) = 11

The common difference (d) = 7

So using the *n*th term formular

T_{n} = a+(n-1)d

Hope you have not forgotten that?

Here we are told to solve for n (number of terms)

The last term in the problems means that we are giving to equate the value to 144

T_{n} = a+(n-1)d = 144

Substitute the value for a and d

11 + (n-1)x7 = 144

We then solve the bracket first right? Yes!

11+nx7-1×7=144

11+7n-7=144

Collect like terms

11-7+7n=144

4+7n=144

Take 4 to the right hand side but note the change of sign

7n=144-4

7n=140

Divide both sides by 7

n = 140/7

n = 20

Therefore, the number of terms it has is 20.

Example 4: The 9th term of an AP is 52 while the 12th term is 70. Find the 20th term.

Solution

In this problem, we are to make equation 1 and 2 and then solve it simultaneously

Using the *n*th term formular

T_{n} = a+(n-1)d remember n is the …. Of term i.e 7 and 12 respectively

T_{9} = 52 and T_{12} = 70

T_{9} = a+(9-1)d = 52 —– (1)

T_{12} = a+(12-1)d = 70

a+8d = 52 —– (1)

a+11d = 70 — (2)

Subtract equation 1 from 2 that means

a-a + 11d -8d = 70-52 (2-1)

0 + 3d=18

3d = 18

Divide both sides by the coefficient of d which is 3

d = 18/3

d = 6

Put d = 6 into equation 1 to obtain

a + 8d = 52

a + 8×6= 52

a + 48 = 52

Collect like terms i.e 48 cross the equal sign to the right hand side

a=52 -48

a=4

The 20th term means n=20

T_{20} = a+(20-1)d

= 4+19×6

4 + 144 = 118

Therefore the 20th term is 118

**CONCLUSION**:

With this foundation I have laid using these examples, I strongly believe you can strive other challenging problems in *n*th term of A.P

Make sure you solve the formula because in the subsequent problems involving A.P you are going to need it. Watch out for other topics in mathematics. Trust me it’s amazing, you will love it.