Some students run away from Geometric Progression problems, simply because they find it difficult to differentiate it from Arithmetic Progression.

They believe the common difference is the same thing as the common ratio so they get tired when solving and they make a lot of mistakes.

In this lesson, I will show you the difference between common difference used in A.P and common ratio used in G.P. and as well give you a breakdown of solution to problems in the *n*th term of a G.P.

At the end of the lesson, you will come to discover how simple problems involving the *n*th term of a G.P. are.

A Geometric Progression also known as exponential Sequence) is a Sequence in which any two consecutive terms differ by a constant factor.

The common ration is densted by ‘r’ and the first term ‘a’. but in A.P common difference is used if a set of numbers 6,18,54,162…are given

The common ration = 18/6 = 3, 54/18 = 3, 162/54.

While the common difference is subtraction i.e 18-6, 54-18….

Common ratio is division, since the common ratio ‘r’ is known and the first term ‘a’ is also known, the process of finding the second term is by multiplying the first term by the common ratio.

That is a, ar, ar^{2}, ar^{3}…. ar^{n-1}

Therefore, an *n*th term of a G.P. is given as

T_{n} = ar^{n-1}

**N/B:** T_{n} = *n*th term, r=common ration, a = first term

Put this formula down and also save it in your brain. You will really need it.

Having known some important things about *n*th term of a G.P. like ho to obtain the common ratio and the formula for *n*th term of a G.P. let me eliminate your fear of G.P problems by solving some examples.

Hope we are good to go!!

**Solutions to Geometric Progression Questions**

** Example 1:** the 6th term of a G.P is 1215. Given that the common ration is 3, find the 9th term.

**Solution**

In this problem common ratio is r=3, n=6

The 6th term = 1215

Using the formular for the *n*th term of G.P T_{n} = ar^{n-1} we substitute what is given in order to obtain the first term ‘a’

T_{6} = ax3^{6-1} = 1215

ax3^{5} = 1215

rem: 3^{5} = 3x3x3x3x3

** Example 2:** The third term of a GP is 9 and the fifth term is 16. Find the 4th term.

**Solution**.

In the above problem given, the third term that is T_{3} = 9 and T_{5} = 16

We are required to find two equations and then solve them simultaneously.

Using the formular T_{n} = ar^{n-1}

T_{9} = ar^{3-1} = 9

ar^{2} = 9 — (1)

T_{5} = ar^{5-1} = 16

ar^{4} = 16 — (2)

Divide equation two by 1

ar^{4}/ar^{2} = 16/9

** Examples 3**: if 3,A,B, 192 are consecutive terms of a GP find the value of A and B.

**Solution**

In the problem given,

The first term a=3

The second term is A

The third term is B

The fourth term is 192

That means

a=3, T_{4} = 192

using the formular for *n*th term of G.P

T_{n} = ar^{n-1}

T_{4} = 3xr^{4-1} = 192

Multiply 3 by r^{3}

3r^{3} = 192

Divide both sides by 3

3r^{3}/3 = 192/3

r^{3} = 64

Take cube rook of both sides

**N/B:** Cube rook can be express as 1/3 (2) Cube root of a number means a number that can multiply itself three times to give that number e.g 8, the cube rook of 8 is 2 means 2x2x2 = 8 (3) Cube root of 64 = 4

How? 4x4x4 = 64

r^{3×1/3} = 4

r = 4

Since we both have gotten a and 8, let solve A and B

** Example 4**: the first and third term of a GP are 5 and 80, respectively, what is the 4

^{th}term?

**Solution**

In the problem above

The first term a =5

The third term T_{3} = 80

Using the formula

T_{n} = ar^{n-1}

T_{3} = ar^{3-1 } = ar^{2} = 80 —– (1)

Put a=5 into equation (1)

5xr^{2} = 80

Multiply 5 by r^{2}

5r^{2} = 80

Divide both sides by 5

The 4th term, n=4

T_{4} = ar^{4-1 }= ar^{3}

Put a=5 and r=4

T_{4} = 5×4^{3 }= 5x4x4x4

T_{4} = 5×64

T_{4} = 320

Recall: Second Term is A, that is T_{2} = A

T_{2} = ar^{2-1} = A

A = ar

Substitute a=3 and r=4

A = 3×4 = 12

Therefore A=12

Again, third Term is B

T_{3} = ar^{3-1} = B

B=ar^{2}

B= 3x(4) = 3x4x4 = 3×16

B = 48

Therefore A=12 and B=48

__Example 5:__

A G.P has its 3rd and 7th term as 45 and 3645 respectively what is the product of its 2^{nd} and 4^{th} term?

Solution

Form the problem

The 3rd term T_{3} = 45

The 7th term T_{7} = 3645

Using the formular

T_{n} = ar^{n-1}

T_{3} = ar^{3-1 } = 45

ar^{2} = 45 —– (1)

T_{7} = ar^{7-1 } = 3645

ar^{6} = 3645 —– (2)

divide equation 2 by 1

ar^{6}/ar^{2} = 3645/45

rxrxrxrxrxr/rxr = 81 How?

Its because rxrxrxr = r^{4}

r^{4} = 81

substitute r = 3 into equation 1

ar^{2}= 45

a x (3)^{2} = 45

ax3x3 = 45,axa = 45

9a = 45

Divide both sides by 9

9a/9a = 45/9 a =5

Therefore a = 5 and r =3

And term T_{2}

T_{2} = ar^{2-1} = ar

Substitute a = 5 and r=3

T_{2} = 5×3 = 15

T_{4} = ar^{3-1} = ar^{2}

T_{4} = 5×3^{2} = 3×3 = 5 x 9

= 45

Product means multiplication so the product of the 2nd and 4th terms is T_{2}xT_{4} = 15×45 = 673

No doubt you have learnt a lot. What you have learned today point to the fact that you can solve as many as possible problems involving *n*th term of G.P. and that problems involving *n*th term of G.P are not difficult and should not scare you. I believe the fear has been eliminated.

**See: **

Remember to take note of the formulas for today’s lesson.