# Geometric Progression : Basic Concept, Formula for Grade 11 Math

In this post we will try to understand the basic concept of geometric progression. Apart from the concept we will also try to understand different formulas and solved examples which are relevant from the point of view of Grade 11 Mathematics syllabus for CBSE/NCERT or American Math curriculum

## What is Geometric Progression?

Consider the following series
a,\ ar,\ ar^{2} ,\ ar^{3} ,\ ar^{4} ,\ ar^{5} ,\ ar^{6} ,\ ar^{7} \ .\ .\ .\ .\ .\ .\ .

This series follows the rule that the ratio of every term with the preceding term is r

For Example
\frac{ar}{a} \ =\ r \\\ \\ \frac{ar^{2}}{ar} \ =\ r \\\ \\ \frac{ar^{3}}{ar^{2}} \ =\ r \\\ \\ \frac{ar^{4}}{ar^{3}} \ =\ r \\\ \\

You can observe that the value of each ratio is same ( r ).
Hence if any sequence which follows the pattern such that the ratio of element with the preceding element is same throughout the series then the sequence is known to be in Geometric Progression

Example 01
2, 4, 8, 16, 32, 64, 128 . . . . . . . . . . . .

This series is Geometric Progression as:
\frac{4}{2} \ =\ 2 \\\ \\ \frac{8}{4} \ =\ 2 \\\ \\ \frac{16}{8} \ =\ 2 \\\ \\ \frac{32}{16} \ =\ 2

You can see that the ratio of each element with the preceding element is same (r = 2)

Example 02
\frac{1}{9} ,\ \frac{-1}{27} ,\ \frac{1}{81} ,\ \frac{-1}{243} ,\ \frac{-1}{729} ,\ .\ \ \ .\ \ .\ .\ .\ \ .\ .\

This series is again of Geometric Progression as:
\frac{\frac{-1}{27}}{\frac{1}{9}} \ =\ -\frac{1}{3} \\\ \\ \frac{\frac{1}{81}}{\frac{-1}{27}} \ =-\ \frac{1}{3}\\\ \\ \frac{\frac{-1}{243}}{\frac{1}{81}} \ =-\ \frac{1}{3}\

Again the ratio of each element with the preceding element is the same (r= -1/3)

## Formula for Geometric Progression (GP)

1. Finding nth term of GP

tn\ =\ ar^{n-1}\

Where
a = First Term
r = Common Ratio

2. Sum to nth term of GP

Sn\ =\ \frac{a\left( r^{n} -1\right)}{r-1}\
Use this formula when, r > 1

Sn\ =\ \frac{a\left( 1-r^{n}\right)}{1-r}\
Use this formula when r < 1

3. Geometric Mean

If two numbers are given a and b.

The geometric of these two number is given as: Gm\ =\sqrt{a\ b\ } \

The Gm calculated is such that the series: a, Gm, b is in Geometric Progression

## Solved Questions of Geometric Progression

(01) Find the 10th term of the following series
5, 10, 20, 40, 80, 160 . . . . . . . .

Solution
You can observe the given series is in Geometric Progression as ratio of each element with the preceding element is 2.

So we will apply the formula
Tn\ =\ a\ r^{n-1}\

Where;
a = First term of series
r = common ratio
n = 10

Tn\ =\ 5\ \times \ 2^{10-1} \\\ \\ {Tn\ =\ 5\ \times \ 2^{9}} \\\ \\

Hence 5 \times \ 2^{9} is the 10th term of the given Geometric Progression

(02) Which term of the G.P., 2,8,32, … up to n terms is 131072?

Solution
In the given Geometric Progression
a = 2
r = 8/2 = 4
Tn = 131072

Using the formula
Tn\ =\ a\ r^{n-1}\

131072\ =\ 2\ \times \ 4^{n-1}\\\ \\ 65536\ =4^{n-1}\\\ \\ 4^{8} \ =\ 4^{n-1}\\\ \\ 8=n-1\\\ \\ n=9\\\ \\

Hence. 131072 is the 9th term of the G.P

(03) Find the sum of 5th term of given G.P
1+\frac{2}{3} \ +\ \frac{4}{9} +\ \frac{8}{18} \ +\ .\ .\ .\ .\

Solution
In the given Geometric Progression
a = 1
r = 2/3

Using the formula
Sn\ =\ \frac{a\left( 1-r^{n}\right)}{1-r}\\\ \\ S5=\frac{1\ \left( 1-\left(\frac{2}{3}\right)^{n}\right)}{1-\frac{2}{3}} \\\ \\ S5=\frac{\left( 1-\left(\frac{2}{3}\right)^{5}\right)}{1-\frac{2}{3}}\\\ \\ S5\ =\ 3\ \times \ \frac{211}{243}\\\ \\ S5\ =\ \frac{211}{81}