Geometric progression(G.P)

In arithmetic progression the next term is obtained by adding or subtracting the common difference to or from the previous term whereas when we are multiplying OR dividing the previous term with any non-zero constant which is called as the common ratio then the sequence thus obtained is said to be Geometric progression (G.P.). Now the common ratio can be either positive or negative also, the series may be finite series or infinite series

G.P and series

Consider the following series 

1, 2, 4, 8, 16, 32 …….

Here, first term = a = 1 

Now If we carefully observe 1*2 = 2 = second term (a2)

2 *2 = 4 = third term (a3) 

4*2 = 8 = fourth term (a4)

8*2= 16 = fifth term and so on 

The constant term 2 that we are multiplying with each term here in this series is called common ratio and is denoted by r. 

Basically, if we write it in terms of a and r our series would look something like below

a, ar, ar2, ar3, ar4 and so on till arn-1 

Now how do we get the common ratio? Well that’s easy, you take a ratio of two consecutive terms. 

Common ratio = ar/a = r i.e., 2 /1 = 2  

Or to express it in a more sophisticated way one can say 

Common ratio = an+1an , where n can be any integer.

The general form is an = arn-1 

Let’s just say I need a 4th term then using an = arn-1 we can find out. Let’s try using this formula, we know that common ratio = r = 2 here in this case. Then a4 = 1 * 24-1 = 8, which is true.

Sum of n terms of G.P

Only 3 cases are possible here. Common ratio (r) can be 1 or less than 1 or greater than 1. 

• r = 1 

sum (s ) = a + ar + ar2 + …. +arn-1 —–(1)

put r = 1 we get, 

s = a + a +a +….. + a 

s = n *a 

• if r ≠1 or r greater than 1

multiply eq (1) by r on both sides we get , 

rs = ar +ar2 + ar3 +…….+ arn-1+ arn ——(2) 

subtract eq ( 1 ) and ( 2 ) , we get 

rs – s = arn – a 

s(r -1) = arn – a

s = a(rn – 1) r-1 

• if r ‹ 1

s = a(-rn+1) -r+1

Geometric mean(GM)

Consider the following series 

5 , 25 , 125 , 625

Common ratio(r) here is 25/5 = 5 

a = 5 

then using the formula an = arn-1, we can get 4th term as 

a4 = 5 * 53 = 5*125 = 625

let try to find sum of all the four terms using the formula s = a(rn – 1) r-1

s = 5(624)/4 = 5*156 = 780

in the above series consider the first 3 terms: 5 , 25 , 125

it can be represented as p , q ,r 

now the series to be in Geometric mean 

it should be in the form of  pq= qr

q2 = pr or q = pr

This is the geometric mean. 

Now coming back to our original series , suppose we need to insert such between 5 and 125 that the series gets into G.P then we will use the just now derived formula to insert such number

And the number would be 125*5 = 25

Relationship between AM and GM

Let a and b be two numbers then, 

AM = a+b2  and GM = ab

AM – GM  = a+b2 –  ab = a+b-2ab2 = a-b22  and this is always greater than 0 

Therefore AM is always equal to or greater than GM

Important results on the sum of special sequences

• Sum of the first n natural numbers

∑n = 1+2+3….+n = n(n+1)2

• Sum of the squares of first n natural numbers

∑n2 = 12+22+32…….+n2 = n(n+1)(2n+1)6

• Sum of cubes of first n natural numbers

∑n3 = 13+23+33….+n3 =[ n(n+1)2 ]2

Example

1. If there are (2n + 1) terms in an A.P., then prove that the ratio of the sum of odd terms and the sum of even terms is (n + 1) : n

Solutions: Let a be the first term and d the common difference of the A.P. Also let S1

be the sum of odd terms of A.P. having (2n + 1) terms. Then

S1 = a1 + a3 + a5 + … + a2n+1

S1  = (n+1)2(a1 + a2n+1)

S1  = (n+1)2[ a+a+2n+1-1d]

= (n + 1) (a + nd)

Similarly, if S2 denotes the sum of even terms, then

S2  = n2 [2a + 2nd] = n (a + nd)

s1s2= n+1n

1. At the end of each year the value of a certain machine has depreciated by 20% of its value at the beginning of that year. If its initial value was Rs 1250, find the value at the end of 5 years

Solutions: After each year the value of the machine is 80% of its value the previous year so at the end of 5 years the machine will depreciate as many times as 5.

Hence, we have to find the 6th term of the G.P. whose first term a1 is 1250 and common

ratio r is 0.8.

Hence, value at the end 5 years = a r5 = 1250 * (0.8)5 = 409.6

Conclusion 

A sequence of numbers is called a Geometric progression if the ratio of any two consecutive terms is always constant. The behaviour of a geometric sequence depends on the value of the common ratio. If the common ratio is:

1. Greater than 0 , the terms will all be the same sign as the initial term.
2. Less than 0 , the terms will alternate between positive and negative.
3. Greater than 1, there will be exponential growth towards positive or negative infinity (depending on the sign of the initial term).