Geometric Progression

Geometric Progression

The series in which the ratio of any two successive numbers is constant is known as geometric progression. For example, consider a series containing the terms 8, 16, 32, 64, and so on. The ratio of any two consecutive numbers i.e(16 )/8=(32 )/16=(64 )/32=2. Similarly, consider a series having 180,60,20, and so on. The ratio of any two consecutive terms in this series is 1/3

From the compound interest that we receive on our deposits in the bank to the growth of population, many examples of geometric progression can be found around us.

Terms of Geometric Progression:

A geometric progression can be specified by two numbers namely the first term and the common ratio.

The first term, denoted by the letter a, is the first term of the geometric progression. For example, in a geometric progression containing the terms 5,15, 45, 135, and so on, the first term a = 5.

The common ratio is the constant ratio of any two consecutive terms in a geometric progression. We can calculate the common ratio r by dividing a term in the GP by its immediate preceding term. So, r =T_n/T_n-1.

In the above-mentioned series, the common ratio r =  15/5= 45/15=3

We can multiply the common ratio to a term in a GP and the resulting product is the immediately succeeding term in the GP.

So, a geometric progression can be expressed in the form of a, ar, ar², ar³, ar⁴……ar^n-1

Where a = the first term
r = common ratio
n = number of terms in the GP

Formulas for Geometric Progression:

n^th term of a Geometric Progression:

Consider a geometric progression having the terms 7, 35, 175, 875, and 4375.

The first term is 7 and the common ratio of the above series is 5.

Now, we can write the different terms in the above series as below.

T₁ = 7  = a = ar^1-1

T₂ = 35 = 7X3 = ar^2-1

T₃ = 175 = 5X3² = ar^3-1

So, the formula to calculate the n^th term of a GP is T_n = ar^n-1

To calculate the n^th term from the end of a GP where the last term is known, we can use the formula  T_n=l/r^n-1

Where l = last term

r = common ratio

n = number of terms from the end of the GP.

Sum of n terms in a geometric progression

The geometric progression can be expressed as a, ar, ar², ar³,…ar^n-1.

Now the common ratio can either be r = 1 or r>1 or r<1.

If r = 1, then S_n = a+a(1)+a(1)²+a(1)³+….+a(1)^n-1 = na.

If r >1, then S_n = (a ( rⁿ-1 ))/(r-1)

And when r < 1, S_n = (a (1- rⁿ  ))/(1-r)

Where,

a = first term

r = common ratio and

n = number of terms in the GP

Sum of an infinite geometric progression

A geometric progression that has an infinite number of terms is called an infinite geometric progression. The sum of an infinite geometric series can be calculated using the formula
S = a /(1-r) where r not equal to 0 and | r | <1

Formula summary

Solved Examples

1. X has Rs. 1,00,000 and he gives 2% of the money remaining with him to charity every day for 10 days. How much cash is left with him at the end of the 10^th day?

Solution:

Mr. X has Rs. 1,00,000 on Day 1. He would be left with 98% of the previous day’s balance after giving away 2% to charity. So he would have Rs. 98,000 on Day 2. Similarly, on Day 3, he would have Rs. 96,040 and so on. We can see that this creates a geometric progression with the first term a = 1,00,000 and the common ratio r = 98,000/1,00,0000.98

To find out how much money is left with him on Day 10, we can use the formula

T_n = ar^n-1

T₁₀ = 100000✕0.98^10-1

T₁₀ = 100000✕0.8337477

T₁₀ = Rs. 83,374.77

So, Mr. X would have Rs. 83,375 as on Day 10.

2. What are the first 10 terms of the GP 5, 20, 80, 320 …?

Solution:

In  the given question, we can see that the first term of the GP a = 5 and the common ratio r = 4. We have to find the first 10 terms of this GP and it is given that T₄ = 320. We know that if we multiply the common ratio with a term in the GP, the resulting product is the immediate succeeding term. So,

T₅ = T₄ x r = 320 x 4 =1,280

T₆ = T₅x r = 1280 x 4 =5,120

T₇ = T₆ x r = 5120 x 4 = 20,480

T₈ = T₇ x r = 20480 x 4 = 81,920

T₉ = T₈ x r = 81920 x 4 = 3,27,680

T₁₀ = T₉ x r = 3,27,680 x 4 = 13,10,720

3. Y joined a job on January 01, 2010, with a salary of Rs. 20,000 p.a. He gets an increment of 10% of the previous salary every year. Calculate the total amount earned by Mr. Y at the end of December 31, 2019.

Solution:

It is given that the salary earned by Mr. Y increases by 10% each year. This means the salary of Mr. Y for any given year is 110% or 1.10 times the salary he received in the previous year. So, we can say, the salary received by Mr. Y during this period forms a geometric progression with the first term 20,000 and the common ratio is 1.10. To calculate how much he had earned by December 2019, we need to calculate the sum of the GP for 10 years.

The sum of GP where r > 1 = S_n =(a ( rⁿ-1 ))/(r-1)

S₁₀ = (20000 ( (1.1)¹⁰-1 ))/(1.1-1)

S₁₀ = (20000 ( 1.5937 ))/0.1

S₁₀ = 3,18,740