Sum of an Infinite Arithmetic Geometric Series

Arithmetic and geometric progression or AGP is a type of progression where every term represents its product of the terms. Hence, both these progressions are summed up together to form AGP. In simple words, arithmetic and geometric series are constructed by multiplying corresponding terms of geometric and arithmetic progression. For example, you can say 13 + 26 + 39 + 412 …… so on. Here the numerator part represents the arithmetic progression, whereas the denominator stands for geometric series.  

The general term of AGP

We can obtain the nth term by multiplying all the corresponding terms of arithmetic and geometric progression. 

In general form, it can be represented as:

a, (a+d)r, (a+2d)r²,………., [a+(n−1)d]rn-1

• Here, a is for the initial value, d is for the common difference, and r is for the ratio of terms. 
• Where (n)th term represents AP. (N)th term can be represented by a T.

Then the formula of AGP would be Tn = [a+(n−1)d]rn-1

What is the Sum of terms of AGP?

 The sum of terms of the initial terms n in the AGP is 

Sn = k=1n[a+(k−1)d]rk-1

Which can be further solved to obtain 

Sn = a-[a+(n-1)d]rn​1-r + dr( 1 – rn-1 )(1-r)2 

The sum of infinity can be represented in AGP as if |r| < 1 

In the formula, the sum of infinity can be written as: 

S = a1- r + dr(1 – r)2

Arithmetic and geometric progression series are usually used in mathematics because their sum is easy to apply. This method can be used for contest problems. 

For example: If the sum of the infinity of series is 1+4x+7x² +10x³+⋯ is 3516. Find the value of x. 

Solution – 

Let S = 1+4x+7x² +10x³ + …   ……. 1

Now multiply x to equation 1,

Then, x S= x+4x² +7x² +…   …….. 2

Subtracting 2 from 1; we get

S – x S = {1+4x+7x² +10x³ + … } – {x+4x² +7x² +… }

(1 – x) S = 1+3x+3x²+3x³ + …

This series is a geometric series with the first term a=3x and the common ratio r=x. So, 

(1 – x) S = 1 + 1 + 2x(1 – x)

⇒ 3516(1−x) – 1 = 1 + 2x(1 – x)

⇒35 (1-x)2= 16 (1+2x)

After solving this equation, we will get x = 15

Calculating The Sum of AGP

If you want to calculate the sum of all the terms in AGP, then you can also do it manually. However, this would take hours of tedious work and complex calculations. Sometimes your solutions would be enormous to even sum up, or they might be even smaller. 

So you can find it easily by using this general approach. Let’s understand it with a simple example.

Question: 1⋅2+2⋅2² +3⋅2³+ …… + 100⋅2¹⁰⁰. Find the sum of these series. 

Solution: If we observe the given equation, we can see that it is too large to calculate manually. So we use the general term. 

Let’s assume the sum of the series is S – then,

S=1⋅2+2⋅2² +3⋅2³+ ……… +100⋅2¹⁰⁰

Now multiply 2 by S so we get, 

2S=1⋅2² +2⋅2³+…… + 99⋅2¹⁰⁰ +100⋅2¹⁰¹

Subtract 2s from S, we obtain – 

S = 1.2 + 2.2² +3.2³ + …. + 100.2¹⁰⁰ – 2S = 0+ 1.2² + 2.2³ +……. + 99.2¹⁰⁰ + 100.2¹⁰¹

After subtraction, we get 

S(1-2) = 1.2 +1.2² +1.2³+….+1.2¹⁰⁰ – 100.2¹⁰⁰ 

We compute these values, 

– S = 2 ( 2¹⁰⁰ – 1 ) – 100.2¹⁰¹ ( since there are first 100 terms in GP) 

S = 100.2¹⁰¹ – 2.2¹⁰⁰ +2 

S= 200.2¹⁰⁰ – 2.2¹⁰⁰ +2 

S = 198.2¹⁰⁰ + 2 

From the above problem, we can say that the important step was to multiply the common ratio and subtract the sequences to further reduce the GP. 

Geometric Progressions

Geometric progression is also known as a geometric sequence. It is a series of equations that differ by a common ratio. For example, we can say 3,6, 12, 24 ….. in this, the common ratio can be taken as 3. So you can find the common ratio in the given series by determining the ratio between any two adjacent ratios. 

For example: 10, 30, 90, 270 … in this progression, the initial term is 10 and the common ratio is 3. 

How to describe the geometric progression? 

Initial Term: Geometric progression starts with an initial term, in the above example, the term is given as 10. 

Common Ratio: Next important terminology is the common ratio. In this sequence we can find the common ratio as said in the above sequence is 3. 

Recursive Formula: The geometric pattern is defined with a recursive formula. It means each term is related to one another or the term before. The next term in an equation is the byproduct of the previous term along with a common ratio. 

Term = previous term x common ratio

More precisely, r¹ we have 

an = an−1× r

Explicit Formula: The recursive formula helps you determine the relationships between the sequence using a common ratio. But, it is also helpful if we determine an explicit formula of the terms in the sequence. This formula will help you in finding any term in a geometric progression. 

The terms are related together by a ratio. We can find the ratio by multiplying. 

Term = initial term and common ratio x …… common ratio (these are the number of steps from the initial value).

So, we can write this as – 

an = a1 x rn-1

Sum of geometric progression 

Sometimes, we need to find the sum of a geometric progression’s terms. When we have a big number of terms to add, it can be tough to do it one at a time. The following problem demonstrates a method that could be expanded into a general technique:

Example – Determine the sum of the first ten terms given in this geometric progression. 

3, 15, 75, 375, 1875, …..

Solution:

A = sum of first ten terms of the given series

A = 3+3⋅5+3⋅5²+ …… +3⋅5⁹ (1)     

Next, multiply the A by 5, we get 

5A = 3⋅5+3⋅5² +3⋅5³+……+ 3⋅5¹⁰ (2) 

Taking (1) – (2) 

A – 5A = (3+3⋅5+3⋅5²+ …… +3⋅5⁹) – (3⋅5+3⋅5² +3⋅5³+……+ 3⋅5¹⁰) 

We get, 

A (1-5) = 3+0+ 0+……..+ 0 -3.5¹⁰

-4A = 3 – 3.5¹⁰

Hence A = 3.5¹⁰-34

In the previous example, we multiplied the geometric progression’s sum by its common ratio and then subtracted the result from the original sum, discovering that all except the first and last terms cancel out. We can now use the same method to determine the sum’s general formula.

Theorem: For finding the geometric progression with initial value a with common ratio r and the sum of first n terms

Sn= a .rn-1r-1            for r ≠1

Sn=a.n                      for r=1

Sum of infinite geometric progression 

Theorem – In this, the initial terms a with common ratio r should satisfy |r| < 1 then the sum of infinite progression is 

S​ = a1-r

Let’s move on to finding the sum of infinitely many terms of a geometric progression now that we know how to calculate the sum of finitely many terms. We’ll start with an example.

Example: Find the geometric series in the given equation. 

5 + 53 + 59 + 527 

Solution: Let the sum be S, 

S = 5 + 53 + 59 + 527 ….. (1) 

Then multiply the given equation S by 13, we get

13S = 53 + 59 + 527 + 581 ….. (2) 

Subtracting equation 1 from 2 gives ……

S – 13S = {5 + 53 + 59 + 527} – {53 + 59 + 527 + 581}

Gives you the results 

S ( 1 – 13 ) = 5+0+0+0+0   

S.23 = 5 

S = 152

Conclusion 

There are an infinite number of terms in an infinite series. A partial sum, Sn, is the sum of the first n terms. The sum to infinity of the series is reached when Sn approaches a limit as n approaches infinity. Infinite arithmetic series has a sum of either + ∞ or – ∞. In this article we learnt different ways to sum an AGP and geometric progression.Hope you liked this article.