Sum of N terms & Infinite Terms of Geometric Progression(GP)-Formula, Solved Examples, Practice Questions, FAQs

📢 Master the Sum of n Terms of a Geometric Progression (GP) – Class 11 Mathematics 📚

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The Sum of n Terms of a Geometric Progression (GP) is an important concept in Class 11 Mathematics, frequently asked in Board Exams & JEE Mains/Advanced. Understanding this topic will help you solve problems related to compound interest, population growth, radioactive decay, and more.

Derivation of Sum of N Terms of Geometric Progression (GP) of Finite Terms

The terms of a GP are represented as a₁, a₂, a₃, a₄, …, a_n.

Expressing all these terms according to the first term a₁, we get

a₁ = a
a₂ = a₁r
a₃ = a₂r = (a₁r)r = a₁r²
a₄ = a₃r = (a₁r²)r = a₁r³
…
a_m = a₁r^m−1
…
Similarly,
a_n = a₁r^{n – 1}

The geometric progression summation is given by

S = a₁ + a₂ + a₃ + … + a_n

S = a₁ + a₁r + a₁r² + a₁r³ + … + a₁rⁿ⁻¹     ….equation (1)

Multiply both sides of Equation (1) by r (common ratio), and we get
S × r= a₁r + a₁r² + a₁r³ + a₁r⁴ + … + a₁rⁿ     ….equation (2)

Subtract Equation (2) from Equation (1)
S – Sr = a₁ – a₁rⁿ
(1 – r)S = a₁(1 – rⁿ)
S_n = a₁(1 – rⁿ)/(1 – r), if r<1

Now, Subtracting Equation (1) from Equation (2) will give
Sr – S = a₁rⁿ – a₁
(r – 1)S = a₁(rⁿ-1)

Hence, Sum of First n Terms of GP is given by:

S_n = a(1 – rⁿ)/(1 – r), if r < 1,  r ≠ 1
S_n = a(rⁿ -1)/(r – 1), if r > 1,  r ≠ 1

Hence,

The formula to find the sum of n terms of GP is:

Sn = a[(rn – 1)/(r – 1)] if r ≠ 1 and r > 1

Where

a is the first term

r is the common ratio

n is the number of terms

If the common ratio is equal to 1, i.e. r = 1, then S_n can be written as:

S_n = a + a(1) + a(1)² + a(1)³ + … + a(1)^n-1

S_n = a + a + a + a + …. + a (n terms)

S_n = na

Thus, the above formulas can be written as:

The sum of first n term of a GP when r > 1

\(\begin{array}{l}\large S_{n}=\frac{a(r^{n}-1)}{(r-1)}\end{array} \)

The sum of first n term of a GP when r < 1

\(\begin{array}{l}\large S_{n}=\frac{a(1-r^{n})}{(1-r)}\end{array} \)

The sum of first n term of a GP when r = 1

\(\begin{array}{l}\large S_{n}=na\end{array} \)

The above formulas can be used to calculate the finite terms of a GP.

Now, the question is how to find the sum of infinite GP. Let’s have a look at the formula given in the next section to know the formula of sum of infinite GP.

Formula for Sum of an Infinite GP of Terms

Consider an infinite GP, a, ar, ar², ar³,…, ar^n-1, arⁿ, …..

Here, a is the first term and r is the common ratio of the GP and the last term is not known.

Thus, the sum of infinite GP is given by the formula:

\(\begin{array}{l}\large S_{n}=\frac{a}{(1-r)}\end{array} \)

The above formulas can be used to calculate the finite terms of a GP.

When |r| < 1, the geometric series converges and the series has a sum.

When |r| > 1 the geometric series does not converge or it is divergent and it has no sum.

Solved Examples of Sum of First n Terms of the GP

Example 1:

Find the sum of GP: 1, 2, 4, 8, and 16.

Given GP is 1, 2, 4, 8 and 16
First term, a = 1
Common ratio, r = 2/1 = 2 > 1
Number of terms, n = 5
Sum of GP is given by;
S_n = a[(rⁿ – 1)/(r – 1)]
S₅ = 1[(2⁵ – 1)/(2 – 1)]
     = 1[(32 – 1)/1]
     = 1[31/2]
     = 1 × 15.5
     = 15.5

Example 2:

Find the sum of first n terms of the GP:

\(\begin{array}{l}1 + \frac{2}{3}+\frac{4}{9}+….\end{array} \)

Solution:

Given GP:

\(\begin{array}{l}1 + \frac{2}{3}+\frac{4}{9}+….\end{array} \)

Here,

First term = a = 1

Common ratio = r = 2/3, i.e. r < 1

Thus, the sum of first n terms is:

\(\begin{array}{l}S_{n}=\frac{a(1-r^{n})}{(1-r)}\end{array} \)

Substituting a = 1 and r = 2/3,

\(\begin{array}{l}S_{n}=\frac{1(1-\frac{2}{3}^{n})}{(1-\frac{2}{3})}\end{array} \)

\(\begin{array}{l}S_{n}=\frac{3(1-\frac{2}{3}^{n})}{(3-2)}\end{array} \)

Therefore,

\(\begin{array}{l}S_{n}=3\left ( 1-\frac{2}{3}^{n} \right )\end{array} \)

Example 3:

Find the sum of the first 6 terms of a GP whose first term is 2 and the common difference is 4.

Solution:

Given,

First term = a = 2,

Common ratio = r = 4

and n = 6

As we know, the sum of first n terms of GP when r > 1

\(\begin{array}{l}S_{n}=\frac{a(r^{n}-1)}{(r-1)}\end{array} \)

Substituting the values,

\(\begin{array}{l}S_{6}=\frac{2(4^{6}-1)}{(4-1)}\end{array} \)

= (2/3) × (4096 – 1)

= (2/3) × 4095

= 2 × 1365

= 2730

Therefore, the sum of the first 6 terms of the given GP is 2730.

Example 4:

Find the sum of the infinite GP: 125, 25, 5,….

Solution:

Given GP is,

125, 25, 5,….

Here,

First term = a = 125

Common ratio = r = 25/125 = 1/5; r < 1

The sum of an infinite GP when r < 1 is:

S_n = a/(1 – r)

= 125/[1 – (1/5)]

= 125/[(5 – 1)/5]

= (125 × 5)/4

= 625/4

= 156.25

Hence, the sum of given infinite GP is 156.25.

Difference Between Geometric Sequence and Geometric Series

Some of the common differences between Geometric Sequences and Series are listed in the following table:

Aspect	Geometric Sequence	Geometric Series
Definition	A sequence of numbers where each term is obtained by multiplying the previous term by a fixed, non-zero number (common ratio).	The sum of terms in a geometric sequence.
General Form	a, ar, ar2, ar3, ar4, . . .	a + ar + ar2 + ar3 + ar4 + . . .
Example	2, 6, 18, 54, . . .	2 + 6 + 18 + 54 + . . .

Practice Problems

1. Which term of the GP, 2, 8, 32, … up to n terms is equal to 131072?

2. Find the sum of the sequence 7, 77, 777, 7777, … to n terms.

3. How many terms of G.P. 3, 3², 3³,… are needed to give the sum 120?

4. If the sum of some terms of G.P. is 315 whose first term and the common ratio are 5 and 2, respectively, then find the last term and the number of terms.

Frequently Asked Questions on Sum of n Terms of GP

Q1

What is the formula for the sum of n terms in GP?

The formula of sum of n terms in GP is given as:
S_n = [a(r^n – 1)]/(r – 1) when |r| > 1
S_n = [a(1 – r^n)]/(1 – r) when |r| < 1
S_n = na when r = 1

Q2

What is the nth term of GP?

The nth term of a GP is denoted by a_n and is calculated using the formula:
a_n = ar^{n-1}
Here, a is the first term, and r is the common ratio of the GP.

Q3

What is the sum of infinite GP?

The sum of infinite GP whose first term a and common ratio r is given by the formula,
S_{n} = a/(1 – r)

Q4

Is it possible to find an infinite sum in a GP?

Yes, we can find the sum of an infinite GP only when the common ratio is less than 1. If the common ratio is greater than 1, there will be no specified sum as we can say that the sum is infinity.

Q5

What is the sum of the first n terms of a GP when the common ratio is one?

If the common ratio is equal to 1, then the GP becomes, a, a, a,…
Thus, the sum of first n terms = a + a + a + … (n terms) = na

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