Sum of First n Terms of GP Formula

What is Sum to n Terms of a GP?

The sum to $n$ terms of a Geometric Progression (GP) is one of the most important concepts in sequences and series. Instead of adding each term individually, mathematicians use a special formula to find the sum of the first $n$ terms quickly. The GP sum formula is widely used in mathematics, finance, economics, population studies, compound interest calculations, and competitive examinations such as JEE, CUET, SSC, and Banking exams.

GP Sum Meaning in Simple Words

In simple words, the sum to $n$ terms of a GP means adding the first $n$ terms of a geometric sequence.

For example, in the GP:

$2, 4, 8, 16, 32$

the sum of the first 5 terms is:

$2+4+8+16+32=62$

Instead of adding each term manually, we can use the GP sum formula.

Definition of Geometric Progression (GP)

A Geometric Progression (GP) is a sequence in which each term after the first is obtained by multiplying the previous term by a fixed non-zero number called the common ratio.

For example:

$2, 4, 8, 16, 32,\dots$

Here, each term is obtained by multiplying the previous term by 2.

Therefore, the common ratio is:

$r=2$

What Does Sum to n Terms Mean?

The sum to $n$ terms means the total obtained by adding the first $n$ terms of a geometric progression.

If:

$a, ar, ar^2, ar^3,\dots$

is a GP, then:

$S_n=a+ar+ar^2+\cdots+ar^{n-1}$

where $S_n$ represents the sum of the first $n$ terms.

Why GP Sum Formula is Important

The GP sum formula helps simplify lengthy calculations and is widely used in practical applications.

Importance of GP Sum Formula

• Saves time in calculations.
• Useful in compound interest problems.
• Applied in population growth models.
• Used in economics and finance.
• Helps solve sequence and series questions quickly.
• Frequently asked in board and competitive examinations.

Basics of Geometric Progression

Before learning the GP sum formula, it is important to understand the basic structure of a geometric progression.

What is a Geometric Progression?

A geometric progression is a sequence where consecutive terms have a constant ratio.

Example:

$3, 6, 12, 24, 48,\dots$

Each term is obtained by multiplying the previous term by 2.

First Term and Common Ratio

Every GP contains two important quantities:

First Term

The first term is denoted by:

$a$

Example:

In $5, 10, 20, 40,\dots$

the first term is:

$a=5$

Common Ratio

The common ratio is denoted by:

$r$

and is calculated as:

$r=\frac{\text{Second Term}}{\text{First Term}}$

Example:

$r=\frac{10}{5}=2$

General Form of a GP

The standard form of a geometric progression is:

$a,\ ar,\ ar^2,\ ar^3,\ ar^4,\dots$

where:

• $a$ = first term
• $r$ = common ratio

Examples of Geometric Progressions

Sum to n Terms of a GP Formula

The GP sum formula allows us to calculate the sum of the first $n$ terms without adding each term separately.

Standard Formula When $r\neq1$

When the common ratio is not equal to 1:

$S_n=\frac{a(r^n-1)}{r-1}$

Alternatively:

$S_n=\frac{a(1-r^n)}{1-r}$

Both formulas are equivalent.

Formula When $r=1$

If every term is the same:

$a,a,a,a,\dots$

then:

$S_n=na$

because the same term is added $n$ times.

Meaning of Variables in the Formula

Derivation of GP Sum Formula

Consider:

$S_n=a+ar+ar^2+\cdots+ar^{n-1}$

Multiply both sides by $r$:

$rS_n=ar+ar^2+ar^3+\cdots+ar^n$

Subtract:

$S_n-rS_n=a-ar^n$

Taking common factors:

$S_n(1-r)=a(1-r^n)$

Therefore:

$S_n=\frac{a(1-r^n)}{1-r}$

This is the standard derivation of the GP sum formula.

How to Find the Sum of n Terms of a GP?

Finding the sum of a GP is straightforward once the first term, common ratio, and number of terms are known.

Step-by-Step Method

Step 1: Identify the first term $a$.

Step 2: Find the common ratio $r$.

Step 3: Determine the number of terms $n$.

Step 4: Apply the GP sum formula.

Step 5: Simplify the result.

Using the GP Sum Formula

For:

$2,4,8,16,32$

we have:

$a=2$

$r=2$

$n=5$

Using:

$S_n=\frac{a(r^n-1)}{r-1}$

Substituting values:

$S_5=\frac{2(2^5-1)}{2-1}$

$=\frac{2(32-1)}{1}$

$=62$

Solved Examples

Example 1: Find the sum of the first 6 terms of:

$3,6,12,24,\dots$

Given:

$a=3$

$r=2$

$n=6$

Using the formula:

$S_6=\frac{3(2^6-1)}{2-1}$

$=3(64-1)$

$=189$

Answer: $189$

Common Mistakes to Avoid

• Using the wrong value of $r$.
• Forgetting the value of $n$.
• Applying the infinite GP formula to finite GP problems.
• Incorrect simplification of exponents.
• Using the formula for $r\neq1$ when $r=1$.

Sum of Finite Geometric Progression

Most GP problems involve a finite number of terms.

Definition of Finite GP

A finite GP contains a limited number of terms.

Example:

$2,4,8,16,32$

contains only 5 terms.

Formula for Finite GP

For $r\neq1$:

$S_n=\frac{a(r^n-1)}{r-1}$

Properties of Finite GP

• Has a fixed number of terms.
• Sum can be calculated exactly.
• Common ratio remains constant.
• May be increasing or decreasing.

Examples of Finite GP

Sum of Infinite Geometric Progression

Infinite geometric progressions continue forever.

What is an Infinite GP?

An infinite GP contains infinitely many terms.

Example:

$1+\frac12+\frac14+\frac18+\cdots$

The sequence never ends.

Formula for Infinite GP

If $|r|<1$:

$S_\infty=\frac{a}{1-r}$

Condition for Convergence

An infinite GP has a finite sum only when:

$|r|<1$

Examples:

• $r=\frac12$ ✓
• $r=\frac13$ ✓
• $r=2$ ✗
• $r=-3$ ✗

Solved Examples

Find:

$1+\frac12+\frac14+\frac18+\cdots$

Here:

$a=1$

$r=\frac12$

Using:

$S_\infty=\frac{1}{1-\frac12}$

$=\frac{1}{\frac12}$

$=2$

Answer: $2$

Applications of GP Sum Formula

The GP sum formula has numerous real-world applications.

Compound Interest Calculations

Compound interest follows a geometric progression because the amount grows by a constant multiplication factor every period.

Population Growth Models

Population growth often follows GP patterns when growth occurs at a fixed percentage rate.

Radioactive Decay Problems

Radioactive substances decay according to geometric progression models with ratios less than 1.

Applications in Finance and Economics

GP formulas are used in:

• Investment analysis
• Loan calculations
• Financial forecasting
• Economic growth models
• Retirement planning

Properties of Geometric Progression

A GP has several unique mathematical properties.

Important Characteristics of GP

• Consecutive terms have a constant ratio.
• Terms grow or decrease exponentially.
• The GP sum formula can be derived algebraically.
• Infinite GP sums exist only under specific conditions.

Relationship Between Terms

For a GP:

$a,\ ar,\ ar^2,\ ar^3,\dots$

each term is obtained by multiplying the previous term by $r$.

Also:

$T_n=ar^{n-1}$

Common Ratio Properties

If:

$r>1$

the GP increases.

If:

$0<r<1$

the GP decreases.

If:

$r<0$

the signs alternate.

Growth and Decay Patterns

Geometric progressions model both growth and decay.

These properties make geometric progressions one of the most important topics in algebra, sequences and series, finance, economics, engineering, and competitive mathematics.

Best Books for Geometric Progression

A solid understanding of geometric progression and sequence-and-series concepts is important for mastering GP sum formulas and related applications.

Best Books for Sequence and Series

Shortcut Tips and Tricks for GP Questions

A few important shortcuts can help solve geometric progression questions much faster.

Shortcut Tips and Tricks

Important Formula Table

This formula sheet contains the most important geometric progression formulas.

Important GP Formulas

Solved Examples Based on the Sum of n Terms of GP

Example 1: Let ${a_k}$ and ${b_k}$, $k \in \mathbb{N}$, be two G.P.s with common ratios $r_1$ and $r_2$ respectively such that $a_1=b_1=4$ and $r_1<r_2$. Let $c_k=a_k+b_k$, $k\in\mathbb{N}$. If $c_2=5$ and $c_3=\frac{13}{4}$, then find

$\sum_{k=1}^{\infty} c_k-(12a_6+8b_4)$.

[JEE Main 2023]

Solution:

Given,

$a_1=4$ and $b_1=4$

Therefore,

$a_2=4r_1,\quad b_2=4r_2$

Since $c_2=a_2+b_2=5$,

$4r_1+4r_2=5$

$\Rightarrow r_1+r_2=\frac{5}{4}\quad ...(1)$

Also,

$c_3=a_3+b_3=\frac{13}{4}$

$4r_1^2+4r_2^2=\frac{13}{4}$

$\Rightarrow r_1^2+r_2^2=\frac{13}{16}\quad ...(2)$

Using

$(r_1+r_2)^2=r_1^2+r_2^2+2r_1r_2$

we get

$\left(\frac{5}{4}\right)^2=\frac{13}{16}+2r_1r_2$

$\frac{25}{16}-\frac{13}{16}=2r_1r_2$

$\frac{12}{16}=2r_1r_2$

$\Rightarrow r_1r_2=\frac{3}{8}$

Using (1),

$r_1+\frac{3}{8r_1}=\frac{5}{4}$

$8r_1^2+3=10r_1$

$8r_1^2-10r_1+3=0$

$(4r_1-3)(2r_1-1)=0$

Therefore,

$r_1=\frac{3}{4}$ or $r_1=\frac{1}{2}$

Since $r_1<r_2$,

$r_1=\frac{1}{2}$ and $r_2=\frac{3}{4}$

Now,

$\sum_{k=1}^{\infty} c_k=\sum_{k=1}^{\infty} a_k+\sum_{k=1}^{\infty} b_k$

$=\frac{4}{1-\frac{1}{2}}+\frac{4}{1-\frac{3}{4}}$

$=8+16$

$=24$

Also,

$a_6=4\left(\frac{1}{2}\right)^5=\frac{1}{8}$

$b_4=4\left(\frac{3}{4}\right)^3=\frac{27}{16}$

Therefore,

$\sum_{k=1}^{\infty} c_k-(12a_6+8b_4)$

$=24-\left(12\times\frac{1}{8}+8\times\frac{27}{16}\right)$

$=24-\left(\frac{3}{2}+\frac{27}{2}\right)$

$=24-15$

$=9$

Hence, the required answer is $9$.

Example 2: The sum of 20 terms of the series $2^2-3^2+2\cdot4^2-5^2+2\cdot6^2-7^2+\cdots$ is equal to:

[JEE Main 2023]

Solution:

Let

$S=(2^2-3^2+4^2-5^2+\cdots \text{ up to 20 terms})+(2^2+4^2+6^2+\cdots \text{ up to 10 terms})$

Therefore,

$S=-(2+3+4+\cdots+21)+4(1^2+2^2+3^2+\cdots+10^2)+1$

Using

$1+2+\cdots+21=\frac{21\times22}{2}=231$

and

$1^2+2^2+\cdots+10^2=\frac{10\times11\times21}{6}=385$

Hence,

$S=1-231+4(385)$

$=1-231+1540$

$=1310$

Hence, the required answer is $1310$.

Example 3: Let $A_1$ and $A_2$ be two arithmetic means and $G_1,G_2,G_3$ be three geometric means between two distinct positive numbers. Find $G_1^4+G_2^4+G_3^4+G_1^2G_3^2$.

[JEE Main 2023]

Solution:

Since $a,A_1,A_2,b$ are in A.P.,

$d=\frac{b-a}{3}$

$A_1=\frac{2a+b}{3}$

$A_2=\frac{a+2b}{3}$

Therefore,

$A_1+A_2=a+b$

Also, $a,G_1,G_2,G_3,b$ are in G.P.

$r=\left(\frac{b}{a}\right)^{1/4}$

$G_1=(a^3b)^{1/4}$

$G_2=(a^2b^2)^{1/4}$

$G_3=(ab^3)^{1/4}$

Thus,

$G_1^4=a^3b$

$G_2^4=a^2b^2$

$G_3^4=ab^3$

and

$G_1^2G_3^2=a^2b^2$

Hence,

$G_1^4+G_2^4+G_3^4+G_1^2G_3^2$

$=a^3b+a^2b^2+ab^3+a^2b^2$

$=ab(a+b)^2$

Hence, the required answer is $ab(a+b)^2$.

Example 4: The 4th term of a G.P. is 500 and its common ratio is $\frac{1}{m}$, where $m\in\mathbb{N}$. Let $S_n$ denote the sum of the first $n$ terms of this G.P. If $S_6>S_5+1$ and $S_7<S_6+\frac{1}{2}$, then find the number of possible values of $m$.

[JEE Main 2023]

Solution:

Given,

$T_4=500$

and

$r=\frac1m$

Since

$T_4=ar^3$

$a\left(\frac1m\right)^3=500$

$\Rightarrow a=500m^3$

Now,

$S_n-S_{n-1}=T_n$

Therefore,

$S_n-S_{n-1}=500m^{4-n}$

Using

$S_6>S_5+1$

$\Rightarrow S_6-S_5>1$

$\Rightarrow500m^{-2}>1$

$\Rightarrow m^2<500 \quad ...(1)$

Also,

$S_7<S_6+\frac12$

$\Rightarrow S_7-S_6<\frac12$

$\Rightarrow500m^{-3}<\frac12$

$\Rightarrow m^3>1000 \quad ...(2)$

From (2),

$m>10$

From (1),

$m<\sqrt{500}\approx22.36$

Thus,

$10<m\le22$

Possible values are

$11,12,13,\ldots,22$

Number of values

$=12$

Hence, the required answer is $12$.

Example 5: For $k\in\mathbb{N}$, if the sum of the series $1+\frac4k+\frac8{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\cdots$ is 10, then find the value of $k$.

[JEE Main 2023]

Solution:

Given,

$1+\frac4k+\frac8{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\cdots=10$

Therefore,

$\frac4k+\frac8{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\cdots=9$

Let this sum be $S$.

Then

$S=9$

Also,

$\frac{S}{k}=\frac4{k^2}+\frac8{k^3}+\frac{13}{k^4}+\frac{19}{k^5}+\cdots$

Subtracting,

$S-\frac{S}{k}$

$=\frac4k+\frac4{k^2}+\frac5{k^3}+\frac6{k^4}+\cdots$

Hence,

$9\left(1-\frac1k\right)$

$=\frac4k+\frac1{k^3}+\frac1{k^4}+\frac1{k^5}+\cdots$

The remaining series is a G.P. with first term $\frac1{k^3}$ and common ratio $\frac1k$.

Therefore,

$\frac1{k^3}+\frac1{k^4}+\frac1{k^5}+\cdots=\frac{\frac1{k^3}}{1-\frac1k}$

Substituting,

$9\left(1-\frac1k\right)=\frac4k+\frac{\frac1{k^3}}{1-\frac1k}$

On simplification,

$9(k-1)^3=4k(k-1)+1$

Solving gives

$k=2$

Hence, the required answer is $2$.

Related Topics to Sum to n Terms of a GP

The following sequence and series topics will help you build a stronger foundation in geometric progressions, arithmetic progressions, infinite series, and mathematical patterns used in advanced mathematics.