Sum of n Terms 

The sum of n terms refers to the total of the first n terms of a sequence, commonly in Arithmetic Progression (AP) or Geometric Progression (GP). It is a fundamental concept in mathematics and helps solve problems involving series and patterns. The formula varies for different types of sequences. In AP, the sum depends on the first term and common difference, while in GP, it depends on the first term and common ratio. It’s widely used in competitive exams like JEE. 

1.0What is the Formula for Sum of n Terms?

The formula for sum of n terms varies based on the type of sequence:

1. Sum of n Terms in Arithmetic Progression (AP)
2. Sum of n Terms in Geometric Progression (GP)
3. Sum to n Terms of Special Series

2.0Sum of n Terms in Arithmetic Progression (AP)

Definition:

An Arithmetic Progression (AP) is a sequence where each term differs from the previous one by a constant difference, called the common difference (d).

Formula for nth Term in AP:

$a_n=a+(n-1)d$

Where:

• a = first term
• d = common difference
• n = number of terms

Sum of N Terms of an AP Formula:

$$\begin{gathered} S_n=\frac{n}{2}[2a+(n-1)d] \\ Or \\ {S}_n=\frac{n}{2}(a+l) \end{gathered}$$

Where l is the last term.

3.0Solved Examples on Sum of N Terms of an AP 

Example 1: Find the sum of first 10 terms of the AP: 3, 6, 9, ...

Solution:
a = 3, d = 6 - 3 = 3, n = 10 

$$\begin{gathered} S_{10}=\frac{10}{2}\left[2\times 3+(10-1)\times 3\right] \\ =5\times [6+27] \\ =5\times 33=165 \\ \text{Answer: 165} \end{gathered}$$

Example 2: Find the sum of first 15 natural numbers.

Here, a = 1, d = 1, n = 15:

$$\begin{gathered} S_{15}=\frac{15}{2}\left[2\times 1+(15-1)\times 1\right] \\ {S}_{15}=\frac{15}{2}\times (2+14) \\ {S}_{15}=\frac{15}{2}\times 16=120 \\ \text{Answer: 120} \end{gathered}$$

Example 3: Find the sum of the first 20 terms of the AP:

7, 13, 19, 25, ...

Solution:
Here,
a = 7 (first term)

d = 13 - 7 = 6 (common difference)

n = 20

Sum of n terms:

$$\begin{gathered} S_n=\frac{n}{2}[2a+(n-1)d] \\ {S}_{20}=\frac{20}{2}\left[2\times 7+(20-1)\times 6\right] \\ =10\times [14+114] \\ =10\times 128=1280 \\ \text{Answer: 1280} \end{gathered}$$

Example 4: The sum of the first n terms of an AP is $S_n=3n^2+5n$. Find its first term and common difference.

Solution:
$$\begin{gathered} Given:S_n=3n^2+5n \\ \text{First term a =}{S}_1: \\ {S}_1=3\times 1^2+5\times 1=3+5=8 \\ \text{Common difference d:} \\ {a}_2=S_2-S_1 \\ {S}_2=3\times 2^2+5\times 2 \\ {S}_2=3\times 4+10 \\ {S}_2=12+10=22 \\ d=a_2-a_1 \\ d=(S_2-S_1)=22-8=14 \end{gathered}$$

Answer: First term a = 8, Common difference d = 14

Example 5: The sum of the first n terms of an AP is $S_n=n^2+2n$. Find its nth term.

Solution:

$$\begin{gathered} \text{nth term:} \\ {a}_n=S_n-S_{n-1} \\ {S}_n=n^2+2n \\ {S}_{n-1}=(n-1{)}^2+2(n-1) \\ {S}_{n-1}=n^2-2n+1+2n-2=n^2-1 \\ Now, \\ {a}_n=n^2+2n-(n^2-1)=2n+1 \\ \text{Answer:}{a}_n=2n+1 \end{gathered}$$

Example 6: Find the sum of all natural numbers from 1 to 100.

Solution:

This is an AP with a = 1, d = 1, n = 100:

$$\begin{gathered} S_{100}=\frac{100}{2}\left[2\times 1+(100-1)\times 1\right] \\ {S}_{100}=50\times [2+99] \\ {S}_{100}=50\times 101=5050 \\ \text{Answer: 5050} \end{gathered}$$

Example 7: A sum of ₹1000 is to be divided among 25 persons such that each person gets ₹2 more than the previous one. Find how much the first and last persons get.

Solution:

$$\begin{gathered} \text{This forms an AP where:} \\ n=25 \\ {S}_{25}=1000 \\ d=2 \\ \text{Sum formula:} \\ {S}_n=\frac{n}{2}[2a+(n-1)d] \\ 1000=\frac{25}{2}[2a+24\times 2] \\ 1000=\frac{25}{2}[2a+48] \\ 1000\times 2=25[2a+48] \\ 2000=25\times (2a+48) \\ 2a+48=\frac{2000}{25}=80 \\ 2a=80-48=32\Rightarrow a=16 \\ \text{First person’s share: ₹16} \\ \text{Last person’s share:} \\ {a}_n=a+(n-1)d \\ {a}_n=16+24\times 2=16+48=64 \\ \text{Answer: First person gets ₹16, Last person gets ₹64} \end{gathered}$$

4.0Sum of n Terms in Geometric Progression (GP)

Definition:

A Geometric Progression (GP) is a sequence where each term is multiplied by a fixed number called the common ratio (r).

Formula for nth Term of GP:

$a_n=a\cdot r^{n-1}$

Sum of n Terms of GP Formula

Case 1: When $r\ne 1$:

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

Case 2: When r = 1:

$S_n=n\times a$

5.0Solved Examples on Sum of N Terms of GP 

Example 1: Find the sum of first 5 terms of the GP: 2, 4, 8, ...

Here, a = 2, r = 2, n = 5:

$$\begin{gathered} S_5=2\cdot \frac{2^5-1}{2-1} \\ {S}_5=2\cdot \frac{32-1}{1} \\ {S}_5=2\cdot 31=62 \\ \text{Answer: 62} \end{gathered}$$

Example 2: Find the sum of the first 5 terms of the GP: $3,6,12,\dots .$

Solution:

$$\begin{gathered} Here, \\ a=3, \\ r=\frac{6}{3}=2, \\ n=5 \\ \text{Using the sum formula for GP (for}r\ne 1) \\ {S}_n=a\cdot \frac{r^n-1}{r-1} \\ {S}_5=3\cdot \frac{2^5-1}{2-1} \\ {S}_5=3\cdot \frac{32-1}{1}=3\cdot 31=93 \\ \text{Answer: 93} \end{gathered}$$

Example 3: Find the sum of first 4 terms of the GP: $81,27,9,\dots .$

Solution:

$$\begin{gathered} Here, \\ a=81, \\ r=\frac{27}{81}=\frac{1}{3} \\ n=4 \\ \text{Applying GP sum formula:} \\ {S}_n=a\cdot \frac{1-r^n}{1-r}(since|r|<1): \\ {S}_4=81\cdot \frac{1-{\left(\frac{1}{3}\right)}^4}{1-\frac{1}{3}} \\ =81\cdot \frac{1-\frac{1}{81}}{\frac{2}{3}} \\ =81\cdot \frac{\frac{80}{81}}{\frac{2}{3}} \\ =81\cdot \frac{80}{81}\cdot \frac{3}{2}=40\cdot 3=120 \\ \text{Answer: 120} \end{gathered}$$

Example 4: The sum of infinite terms of a GP is 8, and the sum of squares of its terms is 192. Find the first term and common ratio.

Solution:

$$\begin{gathered} \text{Let first term a and common ratio r (with |r| < 1):} \\ {S}_{\infty }=\frac{a}{1-r}=8\Rightarrow a=8(1-r) \\ \text{Sum of squares of terms:} \\ {S}_{\text{squares}}=\frac{a^2}{1-r^2}=192 \\ \text{Substitute a = 8(1 - r):} \\ \frac{[8(1-r){]}^2}{1-r^2}=192 \\ \frac{64(1-r{)}^2}{(1-r)(1+r)}=192 \\ \Rightarrow \frac{64(1-r)}{1+r}=192 \\ 64(1-r)=192(1+r) \\ 64-64r=192+192r \\ \Rightarrow 256r=128\Rightarrow r=\frac{1}{2} \\ Now,a=8(1-\frac{1}{2})=8\cdot \frac{1}{2}=4 \\ \text{Answer: First term a = 4, Common ratio }r=\frac{1}{2} \end{gathered}$$

Example 5: Find the sum of the infinite GP: $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\cdots$

Solution:

$$\begin{gathered} Here, \\ a=\frac{1}{2} \\ r=\frac{1}{4}÷\frac{1}{2}=\frac{1}{2}(\text{since }|r|<1) \\ \text{Sum of infinite GP:} \\ {S}_{\infty }=\frac{a}{1-r}=\frac{\frac{1}{2}}{1-\frac{1}{2}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1 \\ \text{Answer: 1} \end{gathered}$$

Example 6: The sum of three numbers in GP is 21, and their product is 216. Find the numbers.

Solution:

$$\begin{gathered} \text{Let the three numbers be}\frac{a}{r},a,ar. \\ \text{Sum:} \\ \frac{a}{r}+a+ar=21 \\ \Rightarrow a\left(\frac{1}{r}+1+r\right)=21\dots \dots i \\ \text{Product:} \\ \frac{a}{r}\cdot a\cdot ar=a^3=216 \\ \Rightarrow a=6 \\ Substitutea=6in(i): \\ 6\left(\frac{1}{r}+1+r\right)=21 \\ \Rightarrow \frac{1}{r}+1+r=\frac{7}{2} \\ \text{Multiply both sides by 2r}: \\ 2+2r+2r^2=7r \\ \Rightarrow 2r^2-5r+2=0 \\ \text{Solve quadratic}: \\ r=\frac{5\pm \sqrt{25-16}}{4}=\frac{5\pm 3}{4} \\ r=2\text{or}\frac{1}{2} \\ Case1:r=2 \\ Numbers:\frac{6}{2}=3,6,6\times 2=12 \\ Case2:r=\frac{1}{2} \\ Number\frac{6}{1/2}=12,6,6\times \frac{1}{2}=3 \\ \text{Both cases give the same numbers}:3,6,12\text{(order may vary)}. \\ \text{Answer: Numbers are 3, 6, 12} \end{gathered}$$

Key Concepts Covered:

• Sum of n terms of GP:

$S_n=a\cdot \frac{r^n-1}{r-1}\text{(for }r\ne 1\text{)}$ 

• Sum of Infinite GP:

$S_{\infty }=\frac{a}{1-r}\text{(for }|r|<1\text{)}$

• Solving quadratic for common ratio
• Application of GP in word problems

6.0Sum to n Terms of Special Series

Some common special series sums often asked in exams include:

1. Sum of First n Natural Numbers:

$S_n=\frac{n(n+1)}{2}$

2. Sum of Squares of First n Natural Numbers:

$S_n=\frac{n(n+1)(2n+1)}{6}$

3. Sum of Cubes of First n Natural Numbers:

$S_n={\left(\frac{n(n+1)}{2}\right)}^2$

Example (Special Series):

Find the sum of cubes of first 4 natural numbers.

$S_4={\left(\frac{4\times 5}{2}\right)}^2=(10{)}^2=100$

Answer: 100

7.0How Do You Find the Sum of n?

It depends on the sequence:

$$\begin{gathered} \text{For AP: Use}{S}_n=\frac{n}{2}[2a+(n-1)d] \\ \text{For GP: Use}{S}_n=a\cdot \frac{r^n-1}{r-1}\text{(if }r\ne 1\text{)} \\ \text{For Special Series: Use respective formulas for}n,n^2,n^3,etc. \end{gathered}$$

What is the Sum of 1 to n?

The sum of first n natural numbers:

$S_n=\frac{n(n+1)}{2}$

Key Formulas Recap:

8.0Related Questions on Sum of n Terms

Q1. What is the formula for sum of n terms in AP?

Ans:  

$S_n=\frac{n}{2}[2a+(n-1)d]$

Q2. What is the formula for sum of n terms in GP?

Ans:  

$S_n=a\cdot \frac{r^n-1}{r-1}\text{(for }r\ne 1\text{)}$

Q3. How do you find the sum of n?

Ans: It depends on the sequence. For the sum of first n natural numbers:

$S_n=\frac{n(n+1)}{2}$

Q4. What is the nth term in AP?

Ans: $a_n=a+(n-1)d$

Q5. What is the nth term of GP?

Ans:  $a_n=a+(n-1)d$

Q6. What is the sum of 1 to n?

Ans:  $S_n=\frac{n(n+1)}{2}$