Arithmetic Progression: Complete Notes for Class 10 CBSE

Table of Contents

1. Introduction to Arithmetic Progression
2. Important Terms and Formulas
3. nth Term of an AP
4. Sum of n Terms
5. Properties of AP
6. Solved Examples
7. CBSE Previous Year Questions

1. Introduction to Arithmetic Progression

An Arithmetic Progression (AP) is a sequence of numbers in which the difference between any two consecutive terms is constant. This constant difference is called the common difference.

Example:

• $2,5,8,11,14,\mathrm{.}\mathrm{.}\mathrm{.}$ (Common difference = 3)
• $10,7,4,1,-2,\mathrm{.}\mathrm{.}\mathrm{.}$ (Common difference = -3)

General Form:

$a,a+d,a+2d,a+3d,\mathrm{.}\mathrm{.}\mathrm{.},a+(n-1)d$

Where:

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

2. Important Terms and Formulas

Common Difference (d)

$d=a_2-a_1=a_3-a_2=a_n-a_{n-1}$

General Formula Summary

3. nth Term of an AP

The nth term (or general term) of an AP is given by:

$\boxed{{\displaystyle a_n=a+(n-1)d}}$

Derivation:

• 1st term: $a_1=a$
• 2nd term: $a_2=a+d$
• 3rd term: $a_3=a+2d$
• 4th term: $a_4=a+3d$
• ...
• nth term: $a_n=a+(n-1)d$

Example 1:

Find the 20th term of the AP: $3,7,11,15,\mathrm{.}\mathrm{.}\mathrm{.}$

Solution:

• First term, $a=3$
• Common difference, $d=7-3=4$
• Number of terms, $n=20$

$a_{20}=a+(n-1)d$
$a_{20}=3+(20-1)\times 4$
$a_{20}=3+76=79$

4. Sum of n Terms of an AP

The sum of first $n$ terms of an AP is:

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

Alternate Formula:

$\boxed{{\displaystyle S_n=\frac{n}{2}[a+l]}}$

where $l=a_n$ (last term)

Derivation:

Let $S_n=a+(a+d)+(a+2d)+\mathrm{.}\mathrm{.}\mathrm{.}+(l-d)+l$ ... (1)

Writing in reverse order:
$S_n=l+(l-d)+(l-2d)+\mathrm{.}\mathrm{.}\mathrm{.}+(a+d)+a$ ... (2)

Adding (1) and (2):
$2S_n=(a+l)+(a+l)+(a+l)+\mathrm{.}\mathrm{.}\mathrm{.}+(a+l)$ [n times]
$2S_n=n(a+l)$
$S_n=\frac{n}{2}(a+l)$

Example 2:

Find the sum of first 30 terms of the AP: $5,9,13,17,\mathrm{.}\mathrm{.}\mathrm{.}$

Solution:

• $a=5$
• $d=9-5=4$
• $n=30$

$S_{30}=\frac{30}{2}[2\times 5+(30-1)\times 4]$
$S_{30}=15[10+116]$
$S_{30}=15\times 126=1890$

5. Important Properties of AP

Property 1: Selection of Terms

When selecting terms in AP:

• 3 terms: $a-d,a,a+d$
• 4 terms: $a-3d,a-d,a+d,a+3d$
• 5 terms: $a-2d,a-d,a,a+d,a+2d$

Property 2:

If $a,b,c$ are in AP, then:
$2b=a+c$

Property 3:

If each term of an AP is increased, decreased, multiplied, or divided by the same non-zero number, the resulting sequence is also an AP.

Property 4:

The sum of terms equidistant from the beginning and end is constant:
$a_1+a_n=a_2+a_{n-1}=a_3+a_{n-2}=\mathrm{.}\mathrm{.}\mathrm{.}$

Property 5:

If $a_n$ is the nth term, then:
$a_n=S_n-S_{n-1}$ (for $n>1$)

6. Solved Examples

Example 3:

Which term of the AP: $21,18,15,\mathrm{.}\mathrm{.}\mathrm{.}$ is $-81$?

Solution:

• $a=21$
• $d=18-21=-3$
• $a_n=-81$

Using $a_n=a+(n-1)d$:
$-81=21+(n-1)(-3)$
$-81=21-3n+3$
$-81=24-3n$
$-105=-3n$
$n=35$

Therefore, $-81$ is the 35th term.

Example 4:

Find the sum of all three-digit natural numbers divisible by 7.

Solution:
Three-digit numbers divisible by 7: $105,112,119,\mathrm{.}\mathrm{.}\mathrm{.},994$

• $a=105$
• $d=7$
• $l=994$

First, find $n$:
$a_n=a+(n-1)d$
$994=105+(n-1)\times 7$
$889=(n-1)\times 7$
$n-1=127$
$n=128$

Now, sum:
$S_n=\frac{n}{2}(a+l)$
$S_{128}=\frac{128}{2}(105+994)$
$S_{128}=64\times 1099=70,336$

Example 5:

The sum of first $n$ terms of an AP is $3n^2+5n$. Find the AP.

Solution:
$S_n=3n^2+5n$

For $n=1$:
$a_1=S_1=3(1{)}^2+5(1)=8$

For $n\ge 2$:
$a_n=S_n-S_{n-1}$
$a_n=(3n^2+5n)-[3(n-1{)}^2+5(n-1)]$
$a_n=3n^2+5n-3(n^2-2n+1)-5n+5$
$a_n=3n^2+5n-3n^2+6n-3-5n+5$
$a_n=6n+2$

Check: For $n=1$, $a_1=6(1)+2=8$ ✓

Therefore, AP is: $8,14,20,26,\mathrm{.}\mathrm{.}\mathrm{.}$ (with $a=8,d=6$)

7. CBSE Previous Year Questions (PYQs)

Q1. (2020, 2 marks)

Find the 11th term from the last term of the AP: $10,7,4,\mathrm{.}\mathrm{.}\mathrm{.},-62$.

Solution:

Method 1: (Converting to AP from the end)

• Last term becomes first term: $a=-62$
• Common difference becomes: $d=3$ (reverse sign)
• 11th term from last: $n=11$

$a_{11}=-62+(11-1)\times 3$
$a_{11}=-62+30=-32$

Method 2: (Finding total terms first)

• $a=10,d=-3,l=-62$
• Total terms: $-62=10+(n-1)(-3)$
• $n=25$
• 11th from last = $(25-11+1)$th from beginning = 15th term
• $a_{15}=10+14(-3)=-32$

Answer: $-32$

Q2. (2019, 3 marks)

If the sum of first 7 terms of an AP is 49 and that of first 17 terms is 289, find the sum of first $n$ terms.

Solution:

Given:
$S_7=49\text{and}{S}_{17}=289$

Using $S_n=\frac{n}{2}[2a+(n-1)d]$:

$S_7=\frac{7}{2}[2a+6d]=49$
$7a+21d=49$
$a+3d=7$ ... (i)

$S_{17}=\frac{17}{2}[2a+16d]=289$
$17a+136d=289$
$a+8d=17$ ... (ii)

Subtracting (i) from (ii):
$5d=10$
$d=2$

Substituting in (i):
$a+6=7$
$a=1$

Therefore:
$S_n=\frac{n}{2}[2(1)+(n-1)(2)]$
$S_n=\frac{n}{2}[2+2n-2]$
$\boxed{{\displaystyle S_n=n^2}}$

Q3. (2018, 4 marks)

The sum of first $n$ terms of three APs are $S_1,S_2,S_3$. The first term of each AP is 1 and common differences are 1, 2, 3 respectively. Prove that $S_1+S_3=2S_2$.

Solution:

For First AP: $a=1,d=1$
$S_1=\frac{n}{2}[2(1)+(n-1)(1)]$
$S_1=\frac{n}{2}[2+n-1]$
$S_1=\frac{n(n+1)}{2}$

For Second AP: $a=1,d=2$
$S_2=\frac{n}{2}[2(1)+(n-1)(2)]$
$S_2=\frac{n}{2}[2+2n-2]$
$S_2=\frac{n\cdot 2n}{2}=n^2$

For Third AP: $a=1,d=3$
$S_3=\frac{n}{2}[2(1)+(n-1)(3)]$
$S_3=\frac{n}{2}[2+3n-3]$
$S_3=\frac{n(3n-1)}{2}$

LHS:
$S_1+S_3=\frac{n(n+1)}{2}+\frac{n(3n-1)}{2}$
$=\frac{n(n+1+3n-1)}{2}$
$=\frac{n\cdot 4n}{2}=2n^2$

RHS:
$2S_2=2n^2$

Therefore, $S_1+S_3=2S_2$ Proved.

Q4. (2017, 3 marks)

How many terms of the AP: $-15,-13,-11,\mathrm{.}\mathrm{.}\mathrm{.}$ are needed to make the sum $-55$?

Solution:

Given: $a=-15,d=-13-(-15)=2,S_n=-55$

Using $S_n=\frac{n}{2}[2a+(n-1)d]$:
$-55=\frac{n}{2}[2(-15)+(n-1)(2)]$
$-110=n[-30+2n-2]$
$-110=n[2n-32]$
$-110=2n^2-32n$
$2n^2-32n+110=0$
$n^2-16n+55=0$

Factorizing:
$n^2-11n-5n+55=0$
$n(n-11)-5(n-11)=0$
$(n-5)(n-11)=0$
$n=5\text{ or }n=11$

Checking:

• For $n=5$: $S_5=\frac{5}{2}[-30+8]=\frac{5}{2}\times (-22)=-55$ ✓
• For $n=11$: $S_{11}=\frac{11}{2}[-30+20]=\frac{11}{2}\times (-10)=-55$ ✓

Answer: Both 5 terms and 11 terms give the sum $-55$.

(This happens because some terms become positive, and their sum cancels out.)

Q5. (2016, 4 marks)

The ratio of the 11th term to the 18th term of an AP is 2:3. Find the ratio of the 5th term to the 21st term. Also, find the ratio of the sum of first 5 terms to the sum of first 21 terms.

Solution:

Given: $\frac{a_{11}}{a_{18}}=\frac{2}{3}$

$\frac{a+10d}{a+17d}=\frac{2}{3}$
$3(a+10d)=2(a+17d)$
$3a+30d=2a+34d$
$a=4d$ ... (i)

Finding ratio of 5th to 21st term:
$\frac{a_5}{a_{21}}=\frac{a+4d}{a+20d}$

Substituting $a=4d$:
$\frac{a_5}{a_{21}}=\frac{4d+4d}{4d+20d}=\frac{8d}{24d}=\frac{1}{3}$

Finding ratio of sums:
$\frac{S_5}{S_{21}}=\frac{\frac{5}{2}[2a+4d]}{\frac{21}{2}[2a+20d]}$
$=\frac{5(2a+4d)}{21(2a+20d)}$

Substituting $a=4d$:
$=\frac{5(8d+4d)}{21(8d+20d)}=\frac{5\times 12d}{21\times 28d}$
$=\frac{60}{588}=\frac{5}{49}$

Answer:

• Ratio of 5th to 21st term = $1:3$
• Ratio of $S_5$ to $S_{21}$ = $5:49$

Q6. (2015, 2 marks)

For what value of $k$ will the consecutive terms $2k+1,3k+3,5k-1$ form an AP?

Solution:

For three terms to be in AP:
$a_2-a_1=a_3-a_2$

$(3k+3)-(2k+1)=(5k-1)-(3k+3)$
$k+2=2k-4$
$6=k$

Verification:

• For $k=6$: Terms are $13,21,29$
• Common difference = $21-13=8$ and $29-21=8$ ✓

Answer: $k=6$

Q7. (2014, 3 marks)

The sum of $n$ terms of two APs are in the ratio $(3n+8):(7n+15)$. Find the ratio of their 12th terms.

Solution:

Let the two APs have:

• First AP: first term = $a_1$, common difference = $d_1$
• Second AP: first term = $a_2$, common difference = $d_2$

$\frac{S_n^{(1)}}{S_n^{(2)}}=\frac{3n+8}{7n+15}$

$\frac{\frac{n}{2}[2a_1+(n-1)d_1]}{\frac{n}{2}[2a_2+(n-1)d_2]}=\frac{3n+8}{7n+15}$

$\frac{2a_1+(n-1)d_1}{2a_2+(n-1)d_2}=\frac{3n+8}{7n+15}$

For the 12th term, we need $n=23$ (because $a_{12}$ is the middle term of 23 terms):

Why 23? Because $a_{12}=\frac{S_{23}-S_{11}}{\text{terms}}$, but easier: $\frac{2a+(n-1)d}{2}$ at $n=23$ gives 12th term coefficient.

Actually, use: $a_n=\frac{S_n-S_{n-1}}{1}$, but better approach:

$\frac{a_{12}^{(1)}}{a_{12}^{(2)}}=\frac{a_1+11d_1}{a_2+11d_2}$

This equals $\frac{2a_1+22d_1}{2a_2+22d_2}$, so put $n-1=22$, i.e., $n=23$:

$\frac{a_{12}^{(1)}}{a_{12}^{(2)}}=\frac{3(23)+8}{7(23)+15}=\frac{69+8}{161+15}=\frac{77}{176}=\frac{7}{16}$

Answer: $7:16$

Practice Questions

Short Answer Type (2-3 marks each)

1. Find the 20th term from the end of the AP: $3,8,13,\mathrm{.}\mathrm{.}\mathrm{.},253$.

2. If the 10th term of an AP is 52 and 17th term is 20 more than the 13th term, find the AP.

3. How many two-digit numbers are divisible by 3?

4. The 4th term of an AP is 11 and the 8th term exceeds twice the 4th term by 5. Find the AP.

Long Answer Type (4 marks each)

5. In an AP, if the 12th term is $-13$ and the sum of its first four terms is 24, find the sum of first 10 terms.

6. Find the sum of all integers between 100 and 1000 which are divisible by 7.

7. The sum of first $m$ terms of an AP is $4m^2-m$. If its nth term is 107, find the value of $n$. Also, find the 21st term.

Important Tips for Exam

✅ Always identify: $a$, $d$, $n$, $a_n$, or $S_n$ from the question

✅ Choose the right formula:

• For finding terms → use $a_n=a+(n-1)d$
• For finding sum → use $S_n=\frac{n}{2}[2a+(n-1)d]$

✅ Common Mistakes to Avoid:

• Miscalculating common difference (always $d=a_2-a_1$)
• Forgetting $(n-1)$ in formulas
• Arithmetic errors in solving quadratics

✅ Word Problems: Read carefully and form equations systematically

✅ Show all steps in board exams for full marks