CBSE Class 10 Maths Chapter Arithmetic Progressions Notes

Arithmetic Progressions is one of the most interesting and practical chapters in CBSE Class 10 Mathematics. This chapter introduces students to number patterns where consecutive terms increase or decrease by a fixed value known as the common difference. From arranging seats in an auditorium to calculating savings plans and analysing patterns in daily life, arithmetic progressions have numerous real-world applications. In this chapter, students learn how to identify an arithmetic progression (AP), find the common difference, determine the nth term, and calculate the sum of a given number of terms. Also check out NCERT Exemplar for Class 10 Maths, NCERT Solutions for Class 10, & Maths formulas prepared and developed by experts at Myclass24. 

The concepts are developed gradually, making it easier for students to understand how mathematical sequences work. Board examination questions often focus on finding missing terms, calculating the nth term, and solving word problems based on AP. A strong understanding of arithmetic progressions helps students improve their algebraic thinking and logical reasoning skills. Since many competitive exams also include sequence-based questions, mastering this chapter provides long-term benefits beyond Class 10. Regular practice of CBSE Class 10 Mathematics formulas and application-based problems enables students to solve questions quickly and accurately, making Arithmetic Progressions an important scoring chapter in the CBSE Mathematics syllabus.

Arithmetic Progressions: Class 10 Notes, Complete Guide with Formulas and Applications

Class 10 Arithmetic Progressions Notes provide a clear understanding of sequences and series, focusing on the formula for the nth term, the sum of n terms, and real-life applications. These notes help students learn step-by-step techniques for solving AP problems efficiently. Each concept is explained with examples and solved exercises, making it easier to grasp patterns and formulas. Understanding arithmetic progressions is crucial for algebra, problem-solving, and competitive exams. The notes also include tips for identifying AP sequences, applying formulas quickly, and avoiding common mistakes. By practicing these notes, students can improve calculation speed, accuracy, and conceptual clarity. 

What is an Arithmetic Progression?

An Arithmetic Progression (AP) is a sequence of numbers in which each term after the first is obtained by adding a fixed constant to the preceding term. This constant is called the common difference, denoted by d.

For example, the sequences 3, 6, 9, 12,... and 4, 8, 12, 16,... are arithmetic progressions where each successive term increases by a fixed amount (3 and 4 respectively).

The general form of an AP can be written as:

a, a + d, a + 2d, a + 3d, ..., a + (n-1)d

where:

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

Key Concepts in Arithmetic Progressions

Common Difference

The common difference is calculated by subtracting any term from its succeeding term:

d = a₂ - a₁ = a₃ - a₂ = ... = constant

If d > 0, the AP is increasing; if d < 0, it is decreasing; and if d = 0, all terms are equal.

nth Term of an AP

The nth term (also called the general term) of an arithmetic progression is given by:

aₙ = a + (n - 1)d

This formula allows you to find any term in the sequence without listing all previous terms. For instance, to find the 18th term of the AP 7, 4, 1, -2, -5,..., you identify a = 7 and d = -3, then calculate a₁₈ = 7 + (18 - 1)(-3) = 7 - 51 = -44.

nth Term from the End

When dealing with a finite AP with last term l, the nth term from the end is:

aₙ (from end) = l - (n - 1)d

This is useful when you need to work backwards through a sequence.

Sum of n Terms of an Arithmetic Progression

The sum of the first n terms of an AP, denoted as Sₙ, can be calculated using two equivalent formulas:

Sₙ = (n/2)[2a + (n - 1)d]

or

Sₙ = (n/2)[a + l]

where l is the last term.

The second formula is particularly useful when the last term is known. For example, to find the sum of the first 20 terms of 1, 4, 7, 10,...:

S₂₀ = (20/2)[2(1) + (20-1)(3)] = 10[2 + 57] = 590

Relationship Between Terms and Sum

An important property: aₙ = Sₙ - Sₙ₋₁

This means any term can be found by subtracting consecutive sums.

Important Properties of Arithmetic Progressions

1. Equidistant Terms: In a finite AP, the sum of terms equidistant from the beginning and end is constant and equals the sum of the first and last terms.
2. Three Terms in AP: If three numbers a, b, c are in AP, then 2b = a + c. This is the condition for b to be the arithmetic mean of a and c.
3. Arithmetic Mean: The arithmetic mean (A.M.) between two numbers a and b is: A = (a + b)/2
4. Linear nth Term: If the nth term of a sequence is a linear expression in n (of the form an + b), then the sequence is always an AP with common difference equal to the coefficient of n.
5. Effect of Operations:
   • Adding or subtracting a constant from each term preserves the AP with the same common difference
   • Multiplying or dividing each term by a non-zero constant k creates a new AP with common difference kd or d/k respectively

Selection of Terms in AP

When solving problems involving multiple terms in AP, it's often convenient to select terms symmetrically:

This approach simplifies calculations because the middle term (or average of middle terms) equals a, and the sum is easy to compute.

Consolidated Formula Table

Practical Applications and Problem-Solving

Finding Unknown Terms

Many problems involve finding missing terms or verifying if a number belongs to an AP. For instance, to check if 184 is a term of the AP 3, 7, 11,...:

Set aₙ = 184, a = 3, d = 4: 184 = 3 + (n-1)(4) 181 = 4n - 4 185 = 4n n = 185/4 = 46.25

Since n is not a natural number, 184 is not a term of this AP.

Applications in Real Life

Arithmetic progressions appear in numerous practical contexts:

• Financial Planning: Regular savings with fixed monthly deposits follow an AP
• Construction: Stacking materials in rows (like logs or bricks) often follows AP patterns
• Depreciation: Linear depreciation of assets over time
• Time-based Patterns: Events occurring at regular intervals

Solving Complex Problems

When dealing with problems where relationships between terms are given (such as "the sum of 4th and 8th terms is 24"), set up equations using the general term formula and solve simultaneously to find a and d.

Study Tips for Mastering Arithmetic Progressions

1. Identify the Pattern: Always verify the common difference is constant before applying AP formulas
2. Choose the Right Formula: Decide whether to use term formulas or sum formulas based on what's given and what's required
3. Work Systematically: Write down known values (a, d, n, l, Sₙ) before solving
4. Check Your Answer: Verify results by substituting back into original conditions
5. Practice Variations: Work through problems involving terms from the end, missing terms, and sum-based questions

Class 10 Arithmetic Progressions Notes, Solved example & Questions

SEQUENCE & SERIES

Sequence: A sequence is an arrangement of number in a definite order, according to a definite rule.

Terms: Various numbers occurring in a sequence are called terms or element.

Consider the following lists of number:

3, 6, 9, 12, ……………

4, 8, 12, 16, ……………

–3, –2, –1, 0, ……………

In all the list above, we observe that each successive terms are obtained by adding a fixed number to the preceding terms. Such list of numbers is said to form on Arithmetic Progression (AP).

PROGRESSIONS

Those sequence whose terms follow certain patterns are called progression. Generally there are three types of progression.

(i) Arithmetic Progression (A.P.)

(ii) Geometric Progression (G.P.)

(iii) Harmonic Progression (H.P.)

ARTHMETIC PROGRESSION:

A sequence is called an A.P., if the difference of a term and the previous term is always same. i.e. d = t_n+1 – t_n = Constant for all . The constant difference, generally denoted by ‘d’ is called the common difference.

1. Find the common difference of the following A.P. : 1,4,7,10,13,16 ......

Sol. 4 – 1 = 7 – 4 = 10 – 7 = 13 – 10 = 16 – 13 = 3 (constant).

Common difference (d) = 3.

GENERAL FORM OF AN A.P. :

If we denote the starting number i.e. the 1^st number by ‘a’ and a fixed number to the added is ‘d’ then a, a + d, a + 2d, a + 3d, a + 4d, ...... forms an A.P.

1. Find the A.P. whose 1^st term is 10 & common difference is 5.

Sol. Given : First term (a) = 10 & Common difference (d) = 5.

A.P. is 10, 15, 20, 25, 30. ...

n^th TERM OF AN A.P. :

Let A.P. be a, a + d, a + 2d, a + 3d. ...

Then, First term (a₁) = a + 0.d

Second term (a₂) = a + 1.d

Third term (a₃) = a + 2.d

. .

. .

. .

n^th term (a_n) = a + (n - 1) d

a_n = a + (n - 1) d is called the n^th term.

Ques 1: Determine the A.P. whose third term is 16 and the difference of 5^th term from 7^th term is 12.

Sol. Given: a₃ = a + (3 – 1) d = a + 2d = 16 .....(i)

a₇ – a₅ = 12 ....(ii)

(a + 6d) – (a + 4d) = 12

a + 6d – a – 4d = 12

2d = 12

d = 6

Put d = 6 in equation (i)

a = 16 - 12

a = 4

A.P. is 4, 10, 16, 22, 28, ......

Ques 2. Which term of the sequence 72, 70, 68, 66, ..... is 40 ?

Sol. Here 1^st term x = 72 and common difference d = 70 – 72 = –2

For finding the value of n

a_n = a + (n – 1)d

40 = 72 + (n – 1) (–2)

40 – 72 = –2n + 2

–32 = –2n + 2

–34 = –2n

n = 17

17^th term is 40.

TO FIND nth TERM FROM THE END OF AN A.P.:

Consider the following A.P. a, a + d, a + 2d, … (l – 2d), (l – d),

where l is the last term

last term = - (1 - 1)d

2^nd last term - d = - (2 - 1) d

3^rd last term - 2d = - (3 - 1)d

……………………………

……………………………

nth term from the end = - (n - 1) d

Ques 3:  Find the 5^th term from the end of the AP, 17, 14, 11, ….., -40

Sol. 1^st method

Using

5^th term from the end will be

=  -40 - (5 - 1) x -3

= -40 -4 x -3

=- 40 + 12

= -28

2^nd method

Sequence can be written as -40, -37, ….11, 14, 17

a = -40

= -37 + 40

= 3

n = 5

Using

= a_n = a + (n-1)d

= -40 + 12

= -28

SELECTION OF TERMS IN AN A.P. :

Sometimes we require certain number of terms in A.P. The following ways of selecting terms are generally very convenient.

Ques 4. The sum of three number in A.P. is -3 and their product is 8. Find the numbers.

Sol. Three no. ‘s in A.P. be a – d, a, a + d

a – d + a + a + d = – 3

3a = –3

a = –1

& (a – d) a (a + d) = 8

a(a² – d²) = 8

(–1) (1 – d²) = 8

1 – d² = –8

d² = 9

d = 3

If a = 8 & d = 3 numbers are –4, –1, 2.

If a = 8 & d = – numbers are 2, –1, –4.

SUM OF n TERMS OF AN A.P. :

Let A.P. be a, a + d, a + 2d, a + 3d,...... a + (n - 1)d

Then,S_n = a + (a + d) + ..... + {a + (n - 2) d} + {a + (n - 1) d} .....(i)

also, S_n = {a + (n - 1)d} + {a + (n - 2)d} + ...... + (a + d) + a .....(ii)

Add (i) & (ii)

2S_n = 2a + (n – 1)d + 2a + (n – 1)d + ............. + 2a + (n – 1)d

2S_n = n [2a + (n – 1)d]

where is the last term.

Ques 5. Find the sum of 20 terms of the A.P. 1, 4, 7, 10...

Sol. a = 1, d = 3

Ques 6. Find the sum of all three digit natural numbers. Which are divisible by 7.

Sol. 1^st no. is 105 and last no. is 994.

We can find n as

994 = 105 + (n + 1)7

n = 128

Sum, _S128 = 128/2 [105 + 994]

PROPERTIES OF A.P. :

(A) For any real numbers a and b, the sequence whose n^th term is a_n= an + b is always an A.P. with common difference ‘a’ (i.e. coefficient of term containing n)

(B)If any n^th term of sequence is a linear expression in n then the given sequence is an A.P.

(C)If a constant term is added to or subtracted from each term of an A.P. then the resulting sequence is also an A.P. with the same common difference.

(D)If each term of a given A.P. is multiplied or divided by a non-zero constant K, then the resulting sequence is also an A.P. with common difference Kd or respectively. Where d is the common difference of the given A.P.

(E) In a finite A.P. the sum of the terms equidistant from the beginning and end is always same and is equal to the sum of 1^st and last term.

(F) If three numbers a, b, c are in A.P., then 2b = a + c.

Ques 7. Check whether a_n = 2n² + 1 is an A. P. or not.

Sol. a_n = 2n² + 1

Then a_n+1 = 2 (n + 1)² + 1

a_n+1 – a_n = 2(n² + 2n + 1) + 1 – 2n² – 1

= 2n² + 4n + 2 + 1 - 2n² – 1

= 4n + 2, which is not constant

The above sequence is not an A.P.

ARITHMETIC MEAN (AM)

When three quantities are in A.P., then the middle one is called as arithmetic mean of other two. If ‘a’ and ‘b’ are two numbers and A be the arithmetic mean of a and b, then a, A, b are in A.P.

A - a = b - A

A = a + b /2

Question: Is 184 a term of the sequence 3, 7, 11, .....?

Solution:

Given Information:

• First term (a) = 3
• Common difference (d) = 7 – 3 = 4

Formula for nth term:

a_n = a + (n – 1)d

Step-by-step calculation:

⇒ 184 = 3 + (n – 1) × 4 ⇒ 184 – 3 = (n – 1) × 4 ⇒ 181 = 4n – 4 ⇒ 181 + 4 = 4n ⇒ 185 = 4n ⇒ n = 185/4 ⇒ n = 46.25

Since n is not a natural number (it's not a whole positive number), 184 is NOT a term of the given sequence.

Question: Which term of the sequence 20, 19¼, 18½, 17¾, ..... is the 1st negative term?

Solution:

Given Information:

• First term (a) = 20
• Common difference (d) = 19¼ – 20 = 19.25 – 20 = –¾ = –0.75

Condition for negative term:

Let nth term of the given A.P. be 1st negative term, so a_n < 0

a + (n – 1)d < 0

Step-by-step calculation:

⇒ 20 + (n – 1) × (–¾) < 0 ⇒ 20 – ¾(n – 1) < 0 ⇒ 20 < ¾(n – 1) ⇒ 80/4 < ¾n – ¾ ⇒ 80 < 3n – 3 ⇒ 83 < 3n ⇒ 3n > 83 ⇒ n > 83/3 ⇒ n > 27⅔ ⇒ n > 27.67

Since 28 is the natural number just greater than 27⅔, the 1st negative term is the 28th term.

Problem 3: Proving an Identity with A.P. Terms

Question: If p^th, q^th and r^th term of an A.P. are a, b, c respectively, then show that: a(q – r) + b(r – p) + c(p – q) = 0

Solution:

Given Information:

• a_p = a ⇒ A + (p - 1)D = a ......(1)
• a_q = b ⇒ A + (q - 1)D = b ......(2)
• a_r = c ⇒ A + (r - 1)D = c ......(3)

Where A is the first term and D is the common difference.

Left Hand Side (L.H.S.):

L.H.S. = a(q – r) + b(r – p) + c(p – q)

Substituting values from equations (1), (2), and (3):

= {A + (p – 1)D}(q – r) + {A + (q – 1)D}(r – p) + {A + (r – 1)D}(p – q)

Expanding:

= A(q – r) + (p – 1)D(q – r) + A(r – p) + (q – 1)D(r – p) + A(p – q) + (r – 1)D(p – q)

Grouping terms:

= A[(q – r) + (r – p) + (p – q)] + D[(p – 1)(q – r) + (q – 1)(r – p) + (r – 1)(p – q)]

Simplifying:

= A[q – r + r – p + p – q] + D[...] = A[0] + D[0] = 0 = R.H.S.

Hence proved: a(q – r) + b(r – p) + c(p – q) = 0

Question: If m times the m^th term of an A.P. is equal to n times its n^th term, show that the (m + n)^th term of the A.P. is zero.

Solution:

Let:

• A = first term of the A.P.
• D = common difference of the given A.P.

Given condition:

m × a_m = n × a_n

Step-by-step proof:

⇒ m[A + (m – 1)D] = n[A + (n – 1)D] ⇒ mA + m(m – 1)D = nA + n(n – 1)D ⇒ mA – nA = n(n – 1)D – m(m – 1)D ⇒ A(m – n) = D[n(n – 1) – m(m – 1)] ⇒ A(m – n) = D[n² – n – m² + m] ⇒ A(m – n) = D[–(m² – n²) + (m – n)] ⇒ A(m – n) = D[–(m – n)(m + n) + (m – n)] ⇒ A(m – n) = D(m – n)[–(m + n) + 1] ⇒ A = D[1 – (m + n)] ⇒ A + D(m + n – 1) = 0 ⇒ A + (m + n – 1)D = 0

This is the formula for (m + n)th term:

a_m+n = A + (m + n – 1)D = 0

Therefore, a_m+n = 0

Hence proved: The (m + n)th term of the A.P. is zero.

Question: If the p^th term of an A.P. is q and the q^th term is p, prove that its n^th term is (p + q – n).

Solution:

Given Information:

• a_p = q ⇒ A + (p – 1)D = q ......(i)
• a_q = p ⇒ A + (q – 1)D = p ......(ii)

Subtracting equation (ii) from equation (i):

[A + (p – 1)D] – [A + (q – 1)D] = q – p A + (p – 1)D – A – (q – 1)D = q – p (p – 1)D – (q – 1)D = q – p D[(p – 1) – (q – 1)] = q – p D[p – 1 – q + 1] = q – p D(p – q) = q – p D(p – q) = –(p – q) D = –1

Substituting D = –1 in equation (i):

A + (p – 1)(–1) = q A – p + 1 = q A = q + p – 1 A = p + q – 1

Finding nth term:

a_n = A + (n – 1)D a_n = (p + q – 1) + (n – 1)(–1) a_n = p + q – 1 – n + 1 a_n = p + q – n

Therefore, the nth term is: a_n = p + q – n

Hence proved.

Question: If the m^th term of an A.P. is 1/n and n^th term is 1/m, then show that its (mn)^th term is 1.

Solution:

Given Information:

a_m = 1/n ⇒ A + (m – 1)D = 1/n ......(i) a_n = 1/m ⇒ A + (n – 1)D = 1/m ......(ii)

Subtracting equation (ii) from equation (i):

[(m – 1)D] – [(n – 1)D] = 1/n – 1/m D[(m – 1) – (n – 1)] = (m – n)/mn D[m – n] = (m – n)/mn D = 1/mn

Substituting D = 1/mn in equation (i):

A + (m – 1) × (1/mn) = 1/n A = 1/n – (m – 1)/mn A = [m – (m – 1)]/mn A = 1/mn

Finding (mn)th term:

a_mn = A + (mn – 1)D a_mn = 1/mn + (mn – 1) × (1/mn) a_mn = 1/mn + (mn – 1)/mn a_mn = [1 + mn – 1]/mn a_mn = mn/mn a_mn = 1

Therefore, a_mn = 1

Hence proved.

Question: Write the A.P., when the first term a and common difference d is as follows:

• (i) a = 10, d = 10
• (ii) a = –1, d = ½
• (iii) a = –1.25, d = 0.25

Solution:

(i) When a = 10, d = 10

The required A.P. is:

10, 10 + 10, 10 + 2 × 10, 10 + 3 × 10, …

Answer: 10, 20, 30, 40, …

(ii) When a = –1, d = ½

The required A.P. is:

–1, –1 + ½, –1 + 2 × ½, –1 + 3 × ½, ... –1, –1 + 0.5, –1 + 1, –1 + 1.5, ...

Answer: –1, –½, 0, ½, ...

(iii) When a = –1.25, d = 0.25

The required A.P. is:

–1.25, –1.25 + (0.25), –1.25 + 2 × (0.25), –1.25 + 3 × (0.25), ... –1.25, –1.25 + 0.25, –1.25 + 0.50, –1.25 + 0.75, ...

Answer: –1.25, –1.50, –1.75, –2.0, ...

Question: Find the 18^th term and n^th term for the sequence 7, 4, 1, –2, –5.

Solution:

Given Information:

• First term: a = 7
• Common difference: d = a₂ – a₁ = 4 – 7 = –3

Finding the 18th term:

Using the formula:

a_n = a + (n – 1)d

For n = 18:

a₁₈ = 7 + (18 – 1) × (–3) a₁₈ = 7 + 17 × (–3) a₁₈ = 7 – 51 a₁₈ = –44

Finding the nth term:

a_n = a + (n – 1)d a_n = 7 + (n – 1)(–3) a_n = 7 – 3n + 3 a_n = 10 – 3n

18th term: a₁₈ = –44

nth term: a_n = 10 – 3n

Question: Which term of the A.P. 7, 12, 17, ..... is 87?

Solution:

Given Information:

• First term: a = 7
• Common difference: d = a₂ – a₁ = 12 – 7 = 5
• We need to find n when a_n = 87

Using the formula:

a_n = a + (n – 1)d

Step-by-step calculation:

87 = 7 + (n – 1) × 5 87 – 7 = 5n – 5 80 = 5n – 5 80 + 5 = 5n 85 = 5n n = 85/5 n = 17

Therefore, 87 is the 17th term of the given A.P.

Question: If 3^rd term and 9^th term of an A.P. are 4 and –8 respectively, which term is zero?

Solution:

Let:

• a = first term
• d = common difference

Given Information:

3^rd term: a₃ = a + 2d = 4 ......(i) 9^th term: a₉ = a + 8d = –8 ......(ii)

Subtracting equation (i) from equation (ii):

(a + 8d) – (a + 2d) = –8 – 4
a + 8d – a – 2d = –12
6d = –12
d = –2

Substituting d = –2 in equation (i):

a + 2(–2) = 4
a – 4 = 4
a = 8

Finding which term is zero:

Let a_n = 0

a + (n – 1)d = 0
8 + (n – 1)(–2) = 0
8 – 2(n – 1) = 0
8 – 2n + 2 = 0
10 – 2n = 0
2n = 10
n = 5

Hence, the 5th term is zero.

Question: The sum of the 4^th and 8^th terms of an A.P. is 24 and sum of 6^th & 10^th terms is 44. Find the first three terms of the AP.

Solution:

Using 1st condition:

a₄ + a₈ = 24
(a + 3d) + (a + 7d) = 24
2a + 10d = 24
a + 5d = 12 ......(i)

Using 2nd condition:

a₆ + a₁₀ = 44
(a + 5d) + (a + 9d) = 44
2a + 14d = 44
a + 7d = 22 ......(ii)

Subtracting equation (i) from (ii):

(a + 7d) – (a + 5d) = 22 – 12
2d = 10
d = 5

Putting value of d in equation (i):

a + 5 × 5 = 12
a + 25 = 12
a = 12 – 25
a = –13

Finding the first three terms:

a₁ = a = –13
a₂ = a + d = –13 + 5 = –8
a₃ = a + 2d = –13 + 2(5) = –13 + 10 = –3

Answer: The first three terms are –13, –8, –3

Problem 12: Sum of 25 Terms with Given nth Term

Question: Find the sum of first 25 terms of an A.P. whose nth term is 1 – 4n.

Solution:

Given:

a_n = 1 – 4n

Finding first term (n = 1):

a₁ = 1 – 4(1) = 1 – 4 = –3

Finding 25th term (n = 25):

a₂₅ = 1 – 4(25) = 1 – 100 = –99

Using sum formula:

S_n = (n/2)[a₁ + a_n]

Calculating S₂₅:

S₂₅ = (25/2)[a₁ + a₂₅]
S₂₅ = (25/2)[–3 + (–99)]
S₂₅ = (25/2)[–102]
S₂₅ = 25 × (–51)
S₂₅ = –1275

Answer: S₂₅ = –1275

Question: If the sum of first p terms of an A.P. is ap² + bp, find its common difference.

Solution:

Given:

S_p = ap² + bp

Finding individual terms:

For p = 1:

S₁ = a(1)² + b(1) = a + b
This is the first term: a₁ = a + b

For p = 2:

S₂ = a(2)² + b(2) = 4a + 2b

For p = 3:

S₃ = a(3)² + b(3) = 9a + 3b

Finding second term:

a₂ = S₂ – S₁
a₂ = (4a + 2b) – (a + b)
a₂ = 4a + 2b – a – b
a₂ = 3a + b

Finding common difference:

d = a₂ – a₁
d = (3a + b) – (a + b)
d = 3a + b – a – b
d = 2a

Answer: Common difference d = 2a

Question: The sum of the first 17 terms of an arithmetic sequence is 187. If a₁₇ = –13, find a and d.

Solution:

Using sum formula:

S_n = (n/2)(a + a_n)

Given:

• S₁₇ = 187
• a₁₇ = –13
• n = 17

Substituting values:

187 = (17/2)(a + (–13))
187 = (17/2)(a – 13)
187 × 2/17 = a – 13
374/17 = a – 13
22 = a – 13
a = 22 + 13
a = 35 ......(i)

Finding d using nth term formula:

a_n = a + (n – 1)d
a₁₇ = a + (17 – 1)d
–13 = 35 + 16d [from (i)]
16d = –13 – 35
16d = –48
d = –48/16
d = –3

Answer: a = 35, d = –3

Question: Find the sum of positive integers less than 400.

Solution:

The sequence of positive integers less than 400:

1, 2, 3, 4, 5, ...., 399

Given Information:

• First term: a = 1
• Common difference: d = 2 – 1 = 1
• Last term: a_n = 399

Finding number of terms (n):

a_n = a + (n – 1)d
399 = 1 + (n – 1) × 1
399 = 1 + n – 1
399 = n
n = 399

Finding the sum:

S_n = (n/2)[2a + (n – 1)d]

Or alternatively:

S_n = (n/2)[a + l]

Calculating S₃₉₉:

S₃₉₉ = (399/2)[1 + 399]
S₃₉₉ = (399/2)[400]
S₃₉₉ = (399 × 400)/2
S₃₉₉ = 159600/2
S₃₉₉ = 79800

Answer: The sum of positive integers less than 400 is 79,800