1
The common difference of 7, 11, 15, 19 is —
💡 Hint:
11 − 7.
Explanation:
d = 4.
Class 10 · Maths · Chapter 5 · बिहार बोर्ड (BSEB)CBSE · NCERT 2026-27
Arithmetic Progressions
Arithmetic Progressions
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- 1 An AP and the common difference
- 2 The general form a, a+d, a+2d
- 3 The nth term an = a + (n − 1)d
- 4 Which term is it — finding n
- 5 The sum Sn = n/2 [2a + (n − 1)d]
- 6 The sum with the last term, and a saving
- Chapter winner — every lesson at mastery ★
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📄 PDF download — notes + answer key
An AP and the common difference
समांतर श्रेढ़ी और सार्व अंतर · NCERT 5.2 · Exercise 5.1
The difference is the same every timeNotes
A list of numbers is an arithmetic progression when, after the first term, each term is the previous term plus one fixed number. That number is the common difference d.
d = next term − previous term. If d is positive the list rises, if it is negative the list falls, and if it is zero every term is equal. Exercise 5.1 is this recognition. Stopping after one difference is not enough.
Find two differences in a rowActivity
In 7, 11, 15, 19, 11 − 7 = 4 and 15 − 11 = 4. The third difference is also 4, so d = 4. In 2, 4, 8, 16 the differences are 2 and then 4. They are not equal, so this is not an arithmetic progression.
Worked exampleExample
Question: Check 10, 7, 4, 1. Find d.
Formula: d = next term − previous term.
Substitution: 7 − 10 = −3, 4 − 7 = −3, 1 − 4 = −3.
Answer: It is an arithmetic progression. The exact common difference is d = −3.
10-second revision
- d is the same at every step
- d may be negative or zero
- If two differences differ, it is not an AP
Board tip · BSEBBoard tip
In a BSEB answer show at least two difference lines, do not write only “yes”.
Board tip · CBSEBoard tip
On CBSE do not treat 2, 4, 8, 16 as arithmetic. It is geometric.
Check your understandingall correct = mastery ★
2
2, 4, 8, 16 is an arithmetic progression.
Explanation:
False — the differences are 2, 4, 8, not fixed.
3
The common difference of 10, 7, 4, 1 is ______.
💡 Hint:
7 − 10.
Explanation:
d = −3.
4
In a list with d = 0 —
💡 Hint:
Nothing is added.
Explanation:
Every term stays equal to the first term. This is still an AP.
5
Is −5, −1, 3, 7 an AP? Find d.
Model answer (board format)
−1 − (−5) = 4 and 3 − (−1) = 4. Formula: d = next − previous. Exact d = 4, so it is an AP.
Check your answer:
The general form a, a+d, a+2d
सामान्य रूप a, a+d, a+2d · First, second and third terms
The term number tells how many differencesNotes
If the first term is a, the second is a + d, the third is a + 2d and the fourth is a + 3d. Up to the kth term, (k − 1) differences are added.
So the fifth term is a + 4d, not a + 5d. That slip swaps n and (n − 1). The general form is a, a+d, a+2d, a+3d, ...
Subtract 1 from the term numberActivity
The list with a = 4 and d = 5 is 4, 9, 14, 19, 24. The fifth term is 4 + 4×5 = 24. Four differences are added, not five. Write the list and match the count.
Worked exampleExample
Question: If a = 4 and d = 5, find the fifth term.
Formula: The kth term = a + (k − 1)d.
Substitution: 4 + (5 − 1)×5 = 4 + 4×5 = 4 + 20.
Answer: The exact fifth term is 24. The list is 4, 9, 14, 19, 24.
10-second revision
- The second term is a+d
- The fifth term adds 4 differences
- a + nd is not the fifth term when n = 5
Board tip · BSEBBoard tip
In a BSEB answer write (k − 1) in brackets so that 5 − 1 = 4 is visible.
Board tip · CBSEBoard tip
On CBSE, writing a + 5d as the fifth term loses the mark.
Check your understandingall correct = mastery ★
1
The third term for a = 4, d = 5 is —
💡 Hint:
a + 2d.
Explanation:
4 + 2×5 = 14.
2
The fifth term is a + 5d.
Explanation:
False — the fifth term is a + 4d.
3
The fourth term is a + ______ d.
💡 Hint:
4 − 1.
Explanation:
a + 3d.
4
If a = 6 and d = −2, write the first four terms.
Model answer (board format)
6, 6+(−2)=4, 4+(−2)=2, 2+(−2)=0. The exact terms are 6, 4, 2, 0.
Check your answer:
The nth term an = a + (n − 1)d
nवाँ पद an = a + (n − 1)d · NCERT 5.3 · Exercise 5.2
The general term is the same formulaNotes
The nth term, or the general term, is an = a + (n − 1)d. Exercise 5.2 is on this.
n is a positive integer. Putting a number in place of n gives that term. If two terms are known, two equations appear and a and d come out.
Put n in, and do the bracket firstActivity
If the 10th term is asked, n = 10 and the product is (10 − 1)d = 9d. Writing 10d jumps one term ahead. After the answer, write the first and second terms and check d.
Worked exampleExample
Question: Find the 10th term of 4, 9, 14, ...
Formula: an = a + (n − 1)d.
Substitution: a = 4, d = 5, n = 10. a10 = 4 + 9×5 = 4 + 45.
Answer: The exact 10th term is 49.
10-second revision
- an = a + (n − 1)d
- The 10th term adds 9d
- n is a positive integer
Board tip · BSEBBoard tip
In a BSEB answer write a, d and n before the substitution.
Board tip · CBSEBoard tip
On CBSE do not replace (n − 1) by n. That is the most common loss of marks.
Check your understandingall correct = mastery ★
1
The 10th term of 4, 9, 14, ... is —
💡 Hint:
4 + 9×5.
Explanation:
a10 = 49.
2
an = a + nd is the correct general term of an AP.
Explanation:
False — the correct formula is a + (n − 1)d.
3
If a = 4, d = 5 and n = 10, then (n − 1)d = ______.
💡 Hint:
9 × 5.
Explanation:
45. Then the term is 4 + 45 = 49.
4
When n = 1, an equals —
💡 Hint:
(1 − 1)d = 0.
Explanation:
The first term is a.
5
Two known terms can make two equations for a and d.
Explanation:
True — each term gives a + (n − 1)d.
6
If a = 7 and d = −3, find the 8th term.
Model answer (board format)
an = a + (n − 1)d. a8 = 7 + 7×(−3) = 7 − 21 = −14.
Check your answer:
Which term is it — finding n
कौन-सा पद है — n निकालना · an is given · n is an integer
Set the term equal to the formulaNotes
If a number sits in the progression, it equals an for some n. Write number = a + (n − 1)d and solve for n.
n must come out as a positive integer. If n is a fraction, that number is not a term of this progression. Gather like terms before you divide by d.
Keep n − 1 on its ownActivity
In 3, 8, 13, 18, ..., where does 78 sit? 78 = 3 + (n − 1)×5. First write 75 = (n − 1)×5, then n − 1 = 15. n is not 15 itself.
Worked exampleExample
Question: Which term of 3, 8, 13, 18, ... is 78?
Formula: an = a + (n − 1)d.
Substitution: a = 3, d = 5. 78 = 3 + (n − 1)×5. 75 = (n − 1)×5. n − 1 = 15.
Answer: n = 16. The exact 16th term is 78. Check: 3 + 15×5 = 78.
10-second revision
- Find n − 1 first, then n
- n must be a positive integer
- In the check, put the term back into the formula
Board tip · BSEBBoard tip
In a BSEB answer keep both lines, n − 1 = 15 and n = 16.
Board tip · CBSEBoard tip
On CBSE, if n is a fraction write “not a term”. Do not force it into an integer.
Check your understandingall correct = mastery ★
1
In 3, 8, 13, ..., 78 is —
💡 Hint:
n − 1 = 15.
Explanation:
The 16th term.
2
If n comes out as a fraction, it is still taken as a term number.
Explanation:
False — a term number is a positive integer.
3
In 78 = 3 + (n − 1)×5, n − 1 = ______.
💡 Hint:
75 / 5.
Explanation:
15. So n = 16.
4
For a = 2, d = 3, 2 + (n − 1)×3 = 2 means —
💡 Hint:
(n − 1)×3 = 0.
Explanation:
n − 1 = 0, n = 1. This is the first term itself.
5
Which term of 5, 9, 13, ... is 49?
Model answer (board format)
a = 5, d = 4. 49 = 5 + (n − 1)×4. 44 = (n − 1)×4. n − 1 = 11. n = 12. Check: 5 + 11×4 = 49.
Check your answer:
The sum Sn = n/2 [2a + (n − 1)d]
योग Sn = n/2 [2a + (n − 1)d] · NCERT 5.4 · Exercise 5.3
Finish the bracket, then halve nNotes
The sum of the first n terms is Sn = n/2 [2a + (n − 1)d]. Exercise 5.3 is this. Exercise 5.4 is optional practice.
Apply n/2 at the end. Find the value of the bracket first. The sum of the first n positive integers is the same formula: with a = 1 and d = 1, Sn = n(n + 1)/2.
Check the bracket against a short additionActivity
The sum of the first three terms of 2, 5, 8 is 15. Formula: S3 = 3/2 [4 + 2×3] = 3/2 × 10 = 15. If a short addition matches the formula, a larger n follows the same path.
Worked exampleExample
Question: Find the sum of the first 10 terms of 2, 5, 8, ...
Formula: Sn = n/2 [2a + (n − 1)d].
Substitution: a = 2, d = 3, n = 10. S10 = 10/2 [4 + 9×3] = 5 × [4 + 27] = 5 × 31.
Answer: The exact sum is 155.
10-second revision
- Sn = n/2 [2a + (n − 1)d]
- The bracket comes before n/2
- 1 + 2 + ... + n = n(n + 1)/2
Board tip · BSEBBoard tip
In a BSEB answer add 2a and (n − 1)d on a separate line.
Board tip · CBSEBoard tip
On CBSE the product of n/2 and the bracket should become one number at the end. Leaving it midway is incomplete.
Check your understandingall correct = mastery ★
1
S10 of 2, 5, 8, ... is —
💡 Hint:
5 × 31.
Explanation:
S10 = 155.
2
The sum of the first n positive integers is n(n + 1)/2.
Explanation:
True — it comes from a = 1, d = 1.
3
S10 = 5 × 31 = ______.
💡 Hint:
Five times thirty-one.
Explanation:
155.
4
The sum for a = 1, d = 1, n = 10 is —
💡 Hint:
10×11/2.
Explanation:
S10 = 55.
5
In the sum formula it is fine to multiply n/2 first and leave the bracket unfinished.
Explanation:
False — find the value of the bracket first.
6
Find the sum of the first 8 terms of 7, 11, 15, ...
Model answer (board format)
a = 7, d = 4, n = 8. S8 = 8/2 [14 + 7×4] = 4 × [14 + 28] = 4 × 42 = 168.
Check your answer:
The sum with the last term, and a saving
अंतिम पद वाला योग और बचत · Sn = n/2 (a + l) · rupees
l is the last term, not just any termNotes
If the last term l is known, Sn = n/2 (a + l). This matches the first formula, because l = a + (n − 1)d.
In a money list, write the unit rupees. The month number is n. The amount in the 12th month is an, and the total for the year is Sn. Do not treat them as the same number.
Keep the term and the sum in different columnsActivity
The first month is 100 rupees, and each month is 20 rupees more. The 12th month is an. The total of twelve months is Sn. Find an first, then Sn. Use the last-term formula only when l is already known.
Worked exampleExample
Question: The first term is 5, the last term is 45 and the sum is 400. Find n and d.
Formula: Sn = n/2 (a + l) and d = (l − a)/(n − 1).
Substitution: 400 = n/2 (5 + 45) = 25n. n = 16. d = (45 − 5)/15 = 40/15.
Answer: Exact values n = 16 and d = 8/3. Check: S16 = 16/2 × (5 + 45) = 400.
10-second revision
- Sn = n/2 (a + l)
- d = (l − a)/(n − 1)
- The monthly amount is an; the year total is Sn
Board tip · BSEBBoard tip
In a BSEB answer, if both n and d are asked, write both checks.
Board tip · CBSEBoard tip
On CBSE, dropping the rupee unit in a money answer loses the mark.
Check your understandingall correct = mastery ★
1
If a = 5, l = 45 and S = 400, then n is —
💡 Hint:
25n = 400.
Explanation:
n = 16.
2
The saving in the 12th month and the total of 12 months are the same amount.
Explanation:
False — one is an and the other is Sn.
3
If a = 5, l = 45 and n = 16, then d = ______.
💡 Hint:
40/15.
Explanation:
d = 8/3.
4
A person saves 100 rupees in the first month and 20 rupees more each month. Find the saving in the 12th month and the total for the year.
Model answer (board format)
a = 100, d = 20, n = 12. a12 = 100 + 11×20 = 320 rupees. S12 = 12/2 [200 + 11×20] = 6 × 420 = 2520 rupees.
Check your answer:
❓ Full question bank — with answers and explanations — 64 questions
Pick a type. The 30 lesson checks are separate — each lesson has as many as its topic needs. All correct earns mastery ★.
BSEB modelpractice · not annual60 tried · 0 correctMultiple choiceMCQ120 tried · 0 correctTrue / falseTrue / false80 tried · 0 correctFill in the blanksFill80 tried · 0 correctMatchMatch20 tried · 0 correctAssertion–reasonAssertion–reason50 tried · 0 correctBalancingBalancing40 tried · 0 correctIdentifyIdentify20 tried · 0 correctShort and longVSA, SA, LA110 tried · 0 correctCBSE-stylenot a PYQ60 tried · 0 correct
No question is marked as a verified past paper. The BSEB set is a model for practice. CBSE items are CBSE-style, not a copy of any year’s paper.
Multiple choice
0/12Pick one option. A wrong try brings a hint.
1
d for 10, 7, 4, 1 is —
💡 Hint:
7 − 10.
Explanation:
d = −3.
2
The 10th term for a = 4, d = 5 is —
💡 Hint:
4 + 9×5.
Explanation:
49.
3
In 3, 8, 13, ..., 78 is —
💡 Hint:
n − 1 = 15.
Explanation:
The 16th term.
4
S10 of 2, 5, 8, ... is —
💡 Hint:
5 × 31.
Explanation:
155.
5
If a = 5, l = 45 and Sn = 400, n is —
💡 Hint:
25n = 400.
Explanation:
n = 16.
6
2, 4, 8, 16 is —
💡 Hint:
The difference is not fixed.
Explanation:
Not arithmetic.
7
In Sn = n/2 (a + l), l is —
💡 Hint:
The last.
Explanation:
The last term.
8
The sum of the first 10 positive integers is —
💡 Hint:
10×11/2.
Explanation:
55.
9
The 12th term for a = 100, d = 20 is —
💡 Hint:
100 + 11×20.
Explanation:
320 rupees. 2520 is the total sum.
10
S1 equals —
💡 Hint:
The first term is the first sum.
Explanation:
S1 = a.
11
If a = 5, l = 45 and n = 16 in d = (l − a)/(n − 1), d is —
💡 Hint:
40/15.
Explanation:
8/3.
12
The fifth term is —
💡 Hint:
Four differences.
Explanation:
a + 4d.
True or false
0/81
The common difference may be negative.
Explanation:
True — the list then decreases.
2
Every increasing list is an arithmetic progression.
Explanation:
False — the difference must be fixed.
3
an = a + (n − 1)d.
Explanation:
The correct formula.
4
A fraction can also be a term number.
Explanation:
False — n is a positive integer.
5
Sn = n/2 [2a + (n − 1)d].
Explanation:
The correct sum formula.
6
The amount in the 12th month and the sum of 12 months are equal.
Explanation:
False — an and Sn are different.
7
A list with d = 0 is not an arithmetic progression.
Explanation:
False — equal terms are still an AP.
8
l = a + (n − 1)d.
Explanation:
True — the last term is the general term.
Fill in the blanks
0/81
The symbol of the common difference is ______.
💡 Hint:
d.
Explanation:
The difference.
2
an = a + (n − ______)d.
💡 Hint:
1.
Explanation:
One less.
3
In the list whose S10 is 155, the sum is ______.
💡 Hint:
155.
Explanation:
Five times 31.
4
If a = 5, l = 45 and the sum is 400, n = ______.
💡 Hint:
16.
Explanation:
25n = 400.
5
The sum of the first n positive integers is n(n + ______)/2.
💡 Hint:
1.
Explanation:
n(n+1)/2.
6
100 + 11×20 = ______ rupees.
💡 Hint:
320 rupees.
Explanation:
The 12th month.
7
an = Sn − S(______) when n ≥ 2.
💡 Hint:
n−1.
Explanation:
The previous sum.
8
The fifth term is a + ______ d.
💡 Hint:
4.
Explanation:
5 − 1.
Match
0/21
Match the name with the formula.
Column B: A. Next − previous · B. a + (n − 1)d · C. n/2 [2a + (n − 1)d] · D. n/2 (a + l)
1. nth term
2. Sum of n terms
3. Sum from the last term
4. Common difference
Explanation:
The term has (n−1)d, the sum has n/2 and a bracket, the last-term sum has (a+l), and the difference is a subtraction.
2
Match the exercise with its job.
Column B: A. Optional · B. Recognition and d · C. The nth term · D. The sum
1. 5.1
2. 5.2
3. 5.3
4. 5.4
Explanation:
5.1 is recognition, 5.2 the term, 5.3 the sum, and 5.4 is optional.
Assertion–reason
0/5Check both statements, then see whether the reason explains the assertion.
1
Assertion (A): 10, 7, 4, 1 is an arithmetic progression.
Reason (R): Every difference is −3.
Explanation:
Both are true and R explains A.
2
Assertion (A): an = a + (n − 1)d.
Reason (R): The sum of the first n positive integers is n(n + 1)/2.
Explanation:
Both are true, but R does not explain the term formula.
3
Assertion (A): S10 of 2, 5, 8, ... is 155.
Reason (R): The common difference of this progression is 2.
Explanation:
A is true. R is false, d = 3.
4
Assertion (A): 2, 4, 8, 16 is an arithmetic progression.
Reason (R): In an AP the difference must be fixed.
Explanation:
A is false. R is true.
5
Assertion (A): If a = 5, l = 45 and the sum is 400, there are 16 terms.
Reason (R): 400 = n/2 × 50, so n = 16.
Explanation:
Both are true and R is the calculation.
Coefficient practice
0/4These chemistry lines are not a result of this maths chapter. They are only practice in filling coefficients. A blank means 1.
1
This is coefficient practice and not a result of this maths chapter. In the line where H2 + O2 makes water, Fill coefficients (blank = 1).
H2 + O2 → H2O
Explanation: 2H2 + O2 → 2H2O
2
This is coefficient practice and not a result of this maths chapter. In the line where N2 + H2 makes ammonia, Fill coefficients (blank = 1).
N2 + H2 → NH3
Explanation: N2 + 3H2 → 2NH3
3
This is coefficient practice and not a result of this maths chapter. In the line where C + O2 makes carbon dioxide, Fill coefficients (blank = 1).
C + O2 → CO2
Explanation: C + O2 → CO2
4
This is coefficient practice and not a result of this maths chapter. In the burning line of CH4 + O2, Fill coefficients (blank = 1).
CH4 + O2 → CO2 + H2O
Explanation: CH4 + 2O2 → CO2 + 2H2O
Classify
0/21
Place each list as an AP or not.
7, 11, 15, 19
2, 4, 8, 16
10, 7, 4, 1
3, 3, 3, 3
Explanation:
The first has d = 4, the third d = −3 and the fourth d = 0. The second has changing differences.
2
Place each amount as an or Sn. a = 100 rupees, d = 20 rupees, 12 months.
320 rupees
2520 rupees
100 rupees, first month
The total of twelve months
Explanation:
320 and 100 are terms. 2520 and the twelve-month total are sums.
Very short answer
0/41
Define an arithmetic progression in one line.
Model answer (board format)
A list in which each term is the previous term plus a fixed number d.
Check your answer:
2
Write the formula for the nth term.
Model answer (board format)
an = a + (n − 1)d.
Check your answer:
3
Write both formulas for the sum.
Model answer (board format)
Sn = n/2 [2a + (n − 1)d] and Sn = n/2 (a + l).
Check your answer:
4
If Sn is given, how is an found?
Model answer (board format)
a = S1. For n ≥ 2, an = Sn − S(n − 1).
Check your answer:
Short answer
0/41
Find the 15th term of 6, 10, 14, ...
Model answer (board format)
a = 6, d = 4. a15 = 6 + 14×4 = 6 + 56 = 62.
Check your answer:
2
Find the sum of the first 12 terms of 4, 7, 10, ...
Model answer (board format)
a = 4, d = 3, n = 12. S12 = 12/2 [8 + 11×3] = 6 × [8 + 33] = 6 × 41 = 246.
Check your answer:
3
If Sn = 4n − n², find the first term and the common difference.
Model answer (board format)
S1 = 4 − 1 = 3 = a. S2 = 8 − 4 = 4. a2 = 4 − 3 = 1. d = 1 − 3 = −2.
Check your answer:
4
Find the sum of the first 20 positive integers.
Model answer (board format)
S = n(n + 1)/2 = 20×21/2 = 210.
Check your answer:
Long answer
0/31
An AP has first term 8 and common difference 3. Find the 20th term and the sum of the first 20 terms.
Model answer (board format)
a20 = 8 + 19×3 = 8 + 57 = 65. S20 = 20/2 [16 + 19×3] = 10 × [16 + 57] = 10 × 73 = 730.
Check your answer:
2
The first term is 17 and the last term is 350. d = 9. Find the number of terms and the sum.
Model answer (board format)
350 = 17 + (n − 1)×9. 333 = (n − 1)×9. n − 1 = 37. n = 38. S = 38/2 × (17 + 350) = 19 × 367 = 6973.
Check your answer:
3
Reema saves 50 rupees in the first week and 10 rupees more each week. Find the saving in the 10th week and the total of ten weeks. Also write the difference between an and Sn.
Model answer (board format)
a = 50, d = 10. a10 = 50 + 9×10 = 140 rupees. S10 = 10/2 [100 + 9×10] = 5 × 190 = 950 rupees. 140 is one week; 950 is the total of ten weeks.
Check your answer:
BSEB model paper · practice
0/6This model set is for practice. It is not a question from any year’s annual examination.
1
Which of these is an arithmetic progression?
💡 Hint:
d = 0 is allowed.
Explanation:
5, 5, 5, 5 has d = 0.
2
The 6th term for a = 3, d = 4 is —
💡 Hint:
3 + 5×4.
Explanation:
23.
3
In S = n/2 (a + l), if a = 2, l = 20 and n = 10, then S = ______.
💡 Hint:
110.
Explanation:
5 × 22.
4
Is 20 a term of the AP 4, 8, 12, ...? Find n.
Model answer (board format)
20 = 4 + (n − 1)×4. 16 = (n − 1)×4. n − 1 = 4. n = 5. Yes, the 5th term.
Check your answer:
5
The first term is 9 and the last term is 81. n = 10. Find the sum and d.
Model answer (board format)
S = 10/2 (9 + 81) = 5 × 90 = 450. d = (81 − 9)/9 = 72/9 = 8.
Check your answer:
6
Assertion (A): The sum of the positive integers from 1 to 10 is 55.
Reason (R): Putting n = 10 in n(n + 1)/2 gives 55.
Explanation:
Both are true and R explains A.
CBSE-style questions
0/6These are competency-based practice questions. They are not a copy of any year’s paper.
1
A student writes the 10th term of 4, 9, 14 as 4 + 10×5 = 54. The correct term is —
💡 Hint:
(10 − 1) times.
Explanation:
4 + 9×5 = 49. 54 is one term too far.
2
The fare is 20 rupees on the first day and rises by 5 rupees each day. The fare on the 7th day is —
💡 Hint:
20 + 6×5.
Explanation:
a7 = 20 + 6×5 = 50 rupees. The sum of seven days is a different number.
3
Assertion (A): The first term of Sn = 4n − n² is 3.
Reason (R): S1 = 4 − 1 = 3.
Explanation:
Both are true and R explains A.
4
Assertion (A): The sum Sn is always equal to the nth term.
Reason (R): an is one term and Sn is the total of many terms.
Explanation:
A is false. R is true.
5
A bus-fare list is 15, 18, 21, ... rupees. Find the fare of the 8th ride and the total fare of the first 8 rides.
Model answer (board format)
a = 15, d = 3. a8 = 15 + 7×3 = 36 rupees. S8 = 8/2 [30 + 7×3] = 4 × 51 = 204 rupees.
Check your answer:
6
Sn = n² + n is given. Find a and d, and say why the progression is arithmetic.
Model answer (board format)
S1 = 2 = a. S2 = 6, a2 = 6 − 2 = 4. d = 2. The difference is fixed, so it is arithmetic. an = 2 + (n − 1)×2 = 2n.
Check your answer:
🏛️ Board exam corner — Bihar Board (BSEB)— CBSE
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🧠 What you learned + equation sheet
What you learned
- In an arithmetic progression the common difference d stays the same at every step.
- The general form is a, a+d, a+2d, a+3d, ...
- The nth term is an = a + (n − 1)d. n is a positive integer.
- The sum is Sn = n/2 [2a + (n − 1)d].
- If the last term is l, then Sn = n/2 (a + l).
- If Sn is given, a = S1 and an = Sn − S(n−1).
| What | Keep this |
|---|---|
| First term | a |
| Common difference | d = a(k+1) − ak |
| nth term | an = a + (n − 1)d |
| Sum of n terms | Sn = n/2 [2a + (n − 1)d] |
| Sum from the last term | Sn = n/2 (a + l) |
| Term from the sum | an = Sn − S(n−1), n ≥ 2 |
The notes are original writing. The textbook was used only for activity order and numbers. “Verified” will be used only on a question that has a source page.