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CBSE · Class 10 · Mathematics

Arithmetic Progressions 5.4: Sum of First n Terms of an AP

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Sum using Sₙ = n/2 [2a + (n − 1)d]

From daily life

Adding 1 + 2 + … + 100 one by one is slow. Pair the first with the last, second with second-last: every pair makes 101.

Simple definition

The sum of the first n terms of an AP with first term a and common difference d is Sₙ = n/2 [2a + (n − 1)d].

Formula: Sₙ = n/2 [2a + (n − 1)d]

find a, d, n2a + (n−1)d× n/2= Sₙ

In the figure: 2, 5, 8, … : S₁₀ = 10/2 [4 + 9 × 3] = 5 × 31 = 155.

Live figure

Sum using Sₙ = n/2 [2a + (n − 1)d]

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Rule 1: Find the sum — Read off a, d and n, then substitute.

  • Easy: a = 2, d = 3, n = 10: S = 5 × 31 = 155.
  • Medium: 1 + 2 + … + 100: S = 50 × 101 = 5050.
  • Exam-type: a = 10, d = −2, n = 6: S = 3 [20 − 10] = 30.

Rule 2: Sum of a list — First confirm it is an AP, then find n from the last term.

  • Easy: 7 + 10 + … + 40: n = 12, S = 6 (14 + 33) = 282.
  • Medium: −5 + (−2) + 1 + … (8 terms): S = 4 (−10 + 21) = 44.
  • Exam-type: Each term adds the same d, so the formula works.
Remember

Write 2a + (n − 1)d first, then multiply by n/2.

Common mistake

Using a instead of 2a inside the bracket.

Memory clue

"Half n times double-a plus steps-d."

Practice questions

Easy
  1. 1An AP has first term −1/2 and common difference −7. Find the sum of its first 34 terms.
  2. 2An AP has first term 24 and common difference −3. Find the sum of its first 8 terms.
  3. 3An AP has first term 15/2 and common difference 3. Find the sum of its first 8 terms.
Medium
  1. 4An AP has first term −3 and common difference −7. Find the sum of its first 39 terms.
  2. 5An AP has first term −2 and common difference −3. Find the sum of its first 16 terms.
  3. 6An AP has first term 16 and common difference −8. Find the sum of its first 9 terms.
Challenging
  1. 7Find the sum of the AP 20, 24, 28, …, 96.
  2. 8Find the sum of the AP 14, 17, 20, …, 38.
Show answers (try them yourself first)
  1. 1−3944
  2. 2108
  3. 3144
  4. 4−5304
  5. 5−392
  6. 6−144
  7. 71160
  8. 8234

Sum using first and last term; special sums

From daily life

If you know the first and last seat numbers in a row, you do not need d: the average seat times the count gives the total.

Simple definition

When the last term l is known, Sₙ = n/2 (a + l). Special sums: first n natural numbers n(n + 1)/2; first n odd numbers n²; first n even numbers n(n + 1).

Formula: Sₙ = n/2 (a + l)

a+ l× n/2= Sₙ

In the figure: 8 + 11 + … + 50 (15 terms): S = 15/2 (8 + 50) = 435.

Live figure

Sum using first and last term; special sums

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Rule 1: Using a and l — Count n first, then average a and l.

  • Easy: 8 + 11 + … + 50: n = 15, S = 435.
  • Medium: Multiples of 4 from 4 to 100: n = 25, S = 25/2 × 104 = 1300.
  • Exam-type: 3 + 6 + … + 99: n = 33, S = 33/2 × 102 = 1683.

Rule 2: Special sums — Use the ready results for natural, odd and even numbers.

  • Easy: 1 + 2 + … + 50 = 50 × 51/2 = 1275.
  • Medium: First 15 odd numbers: 15² = 225.
  • Exam-type: First 20 even numbers: 20 × 21 = 420.
Remember

Sₙ = n × (average of first and last term).

Common mistake

Taking n as the last term instead of counting the terms.

Memory clue

"Count, average the ends, multiply."

Practice questions

Easy
  1. 1An AP has 7 terms with first term −15 and last term −21. Find the sum of all its terms.
  2. 2Find the sum of the first 15 positive even numbers.
  3. 3Find the sum of the first 106 natural numbers.
Medium
  1. 4Find the sum of the first 42 natural numbers.
  2. 5Find the sum of the first 10 positive even numbers.
  3. 6Find the sum of the first 12 positive even numbers.
Challenging
  1. 7An AP has 7 terms with first term −12 and last term 0. Find the sum of all its terms.
  2. 8Find the sum of all numbers from 100 to 999 that are divisible by 12.
Show answers (try them yourself first)
  1. 1−126
  2. 2240
  3. 35671
  4. 4903
  5. 5110
  6. 6156
  7. 7−42
  8. 841400

Finding n, a or d from a given sum

From daily life

If the total of a row of seats is known, you can work backwards to find how many seats it has.

Simple definition

Put the known values in Sₙ = n/2 [2a + (n − 1)d] and solve for the unknown. For n you often get a quadratic equation; reject a negative or fractional n.

put valuesget n² eq.solvereject n < 0

In the figure: 3 + 5 + 7 + … = 120 gives n² + 2n − 120 = 0, so n = 10 (−12 rejected).

Rule 1: Find n — Form the quadratic, factorise, keep the positive root.

  • Easy: 3 + 5 + 7 + … = 120: n = 10.
  • Medium: 1 + 2 + 3 + … = 210: n(n + 1) = 420, n = 20.
  • Exam-type: 2 + 4 + 6 + … = 72: n(n + 1) = 72, n = 8.

Rule 2: Find a or d — Substitute n and S, then solve the linear equation.

  • Easy: n = 10, S = 155, a = 2: 31 = 4 + 9d, d = 3.
  • Medium: n = 8, S = 100, d = 3: 25 = 2a + 21, a = 2.
  • Exam-type: Verify by adding a few terms.
Remember

A negative n has no meaning, since n counts terms.

Common mistake

Keeping both roots of the quadratic instead of rejecting the negative one.

Memory clue

"Solve, then throw away the negative n."

Practice questions

Easy
  1. 1How many terms of the AP 20, 27, 34, 41, … must be added to get the sum 432?
  2. 2An AP has 6 terms, first term 1 and sum −69. Find its last term.
  3. 3How many terms of the AP 20, 23, 26, 29, … must be added to get the sum 203?
Medium
  1. 4An AP has 10 terms, first term −1 and sum −415. Find its last term.
  2. 5How many terms of the AP 10, 17, 24, 31, … must be added to get the sum 165?
  3. 6How many terms of the AP 10, 16, 22, 28, … must be added to get the sum 440?
Challenging
  1. 7The first term of an AP is 13 and the sum of its first 12 terms is 750. Find the common difference.
  2. 8The first term of an AP is 15 and the sum of its first 20 terms is −270. Find the common difference.
Show answers (try them yourself first)
  1. 19
  2. 2−24
  3. 37
  4. 4−82
  5. 56
  6. 611
  7. 79
  8. 8−3

Connecting Sₙ with aₙ

From daily life

The score after 10 matches minus the score after 9 matches is exactly the 10th match’s score.

Simple definition

The nth term is the difference of two sums: aₙ = Sₙ − Sₙ₋₁. Also a₁ = S₁. This works even when Sₙ is given as an expression in n.

Formula: aₙ = Sₙ − Sₙ₋₁

Sₙ− Sₙ₋₁= aₙ

In the figure: If Sₙ = 3n² + 2n, then S₁ = 5, S₂ = 16, so a₂ = 11 and d = 6.

Rule 1: Find the nth term — Subtract Sₙ₋₁ from Sₙ and simplify.

  • Easy: Sₙ = 3n² + 2n: aₙ = 6n − 1 (a₁ = 5, a₂ = 11).
  • Medium: Sₙ = n²: aₙ = 2n − 1 (odd numbers).
  • Exam-type: Sₙ = 2n² + n: a₃ = S₃ − S₂ = 21 − 10 = 11.

Rule 2: Build the AP — Find a₁ and a₂, then d = a₂ − a₁.

  • Easy: Sₙ = 3n² + 2n: AP is 5, 11, 17, … with d = 6.
  • Medium: Sₙ = n² gives 1, 3, 5, … with d = 2.
  • Exam-type: Check: sum of first two terms equals S₂.
Remember

The sum up to n minus the sum up to n − 1 leaves only the nth term.

Common mistake

Writing S(n − 1) wrongly as Sₙ − 1 instead of replacing n by n − 1.

Memory clue

"Today’s total minus yesterday’s total is today’s term."

Practice questions

Easy
  1. 1The sum of the first n terms of an AP is Sₙ = 4n² − 3n. Find the common difference.
  2. 2The sum of the first n terms of an AP is Sₙ = 5n² − 4n. Find the second term.
  3. 3The sum of the first n terms of an AP is Sₙ = −2n² + 8n. Find the second term.
Medium
  1. 4The sum of the first n terms of an AP is Sₙ = 4n² − 7n. Find the first term.
  2. 5The sum of the first n terms of an AP is Sₙ = 4n² − 6n. Find the first term.
  3. 6For an AP, S8 = 232 and S7 = 189. Find a8.
Challenging
  1. 7In an AP the 3rd term is 26. Find the sum of its first 5 terms.
  2. 8For an AP, S17 = −408 and S16 = −352. Find a17.
Show answers (try them yourself first)
  1. 18
  2. 211
  3. 32
  4. 4−3
  5. 5−2
  6. 643
  7. 7130
  8. 8−56

Word problems on the sum of an AP

From daily life

Stacked logs, rows of seats and yearly savings add up as an AP. The sum gives the total.

Simple definition

Identify a, d and n, choose between n/2 [2a + (n − 1)d] and n/2 (a + l), compute, and add units.

finda, d, nchooseformulacomputeanswerwith units

In the figure: Logs: 20 in the bottom row, 19 in the next, … down to 1: S = 20/2 (20 + 1) = 210 logs.

Rule 1: Total amount — Use n/2 [2a + (n − 1)d] with the given a, d and n.

  • Easy: Savings ₹100, +₹50 monthly, 12 months: S = 6 (200 + 550) = ₹4500.
  • Medium: Seats: 15 rows, first row 20, +2 each: S = 15/2 (40 + 28) = 510.
  • Exam-type: Logs: 20, 19, …, 1 gives 210.

Rule 2: Find the number of terms — Equate the sum to the total and solve for n.

  • Easy: Prizes ₹500, ₹450, … total ₹2250: n/2 [1000 − 50(n − 1)] = 2250 gives n = 6.
  • Medium: Save ₹10, +₹5 weekly, total ₹135: n = 6.
  • Exam-type: Reject a negative or fractional n.
Remember

Always attach the unit, such as ₹ or logs.

Common mistake

Using the nth-term formula when the total is asked.

Memory clue

"Total asked? Use S. One value asked? Use a-n."

Practice questions

Easy
  1. 1A girl saves ₹58 in the first week and ₹15 more each week than the week before. How much does she save in 29 weeks altogether (in ₹)?
  2. 2A girl saves ₹66 in the first week and ₹5 more each week than the week before. How much does she save in 28 weeks altogether (in ₹)?
  3. 3A girl saves ₹37 in the first week and ₹25 more each week than the week before. How much does she save in 19 weeks altogether (in ₹)?
Medium
  1. 4A girl saves ₹23 in the first week and ₹20 more each week than the week before. How much does she save in 30 weeks altogether (in ₹)?
  2. 5A girl saves ₹37 in the first week and ₹20 more each week than the week before. How much does she save in 23 weeks altogether (in ₹)?
  3. 6Logs are stacked in rows with 21 logs in the bottom row, 20 in the next, and so on, with 5 in the top row. How many logs are there in all?
Challenging
  1. 7The cost of digging the first metre of a well is ₹150 and it rises by ₹40 for every extra metre. What is the cost of digging 8 metres (in ₹)?
  2. 8Logs are stacked in rows with 22 logs in the bottom row, 21 in the next, and so on, with 11 in the top row. How many logs are there in all?
Show answers (try them yourself first)
  1. 17772
  2. 23738
  3. 34978
  4. 49390
  5. 55911
  6. 6221
  7. 72320
  8. 8198