MathsCBSE Class 10 · Arithmetic Progressions

Sum of First n Terms of an AP: Formula, Derivation and Questions

Q: Which sum formula should I use in the exam?

Use n/2 times (first term plus last term) when the last term is known. Otherwise use n/2 times the bracket 2a plus (n minus 1)d. Both give the same answer, so you can use one to check the other.

Q: Do I need to learn the derivation of the sum formula?

Yes. The syllabus asks for the derivation of the sum of the first n terms, so it can be asked as part of a question. Learn the method of writing the sum forwards and backwards and adding.

Q: Can the sum of an AP be negative or zero?

Yes. If the common difference is negative, later terms become negative and can cancel the positive ones. That is also why a sum question can have two valid values of n.

Q: Is Exercise 5.4 tested in the board exam?

Exercise 5.4 is marked optional in the NCERT book and is not assessed. It is still good practice for harder problems.

Updated 13 October 2026 · Fact-checked

The sum of the first n terms of an AP is n/2 times the bracket 2a + (n - 1)d, or n/2 times the first term plus the last term. The board asks you to derive it, find sums of terms and multiples, find a or d from given sums, and find n from a given sum using a quadratic.

Write the AP forwards and backwards: every column totals a + l, so two copies of S make n columns, and S is half of that.

Illustration drawn for QuestionBuddy.

Understand Sum of First n Terms of an AP

An arithmetic progression keeps adding the same common difference $d$ . Often you need the total of its terms, not just one term. Adding 40 terms one by one is slow and error-prone, so you need a formula for $S_{n}$ , the sum of the first $n$ terms.

The idea is to write the sum twice, once forwards and once backwards: $S_{n} = a + (a + d) + \dots + [a + (n - 1) d]$ and $S_{n} = [a + (n - 1) d] + \dots + (a + d) + a$ . Add the two lines column by column. In each column one term has gone up by $d$ while its partner has gone down by $d$ , so every column adds to the same value, $2 a + (n - 1) d$ . There are $n$ columns, so $2 S_{n} = n [2 a + (n - 1) d]$ , giving $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ .

The last term is $l = a + (n - 1) d$ , so the bracket is also $a + l$ . That gives the second form $S_{n} = \frac{n}{2} (a + l)$ . Read it as: number of terms times the average of the first and last terms. Use this form when the last term is known, as in sums of multiples in a range.

The sum formula connects to the $n$ th term formula. To use $S_{n} = \frac{n}{2} (a + l)$ for a range, you first find $n$ from $l = a + (n - 1) d$ . When two sums are given, each gives an equation in $a$ and $d$ , so you solve a pair of linear equations. When the sum is given and $n$ is unknown, the formula gives a quadratic in $n$ .

Conditions: the formula works only for terms of an AP, and $n$ must be a natural number. The terms may be negative or fractions, and $d$ may be negative, so a sum can even be zero or negative. When $n$ comes out of a quadratic, keep only natural-number roots. A special case is $a = 1$ , $d = 1$ , which gives $1 + 2 + \dots + n = \frac{n ( n + 1 )}{2}$ .

Sum of First n Terms of an AP: formulas to remember

nth term: $a_{n} = a + (n - 1) d$; Used in almost every sum question, for example to find $l$ or $n$ before you sum.
Sum of first n terms: $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$; Use when $a$ , $d$ and $n$ are known, or to form equations or a quadratic in $n$ .
Sum using the last term: $S_{n} = \frac{n}{2} (a + l)$; Use when the first and last terms are known. Here $l = a + (n - 1) d$ .
Number of terms from a to l: $n = \frac{l - a}{d} + 1$; Needed before summing a range of multiples. Do not forget the $+ 1$ .
Sum of a block of terms: $a_{m + 1} + \dots + a_{n} = S_{n} - S_{m}$; For terms from the 11th to the 30th, use $S_{30} - S_{10}$ .
nth term from the sum: $a_{n} = S_{n} - S_{n - 1} (n \geq 2)$; The block-sum idea with a single term. For the first term, $a_{1} = S_{1}$ .
Sum of first n natural numbers: $1 + 2 + \dots + n = \frac{n ( n + 1 )}{2}$; Special case with $a = 1$ , $d = 1$ .
Sum of first n odd numbers: $1 + 3 + 5 + \dots + (2 n - 1) = n^{2}$; Special case with $a = 1$ , $d = 2$ . Useful as a quick check.

How to solve Sum of First n Terms of an AP questions

Every sum question reduces to knowing $a$ , $n$ and either $d$ or $l$ . Use this routine on any board question.

1Write down what is given: $a$ , $d$ , $l$ , $n$ or $S_{n}$ , and what is asked.
2If a range is given (for example multiples of 7 between two numbers), find the first and last terms that actually lie in the range.
3Find any missing value: $d = a_{2} - a_{1}$ , or $n$ from $l = a + (n - 1) d$ .
4Choose the form: $\frac{n}{2} (a + l)$ if $l$ is known, otherwise $\frac{n}{2} [2 a + (n - 1) d]$ .
5If sums are given and $a$ , $d$ are unknown, write one equation per sum and solve the pair.
6If $S_{n}$ is given and $n$ is unknown, form the quadratic, factorise, and reject roots that are not natural numbers with a reason.
7Check: substitute your answer back, or recompute with the other form of the formula, and write units.

Sum of First n Terms of an AP in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Sum of the first n terms and its derivation

Imagine a potato race on your school sports day. A bucket sits at the starting point. The first potato is 5 m from the bucket and there are 10 potatoes in a straight line, each 3 m beyond the previous one. A runner must fetch each potato one at a time and drop it in the bucket.

How far does the runner travel in total to put all 10 potatoes in the bucket?

Try it first, then show the solutionSolution

Distances of potatoes from the bucket: $5, 8, 11, \dots$ , an AP with $a = 5$ , $d = 3$ , $n = 10$ .
Last potato: $l = 5 + 9 \times 3 = 32$ m.
For each potato the runner goes there and back, so the distance is twice the potato's distance.
Total $= 2 (5 + 8 + \dots + 32) = 2 \times \frac{10}{2} (5 + 32)$ .
$= 2 \times 5 \times 37 = 370$ m.

Answer: The runner travels 370 m.

Memory hook: Potato race: there and back doubles the AP sum.

Sum of the first n terms and its derivation

Suppose your uncle repays a loan of ₹79,000 in 40 monthly instalments. The first instalment is ₹1,000 and each instalment is a fixed amount more than the one before.

Find the last instalment and the fixed increase in each instalment.

Try it first, then show the solutionSolution

Instalments form an AP with $a = 1000$ , $n = 40$ , $S_{40} = 79000$ .
$S_{40} = \frac{40}{2} (a + l) \Rightarrow 79000 = 20 (1000 + l)$ .
$1000 + l = 3950 \Rightarrow l = 2950$ .
$l = a + 39 d \Rightarrow 2950 = 1000 + 39 d \Rightarrow d = 50$ .
Check: $S_{40} = 20 [2000 + 39 \times 50] = 20 \times 3950 = 79000$ .

Answer: The last instalment is ₹2,950 and each instalment rises by ₹50.

Memory hook: Loan total = number of instalments times the average of first and last.

Tricks to remember Sum of First n Terms of an AP

Short memory aids. Say each one aloud once, then use it on a question the same day.

Forwards plus backwards
Write the sum, then write it reversed. Each column gives first + last, and there are n columns, so twice the sum is n(first + last).
Use it when: Rebuild the derivation in the exam even if you forget the formula.
Count times average
Sum = how many terms x average of first and last.
Use it when: Quick sums of multiples and ranges with the a + l form.
Fence posts
From 12 to 248 in steps of 4 there are 59 gaps but 60 numbers: always add 1.
Use it when: Finding n before summing a range.
Half goes outside
The n/2 multiplies the whole bracket, never just 2a.
Use it when: Avoiding the commonest arithmetic slip.

Sum of First n Terms of an AP: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Sum with a negative common difference · 2 marks · Very short answerBoard level

Find the sum of the first 22 terms of the AP $8, 3, - 2, - 7, \dots$

Try it first, then show the solutionSolution

$a = 8$ , $d = 3 - 8 = - 5$ , $n = 22$ . (½)
$S_{22} = \frac{22}{2} [2 (8) + (22 - 1) (- 5)]$ (½)
$= 11 [16 - 105]$ (½)
$= 11 \times (- 89) = - 979$ (½)

Answer: $S_{22} = - 979$

Example 2: MCQ: nth term from the sum · 1 mark · MCQBoard level

The sum of the first $n$ terms of an AP is $S_{n} = 3 n^{2} + 2 n$ . Its 10th term is
(a) 320
(b) 59
(c) 61
(d) 57

Try it first, then show the solutionSolution

$S_{10} = 3 (100) + 2 (10) = 320$ and $S_{9} = 3 (81) + 2 (9) = 261$ .
$a_{10} = S_{10} - S_{9} = 320 - 261 = 59$ . (1)
Check: $a_{n} = S_{n} - S_{n - 1} = 6 n - 1$ , so $a_{10} = 59$ . Option (a) is $S_{10}$ , the sum, not the term.

Answer: (b) 59

Example 3: MCQ: sum of a block of terms · 1 mark · MCQBoard level

The sum of the 11th to 30th terms of the AP $3, 7, 11, \dots$ is
(a) 1577
(b) 1620
(c) 1830
(d) 2040

Try it first, then show the solutionSolution

$a = 3$ , $d = 4$ . Sum of the 11th to 30th terms $= S_{30} - S_{10}$ .
$S_{30} = 15 [6 + 29 \times 4] = 15 \times 122 = 1830$ and $S_{10} = 5 [6 + 36] = 210$ .
Required sum $= 1830 - 210 = 1620$ . (1)
Check: $a_{11} = 43$ , $a_{30} = 119$ , 20 terms: $\frac{20}{2} (43 + 119) = 1620$ . Option (a) is $S_{30} - S_{11}$ , which wrongly drops the 11th term.

Answer: (b) 1620

Example 4: Assertion-reason: odd and natural numbers · 1 mark · Assertion-reasonBoard level

Assertion (A): The sum of the first 20 odd natural numbers is 400.
Reason (R): The sum of the first $n$ natural numbers is $\frac{n ( n + 1 )}{2}$ .
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true but R is not the correct explanation of A.
(c) A is true but R is false.
(d) A is false but R is true.

Try it first, then show the solutionSolution

A: odd numbers form an AP with $a = 1$ , $d = 2$ , so $S_{20} = 10 [2 + 38] = 400$ . A is true.
R: with $a = 1$ , $d = 1$ , $S_{n} = \frac{n}{2} [2 + (n - 1)] = \frac{n ( n + 1 )}{2}$ . R is true.
R is about natural numbers, not odd numbers (it gives 210 for $n = 20$ ), so it does not explain A. (1)

Answer: (b) Both A and R are true but R is not the correct explanation of A.

Example 5: Derivation of the sum formula (model answer) · 3 marks · Short answerBoard level

Show that the sum of the first $n$ terms of an AP with first term $a$ and common difference $d$ is $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . Hence write $S_{n}$ in terms of the last term $l$ .

Try it first, then show the solutionSolution

Let $S_{n} = a + (a + d) + (a + 2 d) + \dots + [a + (n - 1) d]$ ... (i) (½)
Writing the terms in reverse order: $S_{n} = [a + (n - 1) d] + [a + (n - 2) d] + \dots + a$ ... (ii) (½)
Adding (i) and (ii) term by term, each pair is $2 a + (n - 1) d$ , because one term rises by $d$ as its partner falls by $d$ . (1)
There are $n$ such pairs, so $2 S_{n} = n [2 a + (n - 1) d]$ , giving $S_{n} = \frac{n}{2} [2 a + (n - 1) d]$ . (½)
Since $l = a + (n - 1) d$ , the bracket is $a + l$ , so $S_{n} = \frac{n}{2} (a + l)$ . (½)

Answer: $S_{n} = \frac{n}{2} [2 a + (n - 1) d] = \frac{n}{2} (a + l)$

Example 6: Find the AP from two sums · 3 marks · Short answerHarder board level

The sum of the first 7 terms of an AP is 49 and the sum of its first 17 terms is 289. Find the sum of its first $n$ terms.

Try it first, then show the solutionSolution

$S_{7} = \frac{7}{2} (2 a + 6 d) = 49 \Rightarrow 7 (a + 3 d) = 49 \Rightarrow a + 3 d = 7$ . (1)
$S_{17} = \frac{17}{2} (2 a + 16 d) = 289 \Rightarrow 17 (a + 8 d) = 289 \Rightarrow a + 8 d = 17$ . (½)
Subtracting: $5 d = 10 \Rightarrow d = 2$ , so $a = 1$ . (½)
$S_{n} = \frac{n}{2} [2 + 2 (n - 1)] = \frac{n}{2} \times 2 n = n^{2}$ . (½)
Check: $S_{7} = 49$ and $S_{17} = 289$ . Both match. (½)

Answer: $S_{n} = n^{2}$

Example 7: Using first term, last term and sum · 3 marks · Short answerBoard level

The first and last terms of an AP are 5 and 45, and the sum of all its terms is 400. Find the number of terms and the common difference.

Try it first, then show the solutionSolution

$S_{n} = \frac{n}{2} (a + l) \Rightarrow 400 = \frac{n}{2} (5 + 45)$ . (1)
$400 = 25 n \Rightarrow n = 16$ . (½)
$l = a + (n - 1) d \Rightarrow 45 = 5 + 15 d$ . (½)
$d = \frac{40}{15} = \frac{8}{3}$ . (½)
Check: $S_{16} = 8 [10 + 15 \times \frac{8}{3}] = 8 \times 50 = 400$ . (½)

Answer: 16 terms, common difference $\frac{8}{3}$

Example 8: Two-digit numbers divisible by 3 or 7 · 5 marks · Long answerHarder board level

Find the sum of all two-digit numbers that are divisible by 3 or by 7.

Try it first, then show the solutionSolution

Two-digit multiples of 3: $12, 15, \dots, 99$ ; $n = \frac{99 - 12}{3} + 1 = 30$ . (1)
Their sum $= \frac{30}{2} (12 + 99) = 15 \times 111 = 1665$ . (1)
Two-digit multiples of 7: $14, 21, \dots, 98$ ; $n = \frac{98 - 14}{7} + 1 = 13$ , sum $= \frac{13}{2} (14 + 98) = 13 \times 56 = 728$ . (1)
Numbers divisible by both 3 and 7 are multiples of 21: $21, 42, 63, 84$ ; $n = 4$ , sum $= \frac{4}{2} (21 + 84) = 210$ . (1)
These were counted in both lists, so required sum $= 1665 + 728 - 210 = 2183$ . (1)

Answer: 2183

Example 9: Proof with general terms · 5 marks · Long answerHarder board level

In an AP, the $m$ th term is $\frac{1}{n}$ and the $n$ th term is $\frac{1}{m}$ , where $m \neq = n$ . Show that the sum of the first $mn$ terms is $\frac{mn + 1}{2}$ .

Try it first, then show the solutionSolution

$a + (m - 1) d = \frac{1}{n}$ ... (i) and $a + (n - 1) d = \frac{1}{m}$ ... (ii) (1)
(i) − (ii): $(m - n) d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn}$ . (1)
Since $m \neq = n$ , divide by $m - n$ : $d = \frac{1}{mn}$ . (½)
From (i): $a = \frac{1}{n} - \frac{m - 1}{mn} = \frac{m - m + 1}{mn} = \frac{1}{mn}$ . (1)
$S_{mn} = \frac{mn}{2} [\frac{2}{mn} + (mn - 1) \frac{1}{mn}]$ (½)
$= \frac{mn}{2} \times \frac{2 + mn - 1}{mn} = \frac{mn + 1}{2}$ . Hence proved. (1)

Answer: $S_{mn} = \frac{mn + 1}{2}$ , as required.

Example 10: Proof: sums that swap · 5 marks · Long answerHOTS / competency

The sum of the first $p$ terms of an AP is $q$ and the sum of its first $q$ terms is $p$ , where $p \neq = q$ . Show that the sum of its first $(p + q)$ terms is $- (p + q)$ .

Try it first, then show the solutionSolution

$\frac{p}{2} [2 a + (p - 1) d] = q \Rightarrow 2 a + (p - 1) d = \frac{2 q}{p}$ ... (i), and similarly $2 a + (q - 1) d = \frac{2 p}{q}$ ... (ii) (1)
(i) − (ii): $(p - q) d = \frac{2 q}{p} - \frac{2 p}{q} = \frac{2 ( q ^{2} - p ^{2} )}{pq} = - \frac{2 ( p - q ) ( p + q )}{pq}$ . (1)
Since $p \neq = q$ , divide by $p - q$ : $d = - \frac{2 ( p + q )}{pq}$ . (1)
$S_{p + q} = \frac{p + q}{2} [2 a + (p + q - 1) d] = \frac{p + q}{2} [2 a + (p - 1) d + q d]$ . (1)
Using (i) and $q d = - \frac{2 ( p + q )}{p}$ : $S_{p + q} = \frac{p + q}{2} [\frac{2 q}{p} - \frac{2 ( p + q )}{p}] = \frac{p + q}{2} \times \frac{- 2 p}{p} = - (p + q)$ . Hence proved. (1)

Answer: $S_{p + q} = - (p + q)$ , as required.

Example 11: Case study: cycling practice · 4 marks · Case-basedHarder board level

Suppose Meera trains for a cycle race. She rides 2 km on day 1 and increases her ride by 0.5 km every day.
(i) How far does she ride on day 15? (1 mark)
(ii) On which day does she first ride 12 km? (1 mark)
(iii) After how many days will her total distance be 65 km? (2 marks)

Try it first, then show the solutionSolution

(i) $a = 2$ , $d = 0.5$ : $a_{15} = 2 + 14 (0.5) = 9$ km. (1)
(ii) $2 + (n - 1) (0.5) = 12 \Rightarrow n - 1 = 20 \Rightarrow n = 21$ , so day 21. (1)
(iii) $\frac{n}{2} [4 + 0.5 (n - 1)] = 65 \Rightarrow \frac{n ( n + 7 )}{4} = 65 \Rightarrow n^{2} + 7 n - 260 = 0$ . (½)
$(n + 20) (n - 13) = 0 \Rightarrow n = 13$ or $n = - 20$ . (½)
Days cannot be negative, so $n = 13$ . (½)
Check: $S_{13} = \frac{13}{2} [4 + 6] = 65$ km. (½)

Answer: (i) 9 km (ii) Day 21 (iii) 13 days

Example 12: Case study: auditorium seats · 4 marks · Case-basedHOTS / competency

Suppose your school auditorium has 15 rows of seats. The front row has 40 seats and each row behind it has 2 seats fewer than the row in front of it.
(i) How many seats are in the last row? (1 mark)
(ii) Which row has 20 seats? (1 mark)
(iii) For a charity show, each seat in the front 5 rows costs ₹200 and every other seat costs ₹150. Find the total number of seats and the money collected if every seat is sold. (2 marks)

Try it first, then show the solutionSolution

(i) $a = 40$ , $d = - 2$ : $a_{15} = 40 + 14 (- 2) = 12$ seats. (1)
(ii) $40 + (n - 1) (- 2) = 20 \Rightarrow n - 1 = 10 \Rightarrow n = 11$ , so row 11. (1)
(iii) Total seats: $S_{15} = \frac{15}{2} (40 + 12) = 15 \times 26 = 390$ . (½)
Front 5 rows: $a_{5} = 40 + 4 (- 2) = 32$ , so $S_{5} = \frac{5}{2} (40 + 32) = 180$ seats; the other rows have $390 - 180 = 210$ seats. (½)
Money in rupees $= 180 \times 200 + 210 \times 150 = 36000 + 31500$ . (½)
$= 67500$ , that is ₹67,500. (½)

Answer: (i) 12 seats (ii) Row 11 (iii) 390 seats, ₹67,500

Where students lose marks in Sum of First n Terms of an AP

The slips examiners see most often, and what to write instead.

Mistake: Applying n/2 to only part of the bracket
The formula is copied without brackets.
Wrong working
$S_{10} = \frac{10}{2} \times 2 (3) + 9 (4) = 30 + 36 = 66$
Right working
$S_{10} = \frac{10}{2} [6 + 36] = 5 \times 42 = 210$
Do this: Always write the square bracket and multiply n/2 by the whole of it.
Usually costs: The final answer mark and often the step mark.
Mistake: Sign error with negative d
$(n - 1) d$ is computed without brackets.
Wrong working
$a = 8$ , $d = - 5$ : $S_{22} = 11 [16 + 21 - 5] = 352$
Right working
$S_{22} = 11 [16 + 21 (- 5)] = 11 (- 89) = - 979$
Do this: Substitute d in brackets: $(n - 1) (- 5)$ .
Usually costs: 1 to 2 marks.
Mistake: Wrong count of terms in a range
Forgetting the +1, or using the endpoints of 'between' as terms.
Wrong working
Multiples of 7 between 50 and 300: $a = 56$ , $l = 294$ , $n = \frac{294 - 56}{7} = 34$ , so $S = \frac{34}{2} (56 + 294) = 17 \times 350 = 5950$
Right working
$n = \frac{294 - 56}{7} + 1 = 35$ , so $S = \frac{35}{2} (56 + 294) = 35 \times 175 = 6125$
Do this: List the actual first and last multiples (56 and 294, not 50 and 300, which would give $n = \frac{250}{7}$ , not a whole number), then use $n = \frac{l - a}{d} + 1$ .
Usually costs: The whole answer goes wrong: 5950 instead of 6125.
Mistake: Keeping an impossible root
Both roots of the quadratic are written as answers.
Wrong working
$n = 13$ or $n = - 20$ , so 13 or −20 days
Right working
$n = - 20$ is rejected as the number of terms must be a natural number, so $n = 13$
Do this: Write a one-line reason for rejecting every root.
Usually costs: ½ to 1 mark.

Exam tips

The derivation of the sum formula can be asked directly; practise writing the forwards-backwards proof in 4 to 5 clean lines.
When two sums are given, simplify each equation (for example to $a + 3 d = 7$ ) before solving; it saves time and errors.
For sums of multiples or ranges, show the line finding n separately; examiners give a step mark for it.
In case studies, read whether each part asks for a single value ( $a_{n}$ ) or a total ( $S_{n}$ ) before choosing a formula.
Finish with a check using the other form of the sum formula; it takes seconds and catches sign slips.

Practice questions on Sum of First n Terms of an AP

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 3 marks · Short answerHarder board level
Find the sum of all three-digit numbers that are divisible by 7 but not by 14.
Hint
Divisible by 7 but not by 14 means odd multiples of 7. They form an AP with d = 14.
Answer
Odd three-digit multiples of 7: $105, 119, \dots, 987$ with $d = 14$ . $n = \frac{987 - 105}{14} + 1 = 64$ . Sum $= \frac{64}{2} (105 + 987) = 32 \times 1092 = 34944$ . Check: all three-digit multiples of 7 add to $64 \times 1099 = 70336$ , multiples of 14 add to $32 \times 1106 = 35392$ , and $70336 - 35392 = 34944$ .
Q2 · 3 marks · Short answerHarder board level
How many terms of the AP $24, 21, 18, \dots$ must be taken so that their sum is 78? Explain the double answer.
Hint
Form a quadratic in n; d is negative.
Answer
$\frac{n}{2} [48 - 3 (n - 1)] = 78 \Rightarrow n^{2} - 17 n + 52 = 0 \Rightarrow (n - 4) (n - 13) = 0$ . So $n = 4$ or $13$ . Both work because terms 5 to 13 ( $12, 9, \dots, - 12$ ) add to 0.
Q3 · 3 marks · Short answerHarder board level
The sum of the first 6 terms of an AP is 42. The ratio of its 10th term to its 30th term is $1 : 3$ . Find the first term and the 13th term.
Hint
Cross-multiply the ratio to relate a and d first.
Answer
$3 (a + 9 d) = a + 29 d \Rightarrow a = d$ . $S_{6} = 3 (2 a + 5 d) = 21 a = 42 \Rightarrow a = 2$ , $d = 2$ . $a_{13} = 2 + 24 = 26$ .
Q4 · 5 marks · Long answerHarder board level
An AP has 21 terms. The sum of its three middle terms is 129 and the sum of its last three terms is 237. Find the AP and the sum of all its terms.
Hint
The middle three are the 10th, 11th, 12th terms; their sum is 3 times the 11th term.
Answer
$3 a_{11} = 129 \Rightarrow a + 10 d = 43$ ; $3 a_{20} = 237 \Rightarrow a + 19 d = 79$ . So $d = 4$ , $a = 3$ . AP: $3, 7, 11, \dots$ with last term 83. $S_{21} = \frac{21}{2} (3 + 83) = 903$ .
Q5 · 2 marks · Short answerBoard level
A student finds $1 + 3 + 5 + \dots + 99$ like this: " $n = \frac{99 - 1}{2} = 49$ , so $S = \frac{49}{2} (1 + 99) = 2450$ ." Find the error and give the correct sum.
Hint
Count the terms with the fence-post idea.
Answer
The student forgot the $+ 1$ : $n = \frac{99 - 1}{2} + 1 = 50$ . Correct sum $= \frac{50}{2} (100) = 2500$ , which also equals $5 0^{2}$ .
Q6 · 1 mark · MCQBoard level
If the sum of the first $n$ terms of an AP is $4 n - n^{2}$ , then its third term is
(a) 3
(b) −1
(c) 1
(d) −3
Hint
Use $a_{3} = S_{3} - S_{2}$ .
Answer
(b) −1, since $S_{3} = 12 - 9 = 3$ and $S_{2} = 8 - 4 = 4$ , so $a_{3} = 3 - 4 = - 1$ .
Q7 · 5 marks · Long answerHarder board level
The sum of the first 9 terms of an AP is 144 and the sum of its first 15 terms is 375. Find the AP. Hence find how many terms of the AP are less than 100, and the sum of all these terms.
Hint
Simplify each sum to an equation like a + 4d = 16, then solve a_n < 100 for the largest natural n.
Answer
$\frac{9}{2} (2 a + 8 d) = 144 \Rightarrow a + 4 d = 16$ ; $\frac{15}{2} (2 a + 14 d) = 375 \Rightarrow a + 7 d = 25$ . So $d = 3$ , $a = 4$ ; AP: $4, 7, 10, \dots$ . $a_{n} = 3 n + 1 < 100 \Rightarrow n < 33$ , so 32 terms, the last being 97. Their sum $= \frac{32}{2} (4 + 97) = 16 \times 101 = 1616$ .
Q8 · 3 marks · Short answerHOTS / competency
Suppose a full water tank holds 1200 litres and develops a leak. It loses 100 litres in the first hour, 95 litres in the second hour, and 5 litres less in each hour after that. After how many hours will 975 litres have leaked out, and how much water is then left? Explain why one root of your equation is rejected.
Hint
The amount lost in an hour can never be negative.
Answer
$\frac{n}{2} [200 - 5 (n - 1)] = 975 \Rightarrow n^{2} - 41 n + 390 = 0 \Rightarrow (n - 15) (n - 26) = 0$ . At $n = 26$ the loss in that hour would be $100 - 5 \times 25 = - 25$ litres, which is impossible (the hourly loss reaches 0 in hour 21), so $n = 15$ . Check: $\frac{15}{2} (200 - 70) = 975$ . Water left $= 1200 - 975 = 225$ litres.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.State the sum of the first n terms with a and d.Show answer
$S_{n} = \frac{n}{2} [2 a + (n - 1) d]$
2.State the sum using the last term.Show answer
$S_{n} = \frac{n}{2} (a + l)$
3.Sum of the first 100 natural numbers?Show answer
$\frac{100 \times 101}{2} = 5050$
4.Sum of the first 20 odd numbers?Show answer
$2 0^{2} = 400$
5.How do you find the sum of terms 11 to 30?Show answer
$S_{30} - S_{10}$
6.Why does every column add to the same value in the derivation?Show answer
One term rises by d while its partner falls by d, so each pair equals $2 a + (n - 1) d$ .

Sum of First n Terms of an AP: frequently asked questions

Which sum formula should I use in the exam?

Use n/2 times (first term plus last term) when the last term is known. Otherwise use n/2 times the bracket 2a plus (n minus 1)d. Both give the same answer, so you can use one to check the other.

Do I need to learn the derivation of the sum formula?

Yes. The syllabus asks for the derivation of the sum of the first n terms, so it can be asked as part of a question. Learn the method of writing the sum forwards and backwards and adding.

Can the sum of an AP be negative or zero?

Yes. If the common difference is negative, later terms become negative and can cancel the positive ones. That is also why a sum question can have two valid values of n.

Is Exercise 5.4 tested in the board exam?

Exercise 5.4 is marked optional in the NCERT book and is not assessed. It is still good practice for harder problems.