MathsCBSE Class 10 · Arithmetic Progressions

nth Term of an AP: Formula, Derivation and Questions

Updated 14 October 2026 · Fact-checked

The nth term of an AP with first term a and common difference d is a + (n - 1)d. The board asks you to derive it, find a particular term, decide whether a number is a term, count the terms between two values, find a term from the end, and find the AP from two given conditions.

Each orange block is one d; Term 4 has 3 of them, so Term n has n − 1 of them, giving a + (n − 1)d.

Illustration drawn for QuestionBuddy.

Understand nth Term of an AP

In an AP you start at the first term $a$ and add the same number $d$ again and again. So $a_{1} = a$ , $a_{2} = a + d$ , $a_{3} = a + 2 d$ , $a_{4} = a + 3 d$ . Look at how many $d$ 's each term has: the 4th term has 3 of them. Each step forward adds one $d$ , and reaching position $n$ from position 1 takes $n - 1$ steps. That is why $a_{n} = a + (n - 1) d$ .

This is the derivation the syllabus expects. In an exam, write the first three or four terms in terms of $a$ and $d$ , point out that the number of $d$ 's is one less than the position, and then state the general result.

The formula links four quantities: $a_{n}$ , $a$ , $n$ and $d$ . Know any three and you can find the fourth. To find a term, substitute $n$ . To find which term equals a number, solve for $n$ . To find how many terms an AP has from $a$ to a last term $l$ , solve $l = a + (n - 1) d$ , which gives $n = \frac{l - a}{d} + 1$ .

One condition always holds: $n$ is a position, so it must be a natural number. If solving gives $n = 24.5$ or $n = - 3$ , the number is not a term of the AP. This is how you prove that a number does not belong to an AP.

When two facts about terms are given, write each term as $a + (n - 1) d$ . This gives two linear equations in $a$ and $d$ , which you solve as in Chapter 3. The shortcut $a_{m} - a_{k} = (m - k) d$ gives $d$ quickly from any two terms. Reading the AP backwards, the first term becomes $l$ and the difference becomes $- d$ . So the $n$ th term from the end is $l - (n - 1) d$ .

Sometimes a question gives a formula such as $a_{n} = p n + q$ and asks you to show that the list is an AP. Work out $a_{n + 1} - a_{n} = p (n + 1) + q - (p n + q) = p$ . This answer does not depend on $n$ , so every term differs from the one before by the same amount. The list is an AP with $d = p$ and first term $a_{1} = p + q$ . If the formula has an $n^{2}$ term, such as $a_{n} = n^{2}$ , the difference ( $2 n + 1$ here) changes with $n$ , so the list is not an AP.

This formula also leads into the next topic. The sum formula $S_{n} = \frac{n}{2} (a + l)$ uses $l = a + (n - 1) d$ , so you will need $a_{n}$ there too.

nth Term of an AP: formulas to remember

nth term: $a_{n} = a + (n - 1) d$; $a$ is the first term, $d$ the common difference, $n$ a natural number.
Common difference: $d = a_{k + 1} - a_{k}$; A later term minus the term before it. It can be negative, zero or a fraction.
Difference of two terms: $a_{m} - a_{k} = (m - k) d$; Gives $d$ straight from any two terms.
Number of terms: $n = \frac{l - a}{d} + 1$; Use it when the AP runs from $a$ to the last term $l$ . Do not forget the $+ 1$ .
nth term from the end: $l - (n - 1) d$; Same as the $(m - n + 1)$ th term from the start when the AP has $m$ terms.
Linear form: $a_{n} = d n + (a - d)$; The coefficient of $n$ is $d$ . For example, $a_{n} = 5 n + 2$ has $d = 5$ .

How to solve nth Term of an AP questions

Every question on this topic fits one method: get $a$ and $d$ , then use $a_{n} = a + (n - 1) d$ in whichever direction the question needs.

1Write $a$ and find $d$ as later term minus earlier term. Check two pairs of terms.
2Decide what is unknown: a term $a_{n}$ , a position $n$ , or $a$ and $d$ themselves.
3For a term, substitute $n$ into $a + (n - 1) d$ , with brackets around a negative $d$ .
4For a position or the number of terms, set $a + (n - 1) d$ equal to the given value and solve for $n$ .
5Check that $n$ is a natural number. If it is not, state that the number is not a term.
6For two conditions, write each term as $a + (n - 1) d$ , form two equations and solve the pair.
7For a term from the end, use $l - (n - 1) d$ , or the $(m - n + 1)$ th term from the start.
8Check by substituting your answer back, then write a final statement.

nth Term 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.

nth Term of an AP

Imagine a new metro line where stations are numbered in order. Suppose the fare to the 1st station after boarding is ₹12, and each further station adds ₹3 to the fare, so the fares form an AP.

What is the fare to the 15th station? You have ₹60. Can a fare be exactly ₹60? What is the farthest station you can reach?

Try it first, then show the solutionSolution

$a = 12$ , $d = 3$ .
$a_{15} = 12 + 14 \times 3 = 12 + 42 = 54$ .
Set $12 + (n - 1) 3 = 60$ , so $(n - 1) 3 = 48$ .
$n - 1 = 16$ , so $n = 17$ , which is a natural number.
So the fare to the 17th station is exactly ₹60, and the 18th costs ₹63, which is more than ₹60.

Answer: ₹54 to the 15th station. Yes, ₹60 is the fare to the 17th station, which is the farthest you can reach.

Memory hook: Each station adds one step of $d$ : station $n$ is $n - 1$ steps after the first.

nth Term of an AP

Suppose a rangoli design has concentric rings of dots. The 3rd ring has 26 dots and the 8th ring has 61 dots. Each ring has a fixed number of dots more than the ring inside it.

How many dots does the innermost ring have? How many rings does the design have if the outermost ring has 103 dots?

Try it first, then show the solutionSolution

$a_{8} - a_{3} = 5 d$ , so $61 - 26 = 5 d$ .
$d = 7$ .
$a + 2 (7) = 26$ , so $a = 12$ .
$103 = 12 + (n - 1) 7$ , so $(n - 1) 7 = 91$ .
$n - 1 = 13$ , so $n = 14$ .

Answer: 12 dots in the innermost ring, and 14 rings.

Memory hook: Two rings five apart differ by $5 d$ : the gap in positions times $d$ .

Tricks to remember nth Term of an AP

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

Steps, not stops
Reaching the $n$ th term from the 1st takes $n - 1$ steps. Each step adds $d$ .
Use it when: Stops you from writing $a + n d$ .
Fence posts
There is always one more term than the number of gaps, so $n = \frac{l - a}{d} + 1$ .
Use it when: Counting terms and multiples in a range.
Turn around, flip d
Walking backwards from the end, start at $l$ and use $- d$ .
Use it when: nth term from the end.
Gap times d
$a_{12} - a_{5} = 7 d$ : the difference in positions times $d$ .
Use it when: Finding $d$ from two terms quickly.

nth Term of an AP: solved examples

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

Example 1: Testing whether a number is a term · 2 marks · Very short answerBoard level

Is $- 150$ a term of the AP $11, 8, 5, 2, \dots$ ? Justify your answer.

Try it first, then show the solutionSolution

$a = 11$ , $d = 8 - 11 = - 3$ . (½)
Suppose $a_{n} = - 150$ : $11 + (n - 1) (- 3) = - 150$ . (½)
$- 3 (n - 1) = - 161$ , so $n - 1 = \frac{161}{3}$ . (½)
$n = \frac{164}{3}$ , which is not a natural number, so $- 150$ is not a term. (½)

Answer: No, because $n = \frac{164}{3}$ is not a natural number.

Example 2: Term from the end and middle term · 3 marks · Short answerHarder board level

For the AP $8, 15, 22, \dots, 218$ , find the 12th term from the end and the middle term.

Try it first, then show the solutionSolution

$a = 8$ , $d = 7$ , $l = 218$ . (½)
$218 = 8 + (n - 1) 7$ , so $n - 1 = 30$ and $n = 31$ . (1)
12th from the end $= 218 - 11 \times 7 = 218 - 77 = 141$ . (½)
Check: this is the $(31 - 12 + 1) = 20$ th term, $8 + 19 \times 7 = 141$ . ✓
There are 31 terms (odd), so the middle term is the 16th. (½)
$a_{16} = 8 + 15 \times 7 = 113$ . (½)

Answer: 12th term from the end is 141. The middle term is 113.

Example 3: AP from two conditions · 3 marks · Short answerHarder board level

The 7th term of an AP is 4 times its 2nd term, and its 3rd term is 8. Find the AP and its 20th term.

Try it first, then show the solutionSolution

$a + 6 d = 4 (a + d)$ , so $a + 6 d = 4 a + 4 d$ , which gives $2 d = 3 a$ ... (1) (½)
$a + 2 d = 8$ ... (2) (½)
Put $2 d = 3 a$ from (1) into (2): $a + 3 a = 8$ , so $a = 2$ . Then $2 d = 6$ , so $d = 3$ . (1)
The AP is $2, 5, 8, 11, \dots$ (½)
$a_{20} = 2 + 19 \times 3 = 59$ . Check: $a_{7} = 20 = 4 \times 5 = 4 a_{2}$ ✓ and $a_{3} = 8$ ✓ (½)

Answer: AP: $2, 5, 8, 11, \dots$ with $a = 2$ , $d = 3$ , and $a_{20} = 59$ .

Example 4: Derivation and counting multiples · 5 marks · Long answerHarder board level

Derive the formula for the $n$ th term of an AP with first term $a$ and common difference $d$ . Hence (i) find how many three-digit numbers are divisible by 6, and (ii) find the 15th such number from the largest.

Try it first, then show the solutionSolution

$a_{1} = a$ , $a_{2} = a + d$ , $a_{3} = a + 2 d$ , $a_{4} = a + 3 d$ . (½)
The number of $d$ 's is one less than the position of the term. (½)
So $a_{n} = a + (n - 1) d$ . (1)
Smallest three-digit multiple of 6 is 102 and the largest is 996, with $d = 6$ . (½)
$996 = 102 + (n - 1) 6$ , so $(n - 1) 6 = 894$ . (½)
$n - 1 = 149$ , so $n = 150$ . (1)
15th from the largest $= 996 - 14 \times 6 = 996 - 84 = 912$ . (1)

Answer: $a_{n} = a + (n - 1) d$ . (i) 150 numbers. (ii) 912.

Example 5: Case study: tiles in a corridor · 4 marks · Case-basedHOTS / competency

Suppose a school is tiling a corridor in bands, and each band has a fixed number of tiles more than the band before it. Band 3 has 52 tiles and band 8 has 82 tiles. The last band of the corridor has 196 tiles.
(i) Find the number of tiles in band 1 and the increase from one band to the next. (1)
(ii) How many bands are there in the corridor? (1)
(iii) One supplier claims that a band in this corridor has exactly 250 tiles. Another claims that some band has exactly 245 tiles. Check each claim and justify your answer. (2)

Try it first, then show the solutionSolution

$a_{8} - a_{3} = 5 d$ , so $82 - 52 = 5 d$ and $d = 6$ . Then $a + 2 (6) = 52$ gives $a = 40$ . (1)
$196 = 40 + (n - 1) 6$ , so $(n - 1) 6 = 156$ , $n - 1 = 26$ and $n = 27$ bands. (1)
For 250: $40 + (n - 1) 6 = 250$ gives $n - 1 = 35$ , so $n = 36$ . (½)
Band 36 would come after the last band, band 27, so no band in this corridor has 250 tiles. (½)
For 245: $40 + (n - 1) 6 = 245$ gives $n - 1 = \frac{205}{6}$ . (½)
This is not a whole number, so $n$ is not a natural number and 245 is not a term of the AP at all. (½)

Answer: (i) 40 tiles, increasing by 6 per band (ii) 27 bands (iii) Both claims are false: 250 would be band 36, but there are only 27 bands; 245 is not a term of the AP because $n$ is not a natural number.

Example 6: Showing a list is an AP from its nth term · 3 marks · Short answerBoard level

The $n$ th term of a list of numbers is $a_{n} = 3 + 4 n$ . Show that the list is an AP. Which term of it is 99? Is 150 a term? Justify.

Try it first, then show the solutionSolution

$a_{n + 1} = 3 + 4 (n + 1) = 4 n + 7$ . (½)
$a_{n + 1} - a_{n} = (4 n + 7) - (4 n + 3) = 4$ , which does not depend on $n$ . (½)
So the list is an AP with $a = a_{1} = 7$ and $d = 4$ : $7, 11, 15, \dots$ (½)
$3 + 4 n = 99$ gives $4 n = 96$ , so $n = 24$ . Check: $3 + 96 = 99$ ✓ (½)
$3 + 4 n = 150$ gives $4 n = 147$ , so $n = \frac{147}{4}$ . (½)
This is not a natural number, so 150 is not a term. (½)

Answer: It is an AP with $a = 7$ , $d = 4$ . 99 is the 24th term. 150 is not a term.

Example 7: Finding the AP from sums of terms · 3 marks · Short answerHarder board level

The sum of the 4th and 8th terms of an AP is 24, and the sum of the 6th and 10th terms is 44. Find the first three terms of the AP.

Try it first, then show the solutionSolution

$(a + 3 d) + (a + 7 d) = 24$ , so $2 a + 10 d = 24$ ... (1) (½)
$(a + 5 d) + (a + 9 d) = 44$ , so $2 a + 14 d = 44$ ... (2) (½)
Subtracting (1) from (2): $4 d = 20$ , so $d = 5$ . (½)
From (1): $2 a + 50 = 24$ , so $2 a = - 26$ and $a = - 13$ . (½)
First three terms: $- 13, - 8, - 3$ . (½)
Check: $a_{4} + a_{8} = 2 + 22 = 24$ ✓ and $a_{6} + a_{10} = 12 + 32 = 44$ ✓ (½)

Answer: $- 13, - 8, - 3$

Example 8: HOTS: mth and nth terms given as fractions · 5 marks · Long answerHOTS / competency

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 its $mn$ th term is 1. Hence find its $(m + n)$ th term.

Try it first, then show the solutionSolution

$a + (m - 1) d = \frac{1}{n}$ ... (1) and $a + (n - 1) d = \frac{1}{m}$ ... (2) (1)
Subtracting (2) from (1): $(m - n) d = \frac{1}{n} - \frac{1}{m} = \frac{m - n}{mn}$ . (1)
Since $m \neq = n$ , divide by $m - n$ : $d = \frac{1}{mn}$ . (½)
From (1): $a = \frac{1}{n} - \frac{m - 1}{mn} = \frac{m - ( m - 1 )}{mn} = \frac{1}{mn}$ . (1)
$a_{mn} = \frac{1}{mn} + (mn - 1) \frac{1}{mn} = \frac{1 + mn - 1}{mn} = 1$ . Hence shown. (1)
$a_{m + n} = \frac{1}{mn} + (m + n - 1) \frac{1}{mn} = \frac{m + n}{mn}$ . (½)

Answer: $a = d = \frac{1}{mn}$ , so $a_{mn} = 1$ , and $a_{m + n} = \frac{m + n}{mn}$ .

Where students lose marks in nth Term of an AP

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

Mistake: Using $a + n d$
This counts one step too many.
Wrong working
$a_{20} = 3 + 20 \times 4 = 83$
Right working
$a_{20} = 3 + 19 \times 4 = 79$
Do this: Multiply $d$ by $n - 1$ , one less than the position.
Usually costs: The whole answer and its follow-on marks.
Mistake: Dropping the sign of a negative d
The sign gets lost without brackets.
Wrong working
$a_{10} = 20 + 9 (3) = 47$ for $20, 17, 14, \dots$
Right working
$a_{10} = 20 + 9 (- 3) = - 7$
Do this: Always put a negative $d$ in brackets.
Usually costs: 1 to 2 marks.
Mistake: Forgetting the +1 when counting terms
Students count the gaps instead of the terms.
Wrong working
$n = \frac{98 - 14}{7} = 12$
Right working
$n = \frac{98 - 14}{7} + 1 = 13$
Do this: Use $l = a + (n - 1) d$ and solve fully.
Usually costs: The answer mark, and every later part.
Mistake: Not stating why a number is not a term
Students leave off the reason after finding a fractional $n$ .
Wrong working
$n = 24.5$ . Answer: no.
Right working
$n = 24.5$ , which is not a natural number, so it is not a term.
Do this: Write the natural-number reason in words.
Usually costs: ½ to 1 mark.

Exam tips

In a derivation, show at least three terms written using $a$ and $d$ before writing the general formula.
Label equations (1) and (2) when you find $a$ and $d$ , and write the AP at the end.
For questions about the term from the end, check with the $(m - n + 1)$ th term from the start.
In assertion-reason items, test whether $n$ is a natural number by actually solving.
Substitute your $a$ and $d$ back into both given conditions before moving on.

Practice questions on nth Term 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 · 2 marks · Very short answerBoard level
Find the 31st term of an AP whose 11th term is 38 and 16th term is 73.
Hint
Subtract the two conditions to get d first.
Answer
$a + 10 d = 38$ and $a + 15 d = 73$ give $5 d = 35$ , $d = 7$ , and $a = 38 - 70 = - 32$ . (1) $a_{31} = - 32 + 30 \times 7 = 178$ . (1)
Q2 · 3 marks · Short answerBoard level
Which term of the AP $5, 12, 19, \dots$ is 84 more than its 21st term?
Hint
$84 = (m - 21) d$ .
Answer
$d = 7$ , so $m - 21 = 12$ and $m = 33$ . Check: $a_{33} = 229$ , $a_{21} = 145$ , difference 84. The 33rd term.
Q3 · 3 marks · Short answerHarder board level
How many two-digit numbers leave a remainder 2 when divided by 7? Find the middle one or ones.
Hint
The numbers are $16, 23, \dots, 93$ .
Answer
$93 = 16 + (n - 1) 7$ , so $n = 12$ . There are 12 numbers (even), so the middle ones are the 6th and 7th: $16 + 35 = 51$ and $58$ .
Q4 · 3 marks · Short answerBoard level
The 3rd term of an AP is 16 and the 7th term exceeds the 5th term by 12. Find the AP and check whether 301 is a term.
Hint
$a_{7} - a_{5} = 2 d$ .
Answer
$2 d = 12$ , so $d = 6$ , and $a + 12 = 16$ gives $a = 4$ . AP: $4, 10, 16, 22, \dots$ . Set $4 + (n - 1) 6 = 301$ : $n - 1 = 49.5$ , which is not a natural number, so 301 is not a term.
Q5 · 3 marks · Short answerBoard level
Find the error. A student says: "In $3, 10, 17, \dots$ , the number of terms up to 206 is $\frac{206 - 3}{7} = 29$ ." Correct the working and find the 10th term from the end.
Hint
Count the terms, not the gaps.
Answer
29 is the number of gaps. There is one more term, so $n = 30$ . 10th from the end $= 206 - 9 \times 7 = 143$ .
Q6 · 1 mark · MCQBoard level
The 11th term from the end of the AP $- 5, - \frac{5}{2}, 0, \frac{5}{2}, \dots, 150$ is
(a) 122.5
(b) 125
(c) 127.5
(d) 130
Hint
Find d, then count back 10 steps from the last term.
Answer
(b) 125. $d = \frac{5}{2}$ , and the 11th term from the end is $150 - 10 \times \frac{5}{2} = 125$ .
Q7 · 1 mark · MCQBoard level
If $k + 2$ , $4 k - 6$ and $3 k - 2$ are three consecutive terms of an AP, then $k$ equals
(a) 2
(b) 3
(c) 4
(d) $- 3$
Hint
Twice the middle term equals the sum of the other two.
Answer
(b) 3. $2 (4 k - 6) = (k + 2) + (3 k - 2)$ gives $8 k - 12 = 4 k$ , so $k = 3$ . The terms are $5, 6, 7$ ✓
Q8 · 1 mark · Assertion-reasonBoard level
Assertion (A): 184 is a term of the AP $3, 8, 13, 18, \dots$
Reason (R): A number $x$ is a term of an AP with first term $a$ and common difference $d$ only if $\frac{x - a}{d} + 1$ is a natural number.
(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.
Hint
Solve $3 + (n - 1) 5 = 184$ and look at $n$ .
Answer
(d). R is true, since the position $n = \frac{x - a}{d} + 1$ must be a natural number. For A: $n - 1 = \frac{181}{5}$ , which is not a whole number, so 184 is not a term and A is false.
Q9 · 5 marks · Long answerHarder board level
An AP has 50 terms. Show that $a_{k} + a_{51 - k}$ has the same value for every $k$ from 1 to 50, so terms equidistant from the two ends have the same sum. If $a_{3} + a_{48} = 110$ and $a_{10} = 24$ , find $a$ , $d$ , the last term and the two middle terms.
Hint
Write $a_{k} = a + (k - 1) d$ and $a_{51 - k} = a + (50 - k) d$ .
Answer
$a_{k} + a_{51 - k} = 2 a + (k - 1 + 50 - k) d = 2 a + 49 d$ , which does not depend on $k$ . So $2 a + 49 d = 110$ ... (1). Also $a + 9 d = 24$ , so $2 a + 18 d = 48$ ... (2). Subtracting: $31 d = 62$ , so $d = 2$ and $a = 24 - 18 = 6$ . Last term $a_{50} = 6 + 49 \times 2 = 104$ . With 50 terms the middle terms are the 25th and 26th: $6 + 48 = 54$ and $56$ . Check: $a_{3} + a_{48} = 10 + 100 = 110$ ✓
Q10 · 4 marks · Case-basedHOTS / competency
Suppose your school auditorium has rows of seats, and each row has a fixed number of seats more than the row in front of it. Row 5 has 36 seats, row 12 has 64 seats and the last row has 100 seats.
(i) How many seats are in the first row, and how many seats are added per row? (1)
(ii) How many rows are there? (1)
(iii) The organisers want one row with exactly 70 seats for guests. Is there such a row? Also find the number of seats in the middle row. (2)
Hint
$a_{12} - a_{5} = 7 d$ . Check whether $n$ is a natural number.
Answer
(i) $7 d = 64 - 36 = 28$ , so $d = 4$ and $a = 36 - 16 = 20$ : 20 seats in row 1, 4 more per row. (ii) $100 = 20 + (n - 1) 4$ gives $n - 1 = 20$ , so there are 21 rows. (iii) $20 + (n - 1) 4 = 70$ gives $n - 1 = 12.5$ , not a natural number, so no row has exactly 70 seats (rows 13 and 14 have 68 and 72). With 21 rows the middle row is row 11, which has $20 + 10 \times 4 = 60$ seats.
Q11 · 3 marks · Short answerHarder board level
The 9th term of an AP is zero. Prove that its 29th term is twice its 19th term.
Hint
Use $a_{9} = 0$ to write $a$ in terms of $d$ .
Answer
$a + 8 d = 0$ , so $a = - 8 d$ . Then $a_{29} = - 8 d + 28 d = 20 d$ and $a_{19} = - 8 d + 18 d = 10 d$ . So $a_{29} = 2 \times 10 d = 2 a_{19}$ . Hence proved.
Q12 · 3 marks · Short answerHarder board level
Suppose you save ₹50 in the first week of a year and increase your saving by the same amount every week. In the 10th week you save ₹185. In which week will your weekly saving first be more than ₹400? How much do you save that week?
Hint
Find $d$ first, then solve $a_{n} > 400$ and take the smallest natural $n$ .
Answer
$50 + 9 d = 185$ gives $d = 15$ . You need $50 + 15 (n - 1) > 400$ , so $n - 1 > \frac{350}{15} = 23 \frac{1}{3}$ , and the smallest natural $n$ is 25. Check: week 24 gives $50 + 23 \times 15 = 395$ , that is ₹395, not enough; week 25 gives $50 + 24 \times 15 = 410$ . So the 25th week, with a saving of ₹410.
Q13 · 5 marks · Long answerHarder board level
The sum of the 3rd and 7th terms of an AP is 6 and their product is 8. Find the first term and the common difference, and write the first three terms of each possible AP.
Hint
From the sum, $a + 4 d = 3$ ; write $a_{3} = 3 - 2 d$ and $a_{7} = 3 + 2 d$ .
Answer
$a_{3} + a_{7} = 2 a + 8 d = 6$ , so $a + 4 d = 3$ . (1) Then $a_{3} = a + 2 d = 3 - 2 d$ and $a_{7} = a + 6 d = 3 + 2 d$ . (1) Product: $(3 - 2 d) (3 + 2 d) = 9 - 4 d^{2} = 8$ , so $d^{2} = \frac{1}{4}$ and $d = \pm \frac{1}{2}$ . (1) If $d = \frac{1}{2}$ : $a = 3 - 2 = 1$ , AP $1, \frac{3}{2}, 2, \dots$ (1). If $d = - \frac{1}{2}$ : $a = 5$ , AP $5, \frac{9}{2}, 4, \dots$ . Check: $a_{3} = 2$ , $a_{7} = 4$ (or 4 and 2), sum 6, product 8. (1)
Q14 · 3 marks · Short answerHarder board level
Three numbers in AP have sum 21 and product 231. Find the numbers. Explain why it helps to call them $a - d$ , $a$ and $a + d$ .
Hint
With $a - d$ , $a$ , $a + d$ the d's cancel in the sum.
Answer
Sum: $(a - d) + a + (a + d) = 3 a = 21$ , so $a = 7$ ; the d's cancel, which is why this choice helps. Product: $(7 - d) \times 7 \times (7 + d) = 231$ , so $49 - d^{2} = 33$ and $d = \pm 4$ . The numbers are 3, 7, 11 (or 11, 7, 3).

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.Formula for the nth term of an AP?Show answer
$a_{n} = a + (n - 1) d$
2.Find $d$ for $9, 5, 1, \dots$ Show answer
$d = - 4$
3. $a_{15} - a_{9}$ equals?Show answer
$6 d$
4.Number of terms from $a$ to $l$ ?Show answer
$\frac{l - a}{d} + 1$
5. $d$ for $a_{n} = 4 - 7 n$ ?Show answer
$- 7$
6.nth term from the end?Show answer
$l - (n - 1) d$

nth Term of an AP: frequently asked questions

Do I need to learn the derivation of the nth term formula?

Yes. The syllabus asks for the derivation. Write the first few terms in terms of a and d, note that the number of d's is one less than the position, and state the formula.

How do I know whether a number is a term of an AP?

Set the nth term equal to the number and solve for n. It is a term only if n is a natural number. A fraction or negative value means it is not a term.

Can the common difference be negative or a fraction?

Yes. A negative d gives a decreasing AP, and a fractional d is also allowed. Only n, the position, must be a natural number.

How do I find the nth term from the end?

Start from the last term l and use minus d, giving l minus (n minus 1) times d. You can check it as the (m minus n plus 1)th term from the start.