Chapter 5 · Arithmetic Progressions

5.1 Arithmetic Progressions

An arithmetic progression is a list of numbers in which the same number is added every time. Learn to spot one, find its common difference, write one down from its first term, and test three numbers for being in AP.

Meera opens a savings jar in January with ₹ $200$. Every month she puts in ₹ $50$ more than the month before: ₹ $200$, ₹ $250$, ₹ $300$, ₹ $350$, … Her friend Kabir has a metro card with ₹ $500$ on it, and every ride to school costs ₹ $35$, so after each ride the balance reads ₹ $465$, ₹ $430$, ₹ $395$, … The dates of the Mondays in a month might be $3, 10, 17, 24, 31$. These three lists look different, but they are built in the same way: to get the next number, you add the same fixed amount — $+50$, $-35$ and $+7$.

Lists like these are called arithmetic progressions. They are the simplest number patterns of all, and they are worth studying because a fixed step makes the list completely predictable: once you know where it starts and how big the step is, you know every number in it. In this section you learn to recognise an arithmetic progression, to find its step, to write one down from its first term, and to decide when three numbers are "in AP". The next section turns this into a formula for any term.

Lists with a fixed step

A list of numbers written in a definite order — first, second, third, … — is called a sequence, and its numbers are its terms. We write the terms as $a_1, a_2, a_3, \dots$ (read "a one, a two, a three"), so $a_3$ is the third term. In Meera's list $a_1=200$, $a_2=250$ and $a_3=300$.

📘 Definition 5.1 — Arithmetic progression; common difference
A list of numbers $a_1, a_2, a_3, \dots$ is an arithmetic progression (an AP) if every term after the first is obtained by adding the same fixed number $d$ to the term just before it. This fixed number is the common difference of the AP: $$d=a_2-a_1=a_3-a_2=a_4-a_3=\cdots$$ that is, $d=a_{k+1}-a_k$ for every $k$. The common difference may be positive, negative or zero.

If we call the first term $a$ (so $a=a_1$), the terms are $a$, then $a+d$, then $a+d+d=a+2d$, and so on. Every AP therefore looks like this — the general form of an AP:

$$a,\quad a+d,\quad a+2d,\quad a+3d,\quad \dots$$

For Kabir's metro card, $a=500$ and $d=-35$; for the Mondays, $a=3$ and $d=7$. When $d$ is positive the terms increase, when $d$ is negative they decrease, and when $d=0$ every term is the same (like $4, 4, 4, \dots$ — a perfectly good, if boring, AP).

📘 Definition 5.2 — Finite and infinite APs
An AP with only a limited number of terms is a finite AP; its final term is called the last term and is written $l$. An AP that goes on for ever is an infinite AP; it has no last term, and we show this with three dots at the end.

The Mondays of one month, $3, 10, 17, 24, 31$, form a finite AP with $l=31$. The odd numbers $1, 3, 5, 7, \dots$ form an infinite AP with $a=1$ and $d=2$.

🧭 Problem-solving strategy — is this list an AP?
  1. Simplify every term first (fractions to a common denominator, surds such as $\sqrt{12}=2\sqrt3$).
  2. Work out every difference "later term minus earlier term": $a_2-a_1$, $a_3-a_2$, $a_4-a_3$, …
  3. If all the differences are equal, the list is an AP and that value is $d$. If even one difference is different, it is not an AP.
✏️ Example 1 — AP or not?
Which of these lists are APs? For each AP give the common difference and the next term.
(a) $4, 10, 16, 22, \dots$  (b) $12, 7, 2, -3, \dots$  (c) $1, 3, 6, 10, \dots$  (d) $-1.5, -1.5, -1.5, \dots$  (e) $\sqrt3, \sqrt{12}, \sqrt{27}, \sqrt{48}, \dots$  (f) $\tfrac13, \tfrac56, \tfrac43, \tfrac{11}6, \dots$
(a) $10-4=6$, $16-10=6$, $22-16=6$. All equal: an AP with $d=6$. Next term $22+6=28$.
(b) $7-12=-5$, $2-7=-5$, $-3-2=-5$. An AP with $d=-5$ (a decreasing AP). Next term $-3+(-5)=-8$.
(c) $3-1=2$, $6-3=3$, $10-6=4$. The differences are not equal, so this is not an AP (the steps themselves grow by $1$).
(d) Every difference is $-1.5-(-1.5)=0$. An AP with $d=0$. Next term $-1.5$.
(e) Simplify $\sqrt{12}=\sqrt{4\times3}=2\sqrt3$, $\sqrt{27}=3\sqrt3$, $\sqrt{48}=4\sqrt3$. The list is $\sqrt3, 2\sqrt3, 3\sqrt3, 4\sqrt3, \dots$
(e) Answer Every difference is $\sqrt3$: an AP with $d=\sqrt3$. Next term $5\sqrt3$ $(=\sqrt{75})$.
(f) With denominator $6$: $\tfrac26, \tfrac56, \tfrac86, \tfrac{11}6$. Every difference is $\tfrac36=\tfrac12$: an AP with $d=\tfrac12$. Next term $\tfrac{14}6=\tfrac73$.
⚠️ Common mistakes

1. Subtracting the wrong way round. $d$ is always later term minus earlier term. For $12, 7, 2, \dots$ the common difference is $7-12=-5$, not $5$.

2. Checking only the first two terms. In $2, 4, 8, 16, \dots$ the first step is $2$, but the next is $4$. One unequal step is enough to spoil an AP.

3. Not simplifying first. $\sqrt2, \sqrt8, \sqrt{18}$ does not look like an AP until you write it as $\sqrt2, 2\sqrt2, 3\sqrt2$.

4. Thinking an AP must increase. A negative $d$ gives a decreasing AP, and $d=0$ gives a constant AP. Both are APs.

Interactive 1 — Hop Along the Number Line

Choose the first term $a$ and the common difference $d$. A frog starts at $a$ on the number line (top) and makes hops of the same size $d$ — the landing points are the terms of the AP. Below, each term $a_n$ is plotted above its position $n$: every red step goes $1$ across and $d$ up (or down). Press Play to watch several APs; try a negative $d$ and $d=0$.

34
terms $a_1$ to $a_8$—
every step—
the AP is—

Writing down an AP and recognising one in real life

Two numbers fix an AP completely: the first term $a$ and the common difference $d$. Start at $a$ and keep adding $d$. Take extra care with negative and fractional values of $d$ — this is where most slips happen.

✏️ Example 2 — From $a$ and $d$ to the AP
Write the first four terms of the AP with (a) $a=7$, $d=-3$; (b) $a=-2$, $d=\tfrac14$; (c) $a=0.6$, $d=-1.1$.
(a) $7$, $\;7-3=4$, $\;4-3=1$, $\;1-3=-2$. The AP is $7, 4, 1, -2, \dots$
(b) $-2$, $\;-2+\tfrac14=-\tfrac74$, $\;-\tfrac74+\tfrac14=-\tfrac32$, $\;-\tfrac32+\tfrac14=-\tfrac54$. The AP is $-2, -\tfrac74, -\tfrac32, -\tfrac54, \dots$
(c) $0.6$, $\;0.6-1.1=-0.5$, $\;-0.5-1.1=-1.6$, $\;-1.6-1.1=-2.7$. The AP is $0.6, -0.5, -1.6, -2.7, \dots$
Check In (c), $a+3d=0.6+3(-1.1)=0.6-3.3=-2.7$ ✓ — the fourth term is $a+3d$, as the general form says.
✏️ Example 3 — Which situations give an AP?
In which of these situations does the list of numbers form an AP?
(a) Kabir's metro card balance (in ₹) after $1, 2, 3, \dots$ rides, starting from ₹ $500$ with ₹ $35$ per ride.
(b) The number of bacteria in a dish that starts with $1000$ and doubles every hour, counted every hour.
(c) The height of a stack of $1, 2, 3, \dots$ plastic chairs, if one chair is $82$ cm tall and each extra chair adds $6$ cm.
(d) The amount in an account after $1, 2, 3$ years when ₹ $10\,000$ is invested at $8\%$ per year (i) simple interest (ii) compound interest.
(a) $465, 430, 395, \dots$ Each ride subtracts the same ₹ $35$: an AP with $a=465$, $d=-35$.
(b) $1000, 2000, 4000, 8000, \dots$ The differences are $1000, 2000, 4000$ — not equal. Doubling multiplies by $2$; it does not add a fixed number. Not an AP.
(c) $82, 88, 94, 100, \dots$ cm. Each chair adds the same $6$ cm: an AP with $a=82$, $d=6$.
(d)(i) Simple interest is $8\%$ of ₹ $10\,000$ $=$ ₹ $800$ every year: ₹ $10\,800$, ₹ $11\,600$, ₹ $12\,400$. An AP with $d=800$.
(d)(ii) Compound interest multiplies by $1.08$ each year: ₹ $10\,800$, ₹ $11\,664$, ₹ $12\,597.12$. The differences are ₹ $864$ and ₹ $933.12$ — not equal, so not an AP.
✏️ Example 4 — Terms given by a rule
Write the first four terms of each list and decide whether it is an AP: (a) $a_n=3n-5$  (b) $b_n=n^2+1$.
(a) Terms Put $n=1,2,3,4$: $\;a_1=-2$, $a_2=1$, $a_3=4$, $a_4=7$. The differences are all $3$.
(a) For every $n$ $a_{n+1}-a_n=\left[3(n+1)-5\right]-\left[3n-5\right]=3$, which does not depend on $n$. So the list is an AP with $d=3$ — not only for the first four terms but all the way along.
(b) $b_1=2$, $b_2=5$, $b_3=10$, $b_4=17$. The differences $3, 5, 7$ are not equal: not an AP.

Three numbers in AP

Three numbers $a$, $b$, $c$ (in this order) are "in AP" when the step from $a$ to $b$ equals the step from $b$ to $c$. On a number line, $b$ sits exactly halfway between $a$ and $c$. This simple test is used again and again.

📐 Theorem 5.1 — The middle term of three numbers in AP
Three numbers $a$, $b$, $c$, taken in this order, are in AP if and only if $$2b=a+c,\qquad\text{that is,}\qquad b=\frac{a+c}{2}.$$ The number $\tfrac{a+c}2$ is called the arithmetic mean of $a$ and $c$.

Why it is true. $a$, $b$, $c$ are in AP exactly when the two steps are equal: $b-a=c-b$. Add $a+b$ to both sides and this becomes $2b=a+c$. Every step can be reversed, so the two statements say the same thing.

Interactive 2 — Is the Middle Number Halfway?

Choose the outer numbers $a$ and $c$, then move the middle number $b$. The two gaps $b-a$ (red) and $c-b$ (blue) are drawn on the number line. The three numbers are in AP only when the gaps are equal — exactly when $2b=a+c$. Press Play to slide $b$ and watch for the moment the gaps match.

−681
gap $b-a$—
gap $c-b$—
$2b$ and $a+c$—
✏️ Example 5 — Finding an unknown so that three numbers are in AP
Find $x$ if $2x$, $x+10$ and $3x+2$ are three consecutive terms of an AP. Write down the three terms.
Rule By Theorem 5.1, twice the middle term equals the sum of the outer terms: $2(x+10)=2x+(3x+2)$.
Solve $2x+20=5x+2$, so $18=3x$ and $x=6$.
Terms $2x=12$, $x+10=16$, $3x+2=20$.
Check $16-12=4$ and $20-16=4$: equal steps ✓.
🧭 Problem-solving strategy — choosing unknown terms cleverly
When a problem is about a few terms of an AP whose sum is given, choose the terms symmetrically so that $d$ cancels when you add them:
  • three terms: $\;a-d,\ a,\ a+d$  (sum $3a$);
  • four terms: $\;a-3d,\ a-d,\ a+d,\ a+3d$  (sum $4a$; here the common difference is $2d$).
The sum then gives $a$ at once, and the second condition gives $d$.
✏️ Example 6 — Sum and product of three terms
Three numbers in AP have sum $27$ and product $693$. Find the numbers.
Set up Let the numbers be $a-d$, $a$, $a+d$. Their sum is $3a=27$, so $a=9$.
Product $(9-d)\times9\times(9+d)=693$, so $(9-d)(9+d)=77$, i.e. $81-d^2=77$. Hence $d^2=4$ and $d=2$ or $d=-2$.
Answer $d=2$ gives $7, 9, 11$; $\;d=-2$ gives $11, 9, 7$ — the same three numbers in reverse order.
Check $7+9+11=27$ ✓ and $7\times9\times11=63\times11=693$ ✓.

Exercises

I
Decide whether each list is an AP. If it is, write down the common difference $d$ and the next two terms.
  • 1 $5, 9, 13, 17, \dots$
  • 2 $20, 14, 8, 2, \dots$
  • 3 $2, 6, 18, 54, \dots$
  • 4 $-7, -7, -7, -7, \dots$
  • 5 $\tfrac12, \tfrac54, 2, \tfrac{11}4, \dots$
  • 6 $0.3, 0.33, 0.333, 0.3333, \dots$
  • 7 $\sqrt2, \sqrt8, \sqrt{18}, \sqrt{32}, \dots$
  • 8 $a, \ 2a+1, \ 3a+2, \ 4a+3, \dots$
  • 9 $1^2, 3^2, 5^2, 7^2, \dots$
Check: type a list of numbers (fractions like 5/4 are fine) and see every difference
terms read—
differences—
AP?—
1 Differences $4, 4, 4$: AP, $d=4$; next terms $21, 25$.
2 Differences $-6, -6, -6$: AP, $d=-6$; next terms $-4, -10$.
3 Differences $4, 12, 36$: not an AP (each term is $3$ times the one before).
4 Differences $0$: AP, $d=0$; next terms $-7, -7$.
5 In quarters: $\tfrac24, \tfrac54, \tfrac84, \tfrac{11}4$. Differences $\tfrac34$: AP, $d=\tfrac34$; next terms $\tfrac{14}4=\tfrac72$ and $\tfrac{17}4$.
6 Differences $0.03, 0.003, 0.0003$: not an AP.
7 $\sqrt2, 2\sqrt2, 3\sqrt2, 4\sqrt2$: AP, $d=\sqrt2$; next terms $5\sqrt2=\sqrt{50}$ and $6\sqrt2=\sqrt{72}$.
8 Each difference is $(2a+1)-a=a+1$ (and $(3a+2)-(2a+1)=a+1$, $(4a+3)-(3a+2)=a+1$): AP, $d=a+1$; next terms $5a+4$, $6a+5$.
9 $1, 9, 25, 49$: differences $8, 16, 24$ — not an AP.
II
Write the first four terms. In parts 6–8, also say whether the list is an AP.
  • 1 $a=10$, $d=10$
  • 2 $a=-3$, $d=0$
  • 3 $a=4$, $d=-3$
  • 4 $a=-1$, $d=\tfrac12$
  • 5 $a=-1.25$, $d=-0.25$
  • 6 $a_n=4n+1$
  • 7 $a_n=7-2n$
  • 8 $a_n=2^n$
1 $10, 20, 30, 40$.
2 $-3, -3, -3, -3$.
3 $4, 1, -2, -5$.
4 $-1, -\tfrac12, 0, \tfrac12$.
5 $-1.25, -1.5, -1.75, -2$.
6 $5, 9, 13, 17$. Since $a_{n+1}-a_n=4(n+1)+1-(4n+1)=4$ for every $n$, it is an AP with $d=4$.
7 $5, 3, 1, -1$. $a_{n+1}-a_n=-2$ for every $n$: an AP with $d=-2$.
8 $2, 4, 8, 16$. Differences $2, 4, 8$: not an AP.
III
Missing terms and three numbers in AP.
  • 1 Fill the box so that the numbers are in AP: $2, \ \square, \ 26$.
  • 2 Fill the boxes: $\square, \ 13, \ \square, \ 3$.
  • 3 Fill the boxes: $5, \ \square, \ \square, \ 9\tfrac12$.
  • 4 Fill the boxes: $-4, \ \square, \ \square, \ \square, \ 6$.
  • 5 Find $k$ if $k+2$, $4k-6$, $3k-2$ are three consecutive terms of an AP.
  • 6 Find $x$ if $x$, $2x+3$, $5x-2$ are in AP, and write the three terms.
  • 7 Show that $(a-b)^2$, $a^2+b^2$, $(a+b)^2$ are in AP for all numbers $a$ and $b$.
  • 8 Three numbers in AP have sum $24$ and product $440$. Find them.
  • 9 Three numbers in AP have sum $15$, and the sum of their squares is $83$. Find them.
1 The middle term is the arithmetic mean: $\tfrac{2+26}2=14$.
2 Write the terms as $a, a+d, a+2d, a+3d$. Then $a+d=13$ and $a+3d=3$; subtracting, $2d=-10$, so $d=-5$ and $a=18$. The list is $18, 13, 8, 3$.
3 $a=5$ and $a+3d=9.5$, so $3d=4.5$, $d=1.5$. The list is $5, 6.5, 8, 9.5$.
4 $a=-4$ and $a+4d=6$, so $4d=10$, $d=2.5$. The list is $-4, -1.5, 1, 3.5, 6$.
5 $2(4k-6)=(k+2)+(3k-2)$, so $8k-12=4k$ and $k=3$. Terms $5, 6, 7$ ✓.
6 $2(2x+3)=x+(5x-2)$, so $4x+6=6x-2$ and $x=4$. Terms $4, 11, 18$ (steps $7$, $7$ ✓).
7 Outer terms: $(a-b)^2+(a+b)^2=(a^2-2ab+b^2)+(a^2+2ab+b^2)=2(a^2+b^2)$, which is twice the middle term. By Theorem 5.1 the three are in AP (with common difference $2ab$).
8 Take $a-d, a, a+d$: $3a=24$, $a=8$. Then $8(64-d^2)=440$, $64-d^2=55$, $d=\pm3$. The numbers are $5, 8, 11$ (check: $5\times8\times11=440$ ✓).
9 $3a=15$, $a=5$. $(5-d)^2+25+(5+d)^2=75+2d^2=83$, so $d^2=4$, $d=\pm2$. The numbers are $3, 5, 7$ (check: $9+25+49=83$ ✓).
IV
In which of these situations do the numbers form an AP? Write the first four numbers and, for each AP, give $a$ and $d$ with units.
  • 1 Taxi fare. A taxi charges ₹ $40$ for the first kilometre and ₹ $18$ for each further kilometre. The fares for journeys of $1, 2, 3, \dots$ km.
  • 2 Pumping out a tank. A tank holds $1500$ L. Each hour a pump removes one-fifth of the water left in it. The amount in the tank at the start and after $1, 2, 3$ hours.
  • 3 Stacked cups. One paper cup is $12$ cm tall; each extra cup pushed into the stack adds $1.5$ cm. The heights of stacks of $1, 2, 3, \dots$ cups.
  • 4 Salary. A nurse's starting salary is ₹ $25\,000$ per month with a fixed rise of ₹ $1500$ per month every year. Her monthly salary in years $1, 2, 3, \dots$
  • 5 Savings. ₹ $5000$ is invested at $10\%$ per year compound interest. The amount after $1, 2, 3, \dots$ years.
  • 6 Ladder. The rungs of a ladder are $25$ cm apart and the bottom rung is $30$ cm above the ground. The heights of the rungs, counting from the bottom.
1 ₹ $40, 58, 76, 94, \dots$ — an AP with $a=$ ₹ $40$, $d=$ ₹ $18$.
2 $1500, 1200, 960, 768$ L. Differences $300, 240, 192$ L — not an AP (four-fifths is multiplied each hour).
3 $12, 13.5, 15, 16.5, \dots$ cm — an AP with $a=12$ cm, $d=1.5$ cm.
4 ₹ $25\,000, 26\,500, 28\,000, 29\,500, \dots$ — an AP with $a=$ ₹ $25\,000$, $d=$ ₹ $1500$.
5 ₹ $5500, 6050, 6655, 7320.50$. Differences ₹ $550, 605, 665.50$ — not an AP.
6 $30, 55, 80, 105, \dots$ cm — an AP with $a=30$ cm, $d=25$ cm.
🎯 Exam corner Board · 1 markBoard · 2 marksBoard · 3 marksJEE Main

How this is examined: board papers test the common difference in 1-mark MCQs and Assertion–Reason items, and "find $k$ so that these are in AP" in 2-mark questions; 3-mark questions often ask for three numbers in AP from their sum and product. Write the condition ($d$ = later term − earlier term, or $2b=a+c$) as your first line — it earns the method mark.

E1. Board · 1 mark For $x\ne0$, the common difference of the AP $\dfrac3x, \dfrac{3+2x}x, \dfrac{3+4x}x, \dots$ is  (A) $2x$  (B) $2$  (C) $3$  (D) $\dfrac2x$

Idea $d=\dfrac{3+2x}{x}-\dfrac3x=\dfrac{2x}{x}=2$. (The next step, $\tfrac{3+4x}x-\tfrac{3+2x}x$, is also $2$.)
Trap (A) is the difference of the numerators only — you must keep the denominator $x$, and then it cancels.
Answer (B) $2$.

E2. Board · 1 mark Assertion (A): $-5, -2, 1, 4, \dots$ is an AP with common difference $3$. Reason (R): If the first term of an AP is negative, its common difference must be negative.
(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.

Check A $-2-(-5)=3$, $1-(-2)=3$, $4-1=3$. True.
Check R A itself is a counter-example: first term $-5$, but $d=3\gt0$. The sign of $a$ says nothing about the sign of $d$. False.
Answer (C).

E3. Board · 2 marks For what value of $k$ are $2k+1$, $13$ and $5k-3$ three consecutive terms of an AP?

Condition Consecutive terms are in AP, so $2\times13=(2k+1)+(5k-3)$. (1 mark)
Solve $26=7k-2$, so $7k=28$ and $k=4$. The terms are $9, 13, 17$, with steps $4$ and $4$ ✓. (1 mark)

E4. Board · 3 marks The sum of three numbers in AP is $21$ and their product is $231$. Find the numbers.

Set up Let the numbers be $a-d$, $a$, $a+d$. Sum: $3a=21$, so $a=7$. (1 mark)
Product $7(7-d)(7+d)=231$, so $49-d^2=33$, $d^2=16$, $d=\pm4$. (1 mark)
Answer The numbers are $3, 7, 11$ (or $11, 7, 3$). Check: $3+7+11=21$, $3\times7\times11=231$ ✓. (1 mark)

E5. JEE Main If $x+1$, $3x$ and $4x+2$ are three consecutive terms of an AP, and $x+1$ is its first term, then the fifth term of the AP is ______. (numerical answer)

Find $x$ $2(3x)=(x+1)+(4x+2)$, so $6x=5x+3$ and $x=3$. The terms are $4, 9, 14$, so $a=4$ and $d=5$.
Fifth term In the general form the fifth term is $a+4d=4+20=24$. (List it: $4, 9, 14, 19, 24$.)
Answer $24$. A common slip is $a+5d=29$ — the fifth term needs only four steps.
🔬 Research–Teaching NexusMini project · optional, about 40 minutes

Research question. Do the heights of stacks of identical cups, chairs or books really form an arithmetic progression — and how well can you predict the height of a tall stack from a short one?

Why it matters. Shops, canteens and warehouses stack goods to save space, and designers of stackable chairs, buckets and crates choose how much each extra item adds to the height. If the heights form an AP, two measurements are enough to predict any stack, which is exactly how packaging engineers plan shelves and lorry loads.

What to do.

  1. Collect at least $8$ identical items that nest or stack: paper or plastic cups, steel tumblers, plastic chairs, or copies of the same textbook.
  2. Measure the height of a stack of $1, 2, 3, \dots, 8$ items with a ruler or tape measure, to the nearest millimetre, and record the heights in a table.
  3. Work out every difference between consecutive heights. Are they (nearly) equal? Take their average as your common difference $d$.
  4. Plot height against number of items (by hand or in a spreadsheet). Do the points lie close to a straight line, as in Interactive 1?
  5. Using $a$ and $d$, predict the height of a stack of $15$ items: start at $a$ and add $d$ fourteen times. Then build the stack (if you can) and compare.

Tools: Ruler or tape measure, $8$–$15$ identical cups or books, a spreadsheet or graph paper

Hand in: Your table of heights and differences, the graph, your values of $a$ and $d$ (with units), your prediction for $15$ items, the measured value, and two sentences on why the differences are not all exactly equal.

Going further. Repeat with items that do not nest (flat books) and items that do (cups). Which has the larger common difference compared with the first term, and why does nesting make $d$ so small?

Teacher/parent note: the first cup sets $a$ (its full height) and each extra cup adds only the small rim height, so $d$ is much smaller than $a$. Measurement errors of $1$–$2$ mm make the differences vary slightly; averaging them is the right idea. The prediction for $n$ items is $a+(n-1)d$, which students meet formally in Section 5.2.

🧪 Self-check quiz

0/5 correct

Q1. The common difference of the AP $15, 11, 7, 3, \dots$ is

Q2. Which of these lists is an AP?

Q3. An AP has first term $-2$ and common difference $3$. Its fourth term is

Q4. If $4$, $x$, $18$ are in AP, then $x$ equals

Q5. Which statement about an AP is true?

← Chapter 5 — Arithmetic Progressions5.2 The nth Term of an AP →