Arithmetic Progressions · Class 10 — 169 MCQs with Answers

Q1. An arithmetic progression is a list in which each term after the first is the previous term

Beginner

A plus a fixed number B times a fixed number C squared D halved

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plus a fixed number

Adding a fixed amount is exactly the definition.

Q2. The fixed number added each time is called the

Beginner

A common ratio B common difference C first term D arithmetic mean

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common difference

It is written d and is the same at every step.

Q3. For the AP $3, 8, 13, 18, \ldots$, the common difference is

Beginner

A 5 B −5 C 3 D 8

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5

$8 - 3 = 5$.

Q4. For the AP $20, 17, 14, 11, \ldots$, the common difference is

Beginner

A 3 B 20 C −20 D −3

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−3

$17 - 20 = -3$; later minus earlier.

Q5. The general form of an AP is

Beginner

A $a,\ a + d,\ a + 2d,\ a + 3d,\ \ldots$ B $a,\ ad,\ ad^2,\ ad^3,\ \ldots$ C $a,\ a + 1,\ a + 2,\ \ldots$ D $a,\ 2a,\ 3a,\ \ldots$

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$a,\ a + d,\ a + 2d,\ a + 3d,\ \ldots$

Each term adds one more copy of d.

Q6. In the AP $a,\ a + d,\ a + 2d,\ \ldots$, the third term is

Beginner

A $a + 2d$ B $a + 3d$ C $3a + d$ D $3d$

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$a + 2d$

The multiplier is one less than the position.

Q7. Which list is an AP?

Beginner

A $1, 4, 7, 10$ B $1, 4, 9, 16$ C $2, 4, 8, 16$ D $1, 2, 4, 7$

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$1, 4, 7, 10$

Only the first has a constant difference, 3.

Q8. Is $5, 5, 5, 5$ an AP?

Beginner

A no, because the terms do not change B yes, with $d = 0$ C no, because $d$ must be positive D only if written differently

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yes, with $d = 0$

Zero is a perfectly good fixed number.

Q9. The first four terms of the AP with $a = 2$ and $d = 3$ are

Beginner

A $2, 6, 18, 54$ B $3, 5, 7, 9$ C $2, 3, 4, 5$ D $2, 5, 8, 11$

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$2, 5, 8, 11$

Add 3 to each term in turn.

Q10. The first four terms of the AP with $a = 7$ and $d = -2$ are

Beginner

A $7, 9, 11, 13$ B $-2, 5, 3, 1$ C $7, 5, 3, 1$ D $7, -2, -11, -20$

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$7, 5, 3, 1$

A negative d makes the list decrease by 2 each step.

Q11. An AP with a last term is described as

Beginner

A infinite B finite C constant D increasing

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finite

A finite AP stops; an infinite one is written with three dots.

Q12. Assertion (A): $-4, -1, 2, 5$ is not an arithmetic progression. Reason (R): Each term is obtained by adding 3 to the previous term.

Beginner

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

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A is false but R is true

The difference is a constant 3, so it is an AP and the assertion is false. The reason states that difference correctly.

Q13. The term that follows $8, 15, 22, \ldots$ is

Beginner

A 30 B 28 C 29 D 23

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29

The common difference is 7, so the next term is $22 + 7$.

Q14. In the general form of an AP, the sixth term is

Beginner

A $a + 6d$ B $6a + d$ C $6d$ D $a + 5d$

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$a + 5d$

The multiplier of d is one less than the position.

Q15. A list of numbers is an AP if and only if

Beginner

A every consecutive difference is the same B the terms increase C the terms are whole numbers D the first two terms differ

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every consecutive difference is the same

The condition is on all the differences — nothing is required about size, sign or type of number.

Q16. Rows of a hall hold 20, 24, 28 and 32 seats. These numbers form

Beginner

A an AP with $d = 20$ B an AP with $d = 4$ C a list that is not an AP D an AP with $d = 0$

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an AP with $d = 4$

Each row has 4 more seats than the one in front.

Q17. For the AP $7, 11, 15, 19, \ldots$, write the first term and the common difference, and find the next two terms.

Beginner

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Q18. Determine whether $2, 4, 8, 16, \ldots$ is an AP, showing your working.

Beginner

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Q19. Determine whether $-10, -6, -2, 2, \ldots$ is an AP. If it is, state a and d.

Beginner

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Q20. Write the first four terms of the AP whose first term is $-2$ and whose common difference is $-2$.

Beginner

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Q21. The nth term of an AP is given by

Beginner

A $a + nd$ B $a - (n-1)d$ C $a + (n-1)d$ D $an + d$

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$a + (n-1)d$

Reaching the nth term takes $n-1$ steps of size d.

Q22. In $a_n = a + (n-1)d$, the letter a stands for

Beginner

A the nth term B the common difference C the number of terms D the first term

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the first term

d is the common difference and n the position.

Q23. The second term of an AP equals

Beginner

A $a + 2d$ B $2a$ C $a + d$ D $2d$

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$a + d$

One step from the first term.

Q24. The tenth term of an AP equals

Beginner

A $a + 10d$ B $a + 9d$ C $10a + d$ D $10d$

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$a + 9d$

Nine steps from the first term.

Q25. For the AP with $a = 2$ and $d = 3$, the tenth term is

Beginner

A 32 B 29 C 30 D 23

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29

$2 + 9(3) = 29$.

Q26. For the AP with $a = 7$ and $d = 5$, the sixth term is

Beginner

A 32 B 37 C 35 D 30

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32

$7 + 5(5) = 32$.

Q27. For the AP $2, 7, 12, 17, \ldots$, the eleventh term is

Beginner

A 52 B 57 C 50 D 47

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52

$a = 2$, $d = 5$, so $2 + 10(5) = 52$.

Q28. For the AP with $a = 5$ and $d = -3$, the eighth term is

Beginner

A 16 B −16 C −21 D 29

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−16

$5 + 7(-3) = 5 - 21 = -16$.

Q29. How many steps of size d separate the first term from the seventh?

Beginner

A 7 B 6 C 8 D 1

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6

The first term is free, so seven terms need six steps.

Q30. A number x is a term of an AP only when solving for n gives

Beginner

A any real number B a positive integer C a positive fraction D a negative integer

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a positive integer

Positions are counting numbers, so n must be a positive whole number.

Q31. For the AP $1, 4, 7, \ldots$, the fifteenth term is

Beginner

A 46 B 45 C 43 D 40

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43

$1 + 14(3) = 43$.

Q32. Assertion (A): The fifth term of an AP is $a + 4d$. Reason (R): Reaching the $n$th term of an AP requires $n$ steps of size $d$.

Beginner

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

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A is true but R is false

It takes $n - 1$ steps, which is why the fifth term is $a + 4d$ and not $a + 5d$. The assertion stands, the reason is false.

Q33. The third term of an AP equals

Beginner

A $a + 3d$ B $3a$ C $a + 2d$ D $3d$

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$a + 2d$

Two steps from the first term.

Q34. For the AP with $a = 12$ and $d = 4$, the ninth term is

Beginner

A 48 B 40 C 44 D 52

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44

$12 + 8(4) = 44$.

Q35. Find the 12th term of the AP $4, 9, 14, 19, \ldots$

Beginner

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Q36. Find the 25th term of the AP $30, 27, 24, \ldots$

Beginner

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Q37. Which expression correctly completes $S_n = \frac{n}{2}[\ldots]$ for an AP?

Beginner

A $2a + nd$ B $2a + (n-1)d$ C $a + (n-1)d$ D $a + nd$

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$2a + (n-1)d$

Inside the bracket sit twice the first term and $(n-1)$ copies of d.

Q38. When the last term l is known, the sum is

Beginner

A $\frac{n}{2}(a - l)$ B $\frac{n}{2}(a + l)$ C $n(a + l)$ D $\frac{a + l}{2}$

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$\frac{n}{2}(a + l)$

Each of the $\frac n2$ pairs totals $a + l$.

Q39. The sum $1 + 2 + 3 + \ldots + n$ equals

Beginner

A $\frac{n(n+1)}{2}$ B $\frac{n(n-1)}{2}$ C $n^2$ D $\frac{n^2}{2}$

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$\frac{n(n+1)}{2}$

The sum formula with $a = 1$ and $d = 1$.

Q40. The sum $1 + 2 + \ldots + 100$ equals

Beginner

A 5000 B 10100 C 2550 D 5050

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5050

$\frac{100 \times 101}{2} = 5050$.

Q41. For an AP with $a = 3$, $d = 2$, the sum of the first 5 terms is

Beginner

A 35 B 25 C 45 D 30

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35

$\frac52[6 + 8] = \frac52 \times 14 = 35$.

Q42. For an AP with $a = 1$, $d = 4$, the sum of the first 8 terms is

Beginner

A 112 B 120 C 128 D 104

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120

$4[2 + 28] = 120$.

Q43. In $S_n = \frac{n}{2}[2a + (n-1)d]$, the letter l does not appear because

Beginner

A l is always zero B the formula only works when l is unknown C a and d already determine the last term D l is the same as a

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a and d already determine the last term

$l = a + (n-1)d$, so it is already inside the bracket.

Q44. $a_1$ equals

Beginner

A $S_2 - S_1$ B $S_0$ C $S_1$ D $\frac{S_1}{2}$

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$S_1$

The sum of one term is that term.

Q45. The nth term can be recovered from the sums as

Beginner

A $S_n + S_{n-1}$ B $\frac{S_n}{n}$ C $S_n - S_{n-1}$ D $S_{n+1} - S_n$

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$S_n - S_{n-1}$

Everything up to $a_{n-1}$ appears in both and cancels.

Q46. Adding the natural numbers from 1 to 20 gives

Beginner

A 210 B 200 C 220 D 190

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210

$\frac{20 \times 21}{2} = 210$.

Q47. Which form should you use when the last term is given?

Beginner

A $\frac{n}{2}[2a + (n-1)d]$, since it is more general B either, since they give different answers C neither; list the terms instead D $\frac{n}{2}(a + l)$, since it needs no d

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$\frac{n}{2}(a + l)$, since it needs no d

The short form is quicker and the two always agree.

Q48. Assertion (A): The two sum formulas always give the same answer. Reason (R): The two forms agree only when $d = 0$.

Beginner

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

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A is true but R is false

They agree always, one being got from the other by substitution. The assertion holds, the reason is false.

Q49. An AP has $a = 10$ and $d = -2$. Its first 6 terms total

Beginner

A 40 B 20 C 36 D 30

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30

$3[20 - 10] = 30$.

Q50. Adding the first 9 terms of $1, 3, 5, \ldots$ gives

Beginner

A 72 B 90 C 81 D 64

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81

$\frac92[2 + 16] = \frac92 \times 18 = 81$ — the sum of the first 9 odd numbers.

Q51. Find the sum of the first 12 terms of the AP $3, 7, 11, \ldots$

Beginner

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Q52. Find the sum of the first 25 natural numbers.

Beginner

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Q53. Is $1, 4, 9, 16, \ldots$ an AP?

Intermediate

A yes, with $d = 3$ B no — the differences are 3, 5, 7 C yes, with $d = 5$ D yes, with $d = 1$

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no — the differences are 3, 5, 7

The gaps between the square numbers keep growing.

Q54. Is $2, 4, 8, 16, \ldots$ an AP?

Intermediate

A yes, with $d = 2$ B yes, with $d = 4$ C yes, with $d = 8$ D no — each term is multiplied by 2, not increased by a fixed amount

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no — each term is multiplied by 2, not increased by a fixed amount

Its differences are 2, 4 and 8.

Q55. Is $3, 7, 11, 14, \ldots$ an AP?

Intermediate

A yes, with $d = 4$ B yes, with $d = 3$ C yes, but only for the first three terms D no — the last difference is 3, not 4

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no — the last difference is 3, not 4

Every difference must match, and this one breaks at the end.

Q56. A pump removes a quarter of the air remaining in a cylinder at each stroke. The amounts left form

Intermediate

A an AP with $d = -\frac14$ B an AP with $d = \frac34$ C a list that is not an AP D an AP with $d = 0$

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a list that is not an AP

The amounts are $1, \frac34, \frac9{16}, \ldots$, whose differences shrink each time.

Q57. A taxi charges ₹15 for the first km and ₹8 for each further km. The fares form an AP with

Intermediate

A $a = 8,\ d = 15$ B $a = 15,\ d = 15$ C $a = 15,\ d = 8$ D $a = 23,\ d = 8$

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$a = 15,\ d = 8$

The first fare is ₹15 and every later kilometre adds ₹8.

Q58. For the AP $\frac12, 1, \frac32, 2, \ldots$, the common difference is

Intermediate

A 1 B $\frac12$ C $\frac32$ D 2

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$\frac12$

$1 - \frac12 = \frac12$.

Q59. For the AP $-5, -1, 3, 7, \ldots$, the common difference is

Intermediate

A −4 B 4 C −5 D 6

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4

$-1 - (-5) = -1 + 5 = 4$.

Q60. If $7, x, 19$ are in AP, then x is

Intermediate

A 12 B 14 C 26 D 13

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13

The middle term is the average: $\frac{7 + 19}{2} = 13$.

Q61. If a, b, c are in AP, then b equals

Intermediate

A $\frac{a + c}{2}$ B $\frac{a - c}{2}$ C $ac$ D $a + c$

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$\frac{a + c}{2}$

Equal differences give $b - a = c - b$, so $2b = a + c$.

Q62. Which of these lists is NOT an AP?

Intermediate

A $1, 3, 6, 10$ B $1, 3, 5, 7$ C $10, 8, 6, 4$ D $0, 0, 0, 0$

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$1, 3, 6, 10$

The first has differences 2, 3 and 4 — these are the triangular numbers.

Q63. The cost of digging a well is ₹150 for the first metre and rises by ₹50 for each metre after. The costs form

Intermediate

A an AP with $a = 50,\ d = 150$ B a list that is not an AP C an AP with $d = 0$ D an AP with $a = 150,\ d = 50$

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an AP with $a = 150,\ d = 50$

The costs are 150, 200, 250, 300, …

Q64. Assertion (A): A list whose first two differences are equal must be an AP. Reason (R): An AP requires every consecutive difference to be the same.

Intermediate

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

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A is false but R is true

A is false — $3, 7, 11, 14$ is a counterexample — while R correctly states the requirement.

Q65. Assertion (A): $1, \frac34, \frac9{16}, \frac{27}{64}$ is not an AP. Reason (R): The quantity subtracted at each step gets smaller, so the differences are unequal.

Intermediate

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

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Both A and R are true, and R is the correct explanation of A

Taking a quarter of a shrinking amount removes less each time, exactly as R says.

Q66. For the AP $-2.5, -2, -1.5, -1, \ldots$, the common difference is

Intermediate

A −0.5 B −2.5 C 0.5 D 1

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0.5

$-2 - (-2.5) = 0.5$; a negative first term does not make d negative.

Q67. An AP has $a = 100$ and $d = -10$. Its fourth term is

Intermediate

A 60 B 90 C 70 D 40

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70

The fourth term is $a + 3d = 100 - 30$.

Q68. A taxi charges ₹15 for the first kilometre and ₹8 for each additional kilometre. Write the fares for the first four kilometres and show that they form an AP.

Intermediate

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Q69. If $x, 13, 21$ are three consecutive terms of an AP, find x.

Intermediate

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Q70. Explain why $5, 5, 5, 5$ is an arithmetic progression.

Intermediate

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Q71. A list begins $3, 7, 11, 14$. A student says it is an AP because the first two differences are both 4. Explain the error.

Intermediate

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Q72. (a) Define an arithmetic progression and its common difference. (b) For each list, decide whether it is an AP, showing all the differences: (i) $4, 10, 16, 22$, (ii) $1, 3, 9, 27$, (iii) $-2, -2, -2, -2$. (c) For those that are APs, state a and d.

Intermediate

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1½ marks — (a) the definition and d. 2½ marks — (b) all three sets of differences and verdicts. 1 mark — (c) a and d for both APs.

Q73. (a) Write the general form of an AP with first term a and common difference d. (b) Using it, explain why the difference between the eighth and the fourth term of any AP is $4d$. (c) An AP has first term 6 and its fourth term is 21. Find d and write the first five terms.

Intermediate

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1 mark — (a) the general form. 2 marks — (b) both terms written and the subtraction shown. 2 marks — (c) d = 5 and all five terms.

Q74. Which term of the AP $3, 7, 11, \ldots$ is 39?

Intermediate

A the 9th B the 11th C 39 is not a term D the 10th

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the 10th

$3 + (n-1)4 = 39$ gives $n - 1 = 9$.

Q75. Is 40 a term of the AP $3, 7, 11, \ldots$?

Intermediate

A yes, the 10th B no — n comes out 10.25 C yes, the 11th D no — n comes out negative

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no — n comes out 10.25

$(n-1)4 = 37$ does not divide exactly, so n is not a whole number.

Q76. How many terms are there in the AP $10, 17, 24, \ldots, 206$?

Intermediate

A 29 B 28 C 30 D 27

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29

$10 + (n-1)7 = 206$ gives $n - 1 = 28$, so $n = 29$.

Q77. How many terms are there in the AP $-3, 1, 5, \ldots, 61$?

Intermediate

A 16 B 18 C 17 D 15

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17

$-3 + (n-1)4 = 61$ gives $n - 1 = 16$.

Q78. Computing $\frac{l - a}{d}$ for a finite AP gives

Intermediate

A the number of terms B the number of steps, one less than the number of terms C the common difference D the last term

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the number of steps, one less than the number of terms

There is always one more term than there are gaps between terms.

Q79. The 10th term of an AP is 41 and the 18th is 73. Its common difference is

Intermediate

A 8 B 4 C 32 D 3

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4

Subtracting the two equations gives $8d = 32$.

Q80. The 10th term of an AP is 41 and the 18th is 73. Its first term is

Intermediate

A 4 B 9 C 5 D 41

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5

With $d = 4$: $a + 9(4) = 41$, so $a = 5$.

Q81. In any AP, the difference $a_{18} - a_{10}$ equals

Intermediate

A $8d$ B $18d$ C $10d$ D $28d$

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$8d$

The a cancels, leaving $(18 - 10)d$.

Q82. The 3rd term of an AP is 5 and the 7th is 9. Its first term is

Intermediate

A 5 B 1 C 4 D 3

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3

$4d = 4$ gives $d = 1$, then $a + 2 = 5$.

Q83. The 4th term from the end of the AP $5, 9, 13, \ldots, 81$ is

Intermediate

A 69 B 73 C 65 D 77

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69

$l - 3d = 81 - 12 = 69$.

Q84. A worker's salary starts at ₹8000 and rises ₹500 each year. In the tenth year it is

Intermediate

A ₹13,000 B ₹12,000 C ₹12,500 D ₹13,500

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₹12,500

$8000 + 9(500) = 12500$.

Q85. Assertion (A): In the AP $2, 5, 8, \ldots$, the number 100 is a term. Reason (R): A number is a term only if solving $a + (n-1)d = x$ gives a positive integer n.

Intermediate

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

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A is false but R is true

$2 + (n-1)3 = 100$ gives $n - 1 = \frac{98}{3}$, not a whole number — so A is false, while R correctly states the test.

Q86. Assertion (A): Two given terms are enough to determine every term of an AP. Reason (R): Subtracting the two equations gives d, and substituting back gives a.

Intermediate

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

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Both A and R are true, and R is the correct explanation of A

a and d between them fix the whole AP, and R is exactly how they are found.

Q87. For the AP $100, 90, 80, \ldots$, the seventh term is

Intermediate

A 30 B 50 C 20 D 40

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40

$d = -10$, so $100 + 6(-10) = 40$.

Q88. Which term of the AP $2, 9, 16, \ldots$ is 79?

Intermediate

A the 12th B the 11th C the 13th D 79 is not a term

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the 12th

$2 + (n-1)7 = 79$ gives $n - 1 = 11$.

Q89. How many terms are there in the AP $4, 9, 14, \ldots, 149$?

Intermediate

A 30 B 29 C 31 D 28

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30

$4 + (n-1)5 = 149$ gives $n - 1 = 29$.

Q90. Which term of the AP $5, 11, 17, \ldots$ is 95?

Intermediate

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Q91. Is 200 a term of the AP $3, 8, 13, 18, \ldots$? Justify your answer.

Intermediate

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Q92. How many terms are there in the AP $7, 13, 19, \ldots, 205$?

Intermediate

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Q93. The 6th term of an AP is 19 and the 14th term is 43. Find a and d.

Intermediate

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Q94. Explain why the nth term is $a + (n-1)d$ and not $a + nd$.

Intermediate

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Q95. (a) Write the formula for the nth term of an AP and explain why the multiplier of d is $n - 1$. (b) Find the 31st term of the AP whose 11th term is 38 and whose 16th term is 73. (c) Verify your answer by checking the 16th term.

Intermediate

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1½ marks — (a) the formula and the reason. 2½ marks — (b) the subtraction, d, a and the 31st term. 1 mark — (c) the verification.

Q96. Adding the first 10 terms of $2, 5, 8, \ldots$ gives

Intermediate

A 145 B 165 C 155 D 150

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155

$5[4 + 27] = 155$.

Q97. The sum of $5, 9, 13, \ldots, 81$ is

Intermediate

A 840 B 860 C 880 D 820

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860

There are 20 terms, so $S_{20} = 10(5 + 81) = 860$.

Q98. Totalling the first 15 terms of $8, 5, 2, \ldots$ gives

Intermediate

A −195 B 195 C −165 D 120

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−195

$d = -3$, so $\frac{15}{2}[16 - 42] = \frac{15}{2}(-26) = -195$.

Q99. Setting $S_n$ equal to a given number produces

Intermediate

A a linear equation in n B a cubic equation in n C no equation at all D a quadratic equation in n

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a quadratic equation in n

n appears both outside and inside the bracket.

Q100. Solving for n gives $n = 12$ and $n = -\frac{53}{4}$. The number of terms is

Intermediate

A $-\frac{53}{4}$ B 12 C both D neither

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12

A count of terms must be a positive whole number.

Q101. How many terms of $9, 17, 25, \ldots$ add up to 636?

Intermediate

A 11 B 12 C 13 D 10

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12

$4n^2 + 5n - 636 = 0$ factorises as $(n-12)(4n+53) = 0$.

Q102. If $S_n = 3n^2 + 5n$, then $a_{10}$ equals

Intermediate

A 65 B 59 C 62 D 350

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62

$S_{10} - S_9 = 350 - 288 = 62$.

Q103. If $S_n = 3n^2 + 5n$, then the first term is

Intermediate

A 3 B 5 C 11 D 8

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8

$a_1 = S_1 = 3 + 5 = 8$.

Q104. A contractor's penalties are ₹200, ₹250, ₹300, … The penalty ON the 30th day is

Intermediate

A ₹27,750 B ₹1650 C ₹1500 D ₹1700

Show answer & explanation

₹1650

That asks for $a_{30} = 200 + 29(50) = 1650$, not the total.

Q105. A contractor's penalties are ₹200, ₹250, ₹300, … The TOTAL for 30 days is

Intermediate

A ₹1650 B ₹25,000 C ₹30,000 D ₹27,750

Show answer & explanation

₹27,750

That asks for $S_{30} = 15[400 + 1450] = 27750$.

Q106. Adding every natural number from 1 up to 50 gives

Intermediate

A 1250 B 2550 C 1275 D 1225

Show answer & explanation

1275

$\frac{50 \times 51}{2} = 1275$.

Q107. Assertion (A): $S_{30}$ and $a_{30}$ are different quantities. Reason (R): $S_{30}$ is the total of the first thirty terms, while $a_{30}$ is the thirtieth term alone.

Intermediate

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

Show answer & explanation

Both A and R are true, and R is the correct explanation of A

R gives exactly the distinction A asserts, and confusing them is the commonest error in this topic.

Q108. Assertion (A): When solving $S_n = 78$ for the AP $24, 21, 18, \ldots$, only one value of n is acceptable. Reason (R): The number of terms must be a positive integer.

Intermediate

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

Show answer & explanation

A is false but R is true

A is false — both $n = 4$ and $n = 13$ are positive integers and both are correct — while R is a true statement.

Q109. The sum of the first n odd numbers $1 + 3 + 5 + \ldots$ equals

Intermediate

A $n(n+1)$ B $n^2$ C $\frac{n(n+1)}{2}$ D $2n$

Show answer & explanation

$n^2$

With $a = 1$ and $d = 2$: $\frac n2[2 + 2(n-1)] = \frac n2 (2n) = n^2$.

Q110. If $S_n = n^2 + 4n$, then $a_6$ equals

Intermediate

A 13 B 60 C 15 D 17

Show answer & explanation

15

$S_6 - S_5 = 60 - 45 = 15$.

Q111. Adding the AP $3, 8, 13, \ldots, 78$ gives

Intermediate

A 640 B 660 C 620 D 648

Show answer & explanation

648

There are 16 terms, so $S = 8(3 + 78) = 648$.

Q112. Find the sum of the AP $7, 10, 13, \ldots, 46$.

Intermediate

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Q113. Find the sum of the first 15 terms of the AP $8, 5, 2, \ldots$

Intermediate

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Q114. The sum of the first n terms of an AP is $S_n = 3n^2 + 5n$. Find the first term and the 10th term.

Intermediate

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Q115. A contractor is fined ₹200 for the first day of delay, ₹250 for the second, ₹300 for the third, and so on. Find the total fine for a delay of 30 days.

Intermediate

Show answer & explanation

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Q116. Explain why setting $S_n$ equal to a given number produces a quadratic equation in n.

Intermediate

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Q117. (a) Explain, using pairing, why $S_n = \frac{n}{2}(a + l)$. (b) Show how the second form $S_n = \frac{n}{2}[2a + (n-1)d]$ follows from it. (c) Use whichever form is more convenient to find the sum of $5, 9, 13, \ldots, 81$.

Intermediate

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2 marks — (a) the pairing argument. 1½ marks — (b) the substitution. 1½ marks — (c) n = 20 and the sum.

Q118. For which value of k are $k,\ 2k + 1,\ 4k - 1$ three consecutive terms of an AP?

Advanced

A 3 B 2 C 1 D 5

Show answer & explanation

3

The middle is the average: $2(2k + 1) = k + (4k - 1)$, so $4k + 2 = 5k - 1$ and $k = 3$ (giving 3, 7, 11).

Q119. If the first term of an AP is 5 and the common difference is $-2$, the list is

Advanced

A $5, 7, 9, 11, \ldots$ B $5, 3, 1, -1, \ldots$ C $-2, 3, 8, 13, \ldots$ D $5, -2, -9, -16, \ldots$

Show answer & explanation

$5, 3, 1, -1, \ldots$

Each term is 2 less than the one before.

Q120. Three numbers in AP have sum 21. Their middle term is

Advanced

A 10.5 B 7 C 3 D cannot be found

Show answer & explanation

7

Writing them as $b - d$, b, $b + d$, the sum is $3b = 21$, so $b = 7$ — the d cancels.

Q121. If $2x, x + 10, 3x + 2$ are in AP, then x is

Advanced

A 4 B 10 C 6 D 2

Show answer & explanation

6

The middle is the average: $2(x + 10) = 2x + 3x + 2$, so $2x + 20 = 5x + 2$ and $3x = 18$.

Q122. An AP has $a = 3$ and its third term is 11. Its common difference is

Advanced

A 8 B 4 C 2 D 11

Show answer & explanation

4

The third term is $a + 2d$, so $3 + 2d = 11$ and $2d = 8$.

Q123. In an AP, the difference between the fifth and the second term is 12. The common difference is

Advanced

A 3 B 12 C 4 D 6

Show answer & explanation

4

The fifth term exceeds the second by $3d$, so $3d = 12$.

Q124. Which situation does NOT produce an AP?

Advanced

A the fares of a taxi charging a fixed rate per km B the cost of digging each further metre of a well at a fixed extra cost C the number of seats in rows increasing by 2 each row D the amount in an account earning compound interest

Show answer & explanation

the amount in an account earning compound interest

Compound interest multiplies by a fixed factor each year rather than adding a fixed amount.

Q125. For the AP $a, a + d, a + 2d, \ldots$, the difference between the tenth and the seventh term is

Advanced

A $10d$ B $7d$ C $3d$ D $17d$

Show answer & explanation

$3d$

Their multipliers are 9 and 6, so the gap is $3d$.

Q126. Assertion (A): Any three numbers in AP can be written as $b - d$, $b$, $b + d$. Reason (R): Writing them this way makes their sum $3b$, which removes d from the equation.

Advanced

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

Show answer & explanation

Both A and R are true, and R is the correct explanation of A

That substitution is the standard trick for 'three numbers in AP' problems, and R gives exactly why it helps.

Q127. An AP begins $-3, -1, 1, \ldots$. Its common difference and fifth term are

Advanced

A $d = -2$ and fifth term $-11$ B $d = 2$ and fifth term 7 C $d = 2$ and fifth term 5 D $d = 1$ and fifth term 1

Show answer & explanation

$d = 2$ and fifth term 5

$d = -1 - (-3) = 2$, and continuing gives $-3, -1, 1, 3, 5$.

Q128. A cylinder contains air. A vacuum pump removes a quarter of the air remaining at each stroke. Show that the amounts left do not form an AP.

Advanced

Show answer & explanation

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Q129. Find the value of x for which $2x, x + 10, 3x + 2$ are in AP.

Advanced

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Q130. The sum of three numbers in AP is 27. Find the middle number.

Advanced

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Q131. A shopkeeper stacks tins in rows. The bottom row has 30 tins, and each row above has 2 fewer than the row below. (a) Write the number of tins in the first four rows. (b) Show that these numbers form an AP and state a and d. (c) Explain why this AP must be finite, and say what limits it.

Advanced

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1 mark — (a) the four numbers. 2 marks — (b) all differences and both values. 2 marks — (c) the reason for finiteness and the limiting row.

Q132. (a) Show that if a, b, c are three consecutive terms of an AP, then $b = \frac{a + c}{2}$. (b) Use this to find k so that $k + 2$, $4k - 6$ and $3k - 2$ are consecutive terms of an AP. (c) Write the three terms and verify they form an AP.

Advanced

Show answer & explanation

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2 marks — (a) the derivation from equal differences. 2 marks — (b) the equation and k = 3. 1 mark — (c) the three terms and the check.

Q133. Which term of the AP $21, 18, 15, \ldots$ is $-81$?

Advanced

A the 34th B the 35th C the 36th D $-81$ is not a term

Show answer & explanation

the 35th

$21 + (n-1)(-3) = -81$ gives $(n-1)(-3) = -102$, so $n - 1 = 34$.

Q134. Is 0 a term of the AP $21, 18, 15, \ldots$?

Advanced

A yes, the 7th B no C yes, the 21st D yes, the 8th

Show answer & explanation

yes, the 8th

$21 + (n-1)(-3) = 0$ gives $n - 1 = 7$, so 0 is the 8th term.

Q135. The 11th term of an AP is 38 and the 16th is 73. Its first term is

Advanced

A 32 B −7 C −32 D 7

Show answer & explanation

−32

$5d = 35$ gives $d = 7$; then $a + 10(7) = 38$, so $a = -32$.

Q136. If the 3rd and 9th terms of an AP are 4 and $-8$, the term that equals 0 is

Advanced

A the 6th B the 4th C the 5th D the 7th

Show answer & explanation

the 5th

$6d = -12$ gives $d = -2$ and $a = 8$; then $8 + (n-1)(-2) = 0$ gives $n = 5$.

Q137. How many two-digit numbers are divisible by 3?

Advanced

A 30 B 29 C 31 D 33

Show answer & explanation

30

They form the AP $12, 15, \ldots, 99$: $12 + (n-1)3 = 99$ gives $n = 30$.

Q138. How many multiples of 4 lie between 10 and 250?

Advanced

A 59 B 60 C 61 D 58

Show answer & explanation

60

They form the AP $12, 16, \ldots, 248$: $12 + (n-1)4 = 248$ gives $n = 60$.

Q139. An AP has $a = 7$ and $d = 3$. The first term exceeding 100 is

Advanced

A the 33rd term, 103 B the 32nd term, 100 C the 34th term, 106 D the 31st term, 97

Show answer & explanation

the 33rd term, 103

$7 + (n-1)3 > 100$ needs $n - 1 > 31$, so $n = 33$ and the term is $7 + 96 = 103$.

Q140. If the nth term of an AP is $3n + 2$, its common difference is

Advanced

A 2 B 5 C $n$ D 3

Show answer & explanation

3

Consecutive terms differ by $3(n+1) + 2 - (3n + 2) = 3$; the coefficient of n is always d.

Q141. If the nth term of an AP is $3n + 2$, its first term is

Advanced

A 2 B 3 C 8 D 5

Show answer & explanation

5

Putting $n = 1$ gives $3 + 2 = 5$.

Q142. Assertion (A): In the AP $10, 17, 24, \ldots, 206$ there are 28 terms. Reason (R): The number of terms of a finite AP is $\frac{l - a}{d}$.

Advanced

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

Show answer & explanation

A is false but R is true

Both are false: the count is 29, because $\frac{l-a}{d}$ gives the number of GAPS and one must be added.

Q143. The 2nd term of an AP is 14 and the 6th is 30. Its common difference is

Advanced

A 4 B 16 C 8 D 3

Show answer & explanation

4

$4d = 16$, so $d = 4$.

Q144. If the nth term of an AP is $7 - 2n$, its common difference is

Advanced

A 2 B 7 C 5 D −2

Show answer & explanation

−2

The coefficient of n is the common difference.

Q145. Find the 5th term from the end of the AP $3, 8, 13, \ldots, 128$.

Advanced

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Q146. How many two-digit numbers are divisible by 7?

Advanced

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Q147. The nth term of an AP is $5n - 3$. Find its first term and common difference.

Advanced

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Q148. An AP has first term 7 and common difference 3. Find its first term greater than 100.

Advanced

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Q149. (a) How many three-digit numbers are divisible by 7? (b) Show your working as an AP, stating a, d and l. (c) State the largest such number and the smallest, and explain how the formula gives the count.

Advanced

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1 mark — (a) the answer. 2 marks — (b) a, d, l and the equation solved. 2 marks — (c) both extremes and the explanation of the count.

Q150. ₹1000 is put away earning simple interest at 8% a year. The amount at the end of each year forms a list. (a) Write the amounts at the end of the first four years. (b) Show that they form an AP and state a and d. (c) Find the amount at the end of the 15th year, and say in which year the amount first exceeds ₹2000.

Advanced

Show answer & explanation

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1 mark — (a) the four amounts. 1½ marks — (b) the differences, a and d. 2½ marks — (c) the 15th-year amount and the year it passes ₹2000, with working.

Q151. (a) Explain how to decide whether a given number is a term of an AP. (b) Determine whether $-150$ is a term of the AP $11, 8, 5, 2, \ldots$ (c) Find which term of the same AP is $-148$, and state the term after it.

Advanced

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1 mark — (a) the method and the positive-integer condition. 2 marks — (b) the working and the conclusion. 2 marks — (c) n = 54 and the next term.

Q152. How many terms of $24, 21, 18, \ldots$ add up to 78?

Advanced

A both 4 and 13 B 4 only C 13 only D neither

Show answer & explanation

both 4 and 13

$n^2 - 17n + 52 = 0$ gives $(n-4)(n-13) = 0$; both roots are positive integers, so both are valid.

Q153. A running total can pass through the same value twice when the AP is

Advanced

A decreasing B increasing C constant D infinite

Show answer & explanation

decreasing

With $d < 0$ the later terms are negative and pull the total back down.

Q154. Totalling the first 20 terms of $5, 8, 11, \ldots$ gives

Advanced

A 650 B 690 C 620 D 670

Show answer & explanation

670

$10[10 + 57] = 670$.

Q155. An AP has $a = -6$ and $d = \frac12$. Its first 40 terms total

Advanced

A 140 B 150 C 130 D 120

Show answer & explanation

150

$\frac{40}{2}\left[2(-6) + 39\left(\tfrac12\right)\right] = 20[-12 + 19.5] = 20 \times 7.5 = 150$.

Q156. Every two-digit multiple of 3 added together comes to

Advanced

A 1650 B 1683 C 1620 D 1665

Show answer & explanation

1665

They form $12, 15, \ldots, 99$ with 30 terms, so $S = 15(12 + 99) = 1665$.

Q157. An AP totals $5n^2 + 2n$ by its nth term. Its common difference is

Advanced

A 5 B 2 C 10 D 7

Show answer & explanation

10

$a_1 = S_1 = 7$ and $a_2 = S_2 - S_1 = 24 - 7 = 17$, so $d = 10$.

Q158. Adding the first 8 multiples of 3 gives

Advanced

A 96 B 108 C 120 D 144

Show answer & explanation

108

They are $3, 6, \ldots, 24$, so $S_8 = 4(3 + 24) = 108$.

Q159. A stack has 20 logs in the bottom row, 19 in the next, and so on. The total in 12 rows is

Advanced

A 174 B 180 C 168 D 190

Show answer & explanation

174

$a = 20$, $d = -1$, so $S_{12} = 6[40 - 11] = 6 \times 29 = 174$.

Q160. Assertion (A): $a_n = S_n - S_{n-1}$ for every AP. Reason (R): $S_n$ contains every term up to $a_n$, and $S_{n-1}$ every term up to $a_{n-1}$, so all but the last cancel.

Advanced

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

Show answer & explanation

Both A and R are true, and R is the correct explanation of A

The cancellation R describes is exactly what leaves $a_n$ behind.

Q161. In the formula, dividing by 2 before completing the bracket

Advanced

A risks fraction errors for no benefit B is required C changes the answer D is faster in every case

Show answer & explanation

risks fraction errors for no benefit

Finishing the bracket first usually keeps the arithmetic in whole numbers.

Q162. Five cash prizes totalling ₹500 decrease by ₹30 each. The largest prize is

Advanced

A ₹140 B ₹160 C ₹100 D ₹130

Show answer & explanation

₹160

$S_5 = \frac52[2a + 4(-30)] = 5(a - 60) = 500$ gives $a - 60 = 100$.

Q163. Totalling the AP $9, 17, 25, \ldots$ from its start reaches 636 after how many terms?

Advanced

Show answer & explanation

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Q164. Find the sum of all two-digit numbers divisible by 3.

Advanced

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Q165. Find the sum of the first 40 terms of the AP whose first term is $-6$ and whose common difference is $\frac12$.

Advanced

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Q166. A stack of logs has 20 in the bottom row, 19 in the next, and one fewer in each row above. Find the total number of logs in 12 rows.

Advanced

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Q167. (a) How many terms of the AP $24, 21, 18, \ldots$ give a sum of 78? (b) Explain why your answer has two parts, and verify one of them directly. (c) State the condition under which a sum can be reached by two different numbers of terms.

Advanced

Show answer & explanation

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2 marks — (a) the quadratic and both roots. 2 marks — (b) the explanation and a verification. 1 mark — (c) the condition d < 0.

Q168. An AP totals $S_n = 4n - n^2$ by its nth term. (a) Find the first term and $S_2$. (b) Find the 10th term, using $a_n = S_n - S_{n-1}$. (c) Find a general expression for the nth term and hence the common difference.

Advanced

Show answer & explanation

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1 mark — (a) both values. 1½ marks — (b) the 10th term. 2½ marks — (c) the expansion, $a_n = 5 - 2n$ and d = −2.

Q169. A college awards ₹900 in eight cash prizes, each worth ₹25 lower than the previous one. (a) Taking the first prize as ₹a, write an expression for the total of the eight prizes. (b) Find the value of each prize. (c) Check that your eight prizes add to ₹900, and state the smallest prize.

Advanced

Show answer & explanation

—

1½ marks — (a) the expression. 2 marks — (b) a = 200 and all eight prizes. 1½ marks — (c) the verification and the smallest prize.

Arithmetic Progressions · Class 10 Mathematics: 169 MCQs with answers

Warm-up · 52 questions

Building up · 65 questions

Topper level · 52 questions

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