50 Arithmetic Progression (AP) Problems with Solutions & Formulas

Essential Arithmetic Progression Formulas

Key Formulas

General Term ($n$th term):
$$a_n = a + (n-1)d$$

Sum of $n$ terms:
$$S_n = \frac{n}{2}[2a + (n-1)d]$$

Sum (using last term $l$):
$$S_n = \frac{n}{2}(a + l)$$

Common Difference:
$$d = a_n - a_{n-1}$$

Arithmetic Mean:
$$A = \frac{a+b}{2}$$

Sum of first $n$ natural numbers:
$$S_n = \frac{n(n+1)}{2}$$

Where $a = \text{first term}$, $d = \text{common difference}$, $n = \text{number of terms}$, $l = \text{last term}$.

Part 1: Basic Terms and Common Difference

1 Find the 10th term of the AP: 2, 7, 12, 17...

Solution Here, $a = 2$, $d = 7 - 2 = 5$, $n = 10$.
$$a_{10} = a + (10-1)d$$ $$a_{10} = 2 + 9(5) = 2 + 45 = 47$$ Answer: 47

2 Find the common difference of the AP: $\frac{1}{2}, \frac{3}{2}, \frac{5}{2}...$

Solution $d = a_2 - a_1 = \frac{3}{2} - \frac{1}{2} = \frac{2}{2} = 1$.
Answer: 1

3 Find the 15th term of the AP: 10, 7, 4...

Solution $a = 10$, $d = 7 - 10 = -3$, $n = 15$.
$a_{15} = 10 + 14(-3) = 10 - 42 = -32$.
Answer: -32

4 If $a = -5$ and $d = 2$, find the first 4 terms.

Solution Terms are $a, a+d, a+2d, a+3d$.
$-5, (-5+2), (-5+4), (-5+6)$
Answer: -5, -3, -1, 1

5 Find the $n$th term of the AP: 13, 8, 3, -2...

Solution $a = 13, d = -5$.
$a_n = 13 + (n-1)(-5) = 13 - 5n + 5 = 18 - 5n$.
Answer: $18 - 5n$

6 Which term of the AP 21, 18, 15... is -81?

Solution $a=21, d=-3, a_n = -81$.
$-81 = 21 + (n-1)(-3) \Rightarrow -102 = (n-1)(-3) \Rightarrow 34 = n-1 \Rightarrow n = 35$.
Answer: 35th term

7 Check if 301 is a term of 5, 11, 17, 23...

Solution $a=5, d=6$. Let $a_n = 301$.
$301 = 5 + (n-1)6 \Rightarrow 296 = 6(n-1) \Rightarrow n-1 = 49.33$.
Since $n$ is not an integer, 301 is not a term.
Answer: No

8 Find $k$ if $2k$, $k+10$, and $3k+2$ are in AP.

Solution If $x, y, z$ are in AP, $2y = x + z$.
$2(k+10) = 2k + (3k+2) \Rightarrow 2k + 20 = 5k + 2 \Rightarrow 18 = 3k \Rightarrow k = 6$.
Answer: $k=6$

9 How many two-digit numbers are divisible by 3?

Solution AP: 12, 15, ..., 99. $a=12, d=3, a_n=99$.
$99 = 12 + (n-1)3 \Rightarrow 87 = 3(n-1) \Rightarrow 29 = n-1 \Rightarrow n=30$.
Answer: 30

10 Find the 11th term from the end of the AP: 10, 7, 4... -62.

Solution Reverse the AP: $a = -62, d = +3$ (opposite of original $d=-3$).
$a_{11} = -62 + 10(3) = -62 + 30 = -32$.
Answer: -32

Part 2: Sum of Terms ($S_n$)

11 Find the sum of the first 20 terms of 1, 4, 7, 10...

Solution $a=1, d=3, n=20$.
$S_{20} = \frac{20}{2}[2(1) + 19(3)] = 10[2 + 57] = 10(59) = 590$.
Answer: 590

12 Find the sum of the first 100 positive integers.

Solution $S_n = \frac{n(n+1)}{2}$.
$S_{100} = \frac{100(101)}{2} = 50(101) = 5050$.
Answer: 5050

13 Find the sum of 2 + 4 + 6 + ... + 200.

Solution $a=2, l=200, n=100$.
$S_{100} = \frac{100}{2}(2 + 200) = 50(202) = 10100$.
Answer: 10100

14 Find the sum of the first 15 multiples of 8.

Solution $a=8, d=8, n=15$.
$S_{15} = \frac{15}{2}[2(8) + 14(8)] = \frac{15}{2}[16 + 112] = \frac{15}{2}(128) = 15(64) = 960$.
Answer: 960

15 How many terms of the AP 24, 21, 18... must be taken so that their sum is 78?

Solution $a=24, d=-3, S_n=78$.
$78 = \frac{n}{2}[48 + (n-1)(-3)] \Rightarrow 156 = n(48 - 3n + 3) \Rightarrow 156 = 51n - 3n^2$.
$n^2 - 17n + 52 = 0 \Rightarrow (n-4)(n-13)=0$.
Answer: $n=4$ or $n=13$

16 Find the sum of all odd numbers between 0 and 50.

Solution AP: 1, 3, 5... 49. Here $n = 25$.
$S_{25} = \frac{25}{2}(1 + 49) = \frac{25}{2}(50) = 25 \times 25 = 625$.
Answer: 625

17 The first and last terms of an AP are 1 and 11. If the sum is 36, find $n$.

Solution $S_n = \frac{n}{2}(a+l) \Rightarrow 36 = \frac{n}{2}(1+11) \Rightarrow 36 = 6n \Rightarrow n=6$.
Answer: 6

18 Find the sum of first 22 terms of an AP in which $d=7$ and 22nd term is 149.

Solution $a_{22} = a + 21d \Rightarrow 149 = a + 21(7) \Rightarrow a = 149 - 147 = 2$.
$S_{22} = \frac{22}{2}(2 + 149) = 11(151) = 1661$.
Answer: 1661

19 If $S_n = 3n^2 + n$, find the AP.

Solution $a_1 = S_1 = 3(1)^2 + 1 = 4$.
$S_2 = 3(2)^2 + 2 = 14$. $a_2 = S_2 - S_1 = 14 - 4 = 10$.
$d = 10 - 4 = 6$.
Answer: 4, 10, 16, 22...

20 Find the sum of first 15 terms of an AP whose $n$th term is $a_n = 9 - 5n$.

Solution $a_1 = 4, a_{15} = 9 - 75 = -66$.
$S_{15} = \frac{15}{2}(4 + (-66)) = \frac{15}{2}(-62) = 15(-31) = -465$.
Answer: -465

Part 3: Finding Unknowns (a, d, n)

21 Determine the AP whose 3rd term is 5 and the 7th term is 9.

Solution $a+2d=5$ and $a+6d=9$.
Subtracting: $4d=4 \Rightarrow d=1$. Substituting $d=1$: $a+2=5 \Rightarrow a=3$.
Answer: 3, 4, 5, 6...

22 Which term of the AP 3, 15, 27, 39... will be 132 more than its 54th term?

Solution $d=12$. Let $a_n = a_{54} + 132$.
$a + (n-1)d = a + 53d + 132$.
$(n-1)12 = 53(12) + 132 \Rightarrow (n-1)12 = 636 + 132 = 768$.
$n-1 = 64 \Rightarrow n = 65$.
Answer: 65th term

23 Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?

Solution Difference between corresponding terms of two APs with same $d$ is constant ($a_1 - a_2$).
Thus, difference is always 100.
Answer: 100

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

Solution $(a+3d)+(a+7d)=24 \Rightarrow 2a+10d=24 \Rightarrow a+5d=12$.
$(a+5d)+(a+9d)=44 \Rightarrow 2a+14d=44 \Rightarrow a+7d=22$.
Solving: $2d=10 \Rightarrow d=5, a=-13$.
Answer: -13, -8, -3...

25 If 7 times the 7th term is equal to 11 times the 11th term, find the 18th term.

Solution $7(a+6d) = 11(a+10d) \Rightarrow 7a+42d = 11a+110d \Rightarrow 4a = -68d \Rightarrow a = -17d$.
$a_{18} = a + 17d = -17d + 17d = 0$.
Answer: 0

26 Find the number of terms in AP: 18, 15.5, 13, ..., -47.

Solution $a=18, d=-2.5, a_n=-47$.
$-47 = 18 + (n-1)(-2.5) \Rightarrow -65 = (n-1)(-2.5) \Rightarrow n-1 = 26 \Rightarrow n=27$.
Answer: 27

27 Subba Rao started work at an annual salary of 5000 with an increment of 200 each year. In which year did his income reach 7000?

Solution $a=5000, d=200, a_n=7000$.
$7000 = 5000 + (n-1)200 \Rightarrow 2000 = 200(n-1) \Rightarrow 10 = n-1 \Rightarrow n=11$.
Answer: 11th year

28 The sum of the first $n$ terms is $4n - n^2$. What is the first term? What is the 2nd term?

Solution $S_1 = 4(1) - 1 = 3$ (First term).
$S_2 = 4(2) - 4 = 4$. $a_2 = S_2 - S_1 = 4 - 3 = 1$.
Answer: 1st term = 3, 2nd term = 1

29 Find the sum of all 3-digit numbers divisible by 7.

Solution First: 105, Last: 994. $n = \frac{994-105}{7} + 1 = 128$.
$S = \frac{128}{2}(105 + 994) = 64(1099) = 70336$.
Answer: 70336

30 A sum of 700 is to be used to give 7 cash prizes. If each prize is 20 less than its preceding prize, find the value of each prize.

Solution $n=7, S_7=700, d=-20$.
$700 = \frac{7}{2}[2a + 6(-20)] \Rightarrow 200 = 2a - 120 \Rightarrow 320 = 2a \Rightarrow a=160$.
Answer: 160, 140, 120, 100, 80, 60, 40

Part 4: Word Problems & Higher Order Thinking

31 Ramkali saved 5 in the first week of a year and then increased her weekly savings by 1.75. If in the $n$th week her savings become 20.75, find $n$.

Solution $a=5, d=1.75, a_n=20.75$.
$20.75 = 5 + (n-1)1.75 \Rightarrow 15.75 = (n-1)1.75 \Rightarrow n-1 = 9 \Rightarrow n=10$.
Answer: 10

32 The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of the first 16 terms of the AP.

Solution $a_3 + a_7 = 6 \Rightarrow 2a + 8d = 6 \Rightarrow a + 4d = 3 \Rightarrow a = 3-4d$.
$a_3 \times a_7 = 8 \Rightarrow (a+2d)(a+6d) = 8$.
Substitute $a$: $(3-2d)(3+2d) = 8 \Rightarrow 9 - 4d^2 = 8 \Rightarrow d^2 = 1/4 \Rightarrow d = \pm 1/2$.
If $d=1/2, a=1, S_{16} = 76$. If $d=-1/2, a=5, S_{16} = 20$.
Answer: 76 or 20

33 Find the sum of odd numbers between 0 and 50.

Solution Repeated Concept: $n=25$, Sum = $n^2 = 25^2 = 625$.
Answer: 625

34 A manufacturer of TV sets produced 600 sets in the 3rd year and 700 sets in the 7th year. Assuming uniform increase, find production in 1st year.

Solution $a+2d=600$ and $a+6d=700$.
$4d=100 \Rightarrow d=25$. $a = 600 - 50 = 550$.
Answer: 550 sets

35 In Q34, what is the total production in 7 years?

Solution $S_7 = \frac{7}{2}(550 + 700) = \frac{7}{2}(1250) = 7(625) = 4375$.
Answer: 4375 sets

36 Which term of the AP 121, 117, 113... is its first negative term?

Solution $a=121, d=-4$. We need $a_n < 0$.
$121 + (n-1)(-4) < 0 \Rightarrow 121 - 4n + 4 < 0 \Rightarrow 125 < 4n \Rightarrow n > 31.25$.
Answer: 32nd term

37 The houses of a row are numbered consecutively from 1 to 49. Is there a value of $x$ such that sum of numbers preceding $x$ is equal to sum of numbers following it?

Solution $S_{x-1} = S_{49} - S_x$.
$\frac{x-1}{2}(1 + x-1) = \frac{49}{2}(50) - \frac{x}{2}(1+x)$.
$x(x-1) = 2450 - x(x+1) \Rightarrow x^2 - x = 2450 - x^2 - x \Rightarrow 2x^2 = 2450 \Rightarrow x^2=1225$.
$x = \sqrt{1225} = 35$.
Answer: Yes, x = 35

38 Find the middle term of the AP 6, 13, 20... 216.

Solution $a=6, d=7$. $216 = 6 + (n-1)7 \Rightarrow 210 = 7(n-1) \Rightarrow n=31$.
Middle term is $\frac{31+1}{2} = 16$th term.
$a_{16} = 6 + 15(7) = 6 + 105 = 111$.
Answer: 111

39 If the $p$th term is $q$ and $q$th term is $p$, prove that $n$th term is $(p+q-n)$.

Solution $a+(p-1)d=q$ and $a+(q-1)d=p$.
Subtract: $(p-q)d = q-p \Rightarrow d = -1$.
Substitute: $a - p + 1 = q \Rightarrow a = p+q-1$.
$a_n = (p+q-1) + (n-1)(-1) = p+q-1-n+1 = p+q-n$. (Proved)

40 Split 207 into 3 parts such that these are in AP and the product of the two smaller parts is 4623.

Solution Let parts be $a-d, a, a+d$. Sum $= 3a = 207 \Rightarrow a=69$.
Product: $(69-d)(69) = 4623 \Rightarrow 69-d = 67 \Rightarrow d=2$.
Parts: $67, 69, 71$.
Answer: 67, 69, 71

Part 5: Rapid Fire & Properties

41 Is the sequence $a_n = 2n^2 + 1$ an AP?

Solution $a_1=3, a_2=9, a_3=19$. Differences: $6, 10$. Not constant.
Answer: No

42 If $a, b, c$ are in AP, then $2b = \dots$?

Solution $2b = a + c$ (Arithmetic Mean property).
Answer: $a+c$

43 Find the sum of first $n$ odd natural numbers.

Solution Formula for sum of $n$ odd numbers is $n^2$.
Answer: $n^2$

44 If the common difference is 5, what is $a_{18} - a_{13}$?

Solution $a_{18} - a_{13} = (a+17d) - (a+12d) = 5d$.
$5(5) = 25$.
Answer: 25

45 Find $a$ if $a, a-2, 3a$ are in AP.

Solution $2(a-2) = a + 3a \Rightarrow 2a - 4 = 4a \Rightarrow -4 = 2a \Rightarrow a = -2$.
Answer: -2

46 If the sum of first $p$ terms is equal to the sum of first $q$ terms, find $S_{p+q}$.

Solution Standard property: If $S_p = S_q$, then $S_{p+q} = 0$.
Answer: 0

47 Find the sum of even numbers between 5 and 15.

Solution Numbers: 6, 8, 10, 12, 14. Sum $= 50$.
Answer: 50

48 In an AP, $a_m = n$ and $a_n = m$. Find $a_{m+n}$.

Solution Standard result: $d = -1$, $a = m+n-1$.
$a_{m+n} = a + (m+n-1)d = (m+n-1) - (m+n-1) = 0$.
Answer: 0

49 Three numbers are in AP and their sum is 24. Find the middle term.

Solution Let terms be $a-d, a, a+d$. Sum $= 3a = 24 \Rightarrow a=8$.
Answer: 8

50 If $18, a, b, -3$ are in AP, find $a+b$.

Solution Sum of equidistant terms is constant. $a+b = 18 + (-3) = 15$.
Answer: 15