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

46 Arithmetic Progression Questions for Reference.

46 Arithmetic Progression Questions with Solutions

A complete question bank for Class 10 & 12 Board Prep provided by Omtex Classes.

Part 1: Basic Concepts and Nth Term

Question 1

Find the common difference of the AP: \( 5, 8, 11, 14, \dots \)

Solution: Common difference \( d = a_2 - a_1 \)
\( d = 8 - 5 = 3 \).
Answer: 3

Question 2

Find the 10th term of the AP: \( 2, 7, 12, \dots \)

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

Question 3

Check if the list of numbers \( 2, 4, 8, 16, \dots \) forms an AP.

Solution: \( a_2 - a_1 = 4 - 2 = 2 \)
\( a_3 - a_2 = 8 - 4 = 4 \)
Since the difference is not constant, it is not an AP.

Question 4

Find the first term \( a \) if the common difference is -3 and the 10th term is -25.

Solution: Given \( d = -3, a_{10} = -25 \).
\( a_{10} = a + 9d \)
\( -25 = a + 9(-3) \)
\( -25 = a - 27 \Rightarrow a = 2 \).
Answer: 2

Question 5

Find the general term (\( n \)th term) of the AP: \( 13, 8, 3, -2, \dots \)

Solution: \( a = 13, d = 8 - 13 = -5 \).
\( a_n = 13 + (n-1)(-5) = 13 - 5n + 5 \).
Answer: \( a_n = 18 - 5n \)

Question 6

Which term of the AP \( 3, 8, 13, 18, \dots \) is 78?

Solution: \( a=3, d=5, a_n=78 \).
\( 78 = 3 + (n-1)5 \)
\( 75 = 5(n-1) \Rightarrow 15 = n - 1 \Rightarrow n = 16 \).
Answer: 16th term

Question 7

Find the 20th term from the last term of the AP: \( 3, 8, 13, \dots, 253 \).

Solution: Reverse the AP: \( 253, \dots, 13, 8, 3 \).
New \( a = 253 \), New \( d = -5 \).
\( a_{20} = 253 + (19)(-5) = 253 - 95 = 158 \).
Answer: 158

Question 8

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

Solution: \( a+2d = 5 \) (i)
\( a+6d = 9 \) (ii)
Subtract (i) from (ii): \( 4d = 4 \Rightarrow d = 1 \).
Substitute in (i): \( a + 2(1) = 5 \Rightarrow a = 3 \).
Answer: 3, 4, 5, 6, ...

Question 9

How many two-digit numbers are divisible by 3?

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

Question 10

Find the middle term of the AP: \( 6, 13, 20, \dots, 216 \).

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

Part 2: Finding Unknowns

Question 11

For what value of \( k \) are \( 2k, k+10, \) and \( 3k+2 \) in AP?

Solution: If \( a, b, c \) are in AP, \( 2b = a + c \).
\( 2(k+10) = 2k + (3k+2) \)
\( 2k + 20 = 5k + 2 \)
\( 18 = 3k \Rightarrow k = 6 \).
Answer: 6

Question 12

If \( x+2, 2x, 2x+3 \) are in AP, find \( x \).

Solution: \( 2(2x) = (x+2) + (2x+3) \)
\( 4x = 3x + 5 \Rightarrow x = 5 \).
Answer: 5

Question 13

Find \( a, b, c \) such that the numbers \( a, 7, b, 23, c \) are in AP.

Solution: \( a_2 = 7 \Rightarrow a+d=7 \). \( a_4 = 23 \Rightarrow a+3d=23 \).
Subtracting: \( 2d = 16 \Rightarrow d = 8 \).
\( a = -1, b = 15, c = 31 \).
Answer: a=-1, b=15, c=31

Question 14

Is 301 a term of the AP \( 5, 11, 17, 23, \dots \)?

Solution: \( a=5, d=6 \).
\( 301 = 5 + (n-1)6 \Rightarrow 296 = 6(n-1) \).
\( n-1 = 296/6 = 49.33 \).
Since \( n \) is not an integer, 301 is not a term.

Question 15

Find the value of \( p \) if the numbers \( 2p-1, 3p+1, 11 \) are in AP.

Solution: \( 2(3p+1) = (2p-1) + 11 \)
\( 6p + 2 = 2p + 10 \)
\( 4p = 8 \Rightarrow p = 2 \).
Answer: 2

Part 3: Sum of n Terms (Sn)

Question 16

Find the sum of the first 20 terms of the AP: \( 1, 4, 7, 10, \dots \)

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

Question 17

Find the sum of the first 100 positive integers.

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

Question 18

Find the sum of: \( 34 + 32 + 30 + \dots + 10 \).

Solution: \( a=34, d=-2, l=10 \).
Find \( n \): \( 10 = 34 + (n-1)(-2) \Rightarrow -24 = -2(n-1) \Rightarrow n=13 \).
\( S_{13} = \frac{13}{2}(a+l) = \frac{13}{2}(34+10) = \frac{13}{2}(44) = 13 \times 22 = 286 \).
Answer: 286

Question 19

The sum of the first \( n \) terms of an AP is given by \( S_n = 3n^2 + n \). Find the 2nd term.

Solution: \( a_1 = S_1 = 3(1)^2 + 1 = 4 \).
\( S_2 = 3(2)^2 + 2 = 14 \).
\( a_2 = S_2 - S_1 = 14 - 4 = 10 \).
Answer: 10

Question 20

How many terms of the AP \( 24, 21, 18, \dots \) must be taken so that their sum is 78?

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

Question 21

Find the sum of the first 15 multiples of 8.

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

Question 22

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

Solution: AP: \( 1, 3, 5, \dots, 49 \). \( a=1, d=2, l=49 \).
\( 49 = 1 + (n-1)2 \Rightarrow 48 = 2(n-1) \Rightarrow n=25 \).
\( S_{25} = \frac{25}{2}(1+49) = \frac{25}{2}(50) = 625 \).
Answer: 625

Question 23

If the sum of first 7 terms is 49 and that of 17 terms is 289, find the sum of first \( n \) terms.

Solution: Given pattern: \( S_n = n^2 \).
(Proof: \( S_7=7^2, S_{17}=17^2 \)).
Answer: \( n^2 \)

Question 24

Find the sum of the first 40 positive integers divisible by 6.

Solution: \( a=6, d=6, n=40 \).
\( S_{40} = \frac{40}{2}[12 + 39(6)] = 20(12 + 234) = 20(246) = 4920 \).
Answer: 4920

Question 25

The first and the last terms of an AP are 17 and 350. If the common difference is 9, how many terms are there and what is their sum?

Solution: \( a=17, l=350, d=9 \).
\( 350 = 17 + (n-1)9 \Rightarrow 333 = 9(n-1) \Rightarrow 37 = n-1 \Rightarrow n=38 \).
\( S_{38} = \frac{38}{2}(17+350) = 19(367) = 6973 \).
Answer: n=38, Sum=6973

Part 4: Word Problems

Question 26

Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Solution: AP: \( 5000, 5200, \dots, 7000 \).
\( a=5000, d=200, a_n=7000 \).
\( 7000 = 5000 + (n-1)200 \Rightarrow 2000 = 200(n-1) \Rightarrow 10 = n-1 \Rightarrow n=11 \).
Year = \( 1995 + 10 = 2005 \).
Answer: 2005

Question 27

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

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

Question 28

In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of Class I will plant 1 tree, a section of Class II will plant 2 trees and so on till Class XII. There are three sections of each class. How many trees will be planted?

Solution: Trees per class: \( 3 \times 1, 3 \times 2, \dots, 3 \times 12 \).
AP: \( 3, 6, 9, \dots, 36 \). \( n=12 \).
\( S_{12} = \frac{12}{2}(3+36) = 6(39) = 234 \).
Answer: 234 trees

Question 29

200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed?

Solution: \( S_n = 200, a=20, d=-1 \).
\( 200 = \frac{n}{2}[40 + (n-1)(-1)] \Rightarrow 400 = n(41-n) = 41n - n^2 \).
\( n^2 - 41n + 400 = 0 \). Factors of 400 summing to 41 are 16 and 25.
If \( n=25, a_{25} = 20 - 24 = -4 \) (impossible). So \( n=16 \).
Answer: 16 rows

Question 30

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

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

Part 5: Advanced & Properties

Question 31

Find three numbers in AP whose sum is 24 and whose product is 440.

Solution: Let terms be \( a-d, a, a+d \).
Sum: \( 3a = 24 \Rightarrow a = 8 \).
Product: \( (8-d)(8)(8+d) = 440 \)
\( 64 - d^2 = 55 \Rightarrow d^2 = 9 \Rightarrow d = \pm 3 \).
Terms: \( 5, 8, 11 \).
Answer: 5, 8, 11

Question 32

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.

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

Question 33

Split 207 into three 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 \( 3a = 207 \Rightarrow a=69 \).
Smaller parts: \( (69-d)(69) = 4623 \).
\( 69-d = 67 \Rightarrow d=2 \).
Terms: \( 67, 69, 71 \).
Answer: 67, 69, 71

Question 34

Find the sum of all integers between 100 and 550 which are divisible by 9.

Solution: First term \( 108 \), Last term \( 549 \). \( d=9 \).
\( 549 = 108 + (n-1)9 \Rightarrow 441 = 9(n-1) \Rightarrow 49 = n-1 \Rightarrow n=50 \).
\( S_{50} = \frac{50}{2}(108+549) = 25(657) = 16425 \).
Answer: 16425

Question 35

If the \( m \)th term of an AP is \( 1/n \) and the \( n \)th term is \( 1/m \), show that the \( mn \)th term is 1.

Solution: \( a + (m-1)d = 1/n \) (1)
\( a + (n-1)d = 1/m \) (2)
Subtracting: \( (m-n)d = \frac{m-n}{mn} \Rightarrow d = \frac{1}{mn} \).
Substituting \( d \): \( a = \frac{1}{mn} \).
\( a_{mn} = \frac{1}{mn} + (mn-1)\frac{1}{mn} = \frac{1 + mn - 1}{mn} = 1 \).
Answer: 1 (Proved)

Question 36

The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles.

Solution: Let angles be \( a-d, a, a+d \). Sum is 180.
\( 3a = 180 \Rightarrow a = 60^\circ \).
\( a+d = 2(a-d) \Rightarrow 60+d = 120 - 2d \Rightarrow 3d = 60 \Rightarrow d = 20 \).
Angles: \( 40^\circ, 60^\circ, 80^\circ \).
Answer: 40°, 60°, 80°

Question 37

Find the sum of the integers between 1 and 100 which are not divisible by 3.

Solution: Total sum (1 to 100) = 5050.
Sum of numbers divisible by 3 (\( 3, 6, \dots, 99 \)): \( n=33 \).
\( S_3 = \frac{33}{2}(3+99) = \frac{33}{2}(102) = 33 \times 51 = 1683 \).
Required Sum = \( 5050 - 1683 = 3367 \).
Answer: 3367

Question 38

If 9 times the 9th term of an AP is equal to 13 times the 13th term, find the 22nd term.

Solution: \( 9(a+8d) = 13(a+12d) \)
\( 9a + 72d = 13a + 156d \)
\( -4a = 84d \Rightarrow a = -21d \Rightarrow a + 21d = 0 \).
Since \( a_{22} = a+21d \), answer is 0.
Answer: 0

Question 39

If the sum of first \( m \) terms of an AP is same as the sum of its first \( n \) terms (\( m \neq n \)), find the sum of its first \( (m+n) \) terms.

Solution: \( S_m = S_n \Rightarrow \frac{m}{2}[2a+(m-1)d] = \frac{n}{2}[2a+(n-1)d] \).
Simplify: \( 2a(m-n) + d(m^2-m - n^2+n) = 0 \).
\( 2a(m-n) + d[(m-n)(m+n) - (m-n)] = 0 \).
Divide by \( m-n \): \( 2a + d(m+n-1) = 0 \).
\( S_{m+n} = \frac{m+n}{2}[2a + (m+n-1)d] = \frac{m+n}{2}(0) = 0 \).
Answer: 0

Question 40

The digits of a positive integer having three digits are in AP and their sum is 15. The number obtained by reversing the digits is 594 less than the original number. Find the number.

Solution: Digits \( a-d, a, a+d \). Sum \( 3a=15 \Rightarrow a=5 \).
Number: \( 100(5-d) + 10(5) + (5+d) \).
Reversed: \( 100(5+d) + 10(5) + (5-d) \).
Difference: \( -99(2d) = -594 \Rightarrow 198d = 594 \Rightarrow d = 3 \).
Digits: \( 8, 5, 2 \). Number: 852.
Answer: 852

Part 6: HOTS (Higher Order Thinking Skills)

Question 41

Determine \( k \) so that \( k^2 + 4k + 8, 2k^2 + 3k + 6, \) and \( 3k^2 + 4k + 4 \) are in AP.

Solution: \( 2b = a+c \).
\( 2(2k^2+3k+6) = (k^2+4k+8) + (3k^2+4k+4) \)
\( 4k^2+6k+12 = 4k^2+8k+12 \)
\( 6k = 8k \Rightarrow 2k = 0 \Rightarrow k = 0 \).
Answer: 0

Question 42

The sum of the first \( n \) terms of two APs are in the ratio \( (7n+1):(4n+27) \). Find the ratio of their \( m \)th terms.

Solution: Ratio of sums \( \frac{S_n}{S'_n} = \frac{2a+(n-1)d}{2a'+(n-1)d'} = \frac{7n+1}{4n+27} \).
To find ratio of \( m \)th term (\( \frac{a+(m-1)d}{a'+(m-1)d'} \)), replace \( n \) with \( 2m-1 \).
\( \frac{a_m}{a'_m} = \frac{7(2m-1)+1}{4(2m-1)+27} = \frac{14m-6}{8m+23} \).
Answer: \( (14m-6):(8m+23) \)

Question 43

Find the sum of all two-digit numbers which when divided by 4, yield 1 as remainder.

Solution: AP: \( 13, 17, \dots, 97 \).
\( 97 = 13 + (n-1)4 \Rightarrow 84 = 4(n-1) \Rightarrow 21 = n-1 \Rightarrow n=22 \).
\( S_{22} = \frac{22}{2}(13+97) = 11(110) = 1210 \).
Answer: 1210

Question 44

Solve the equation: \( 1 + 4 + 7 + 10 + \dots + x = 287 \).

Solution: \( a=1, d=3 \). Let total terms be \( n \).
\( 287 = \frac{n}{2}[2 + (n-1)3] \Rightarrow 574 = n(3n-1) \).
\( 3n^2 - n - 574 = 0 \).
Solving for \( n \): \( n = 14 \).
\( x = a_{14} = 1 + 13(3) = 1 + 39 = 40 \).
Answer: x = 40

Question 45

The 4th term of an AP is equal to 3 times the first term and the 7th term exceeds twice the 3rd term by 1. Find the first term and common difference.

Solution: \( a+3d = 3a \Rightarrow 3d = 2a \) (i)
\( a+6d = 2(a+2d) + 1 \Rightarrow a+6d = 2a+4d+1 \Rightarrow 2d = a+1 \).
From (i), \( a = 1.5d \). Substitute: \( 2d = 1.5d + 1 \Rightarrow 0.5d = 1 \Rightarrow d = 2 \).
\( a = 1.5(2) = 3 \).
Answer: a=3, d=2

Question 46

If \( S_1, S_2, S_3 \) are the sums of \( n \) terms of three APs, the first term of each being 1 and the respective common differences being 1, 2, 3, prove that \( S_1 + S_3 = 2S_2 \).

Solution: \( S_1 = \frac{n}{2}[2 + (n-1)1] = \frac{n(n+1)}{2} \).
\( S_2 = \frac{n}{2}[2 + (n-1)2] = n^2 \).
\( S_3 = \frac{n}{2}[2 + (n-1)3] = \frac{n(3n-1)}{2} \).
\( S_1 + S_3 = \frac{n}{2}(n+1 + 3n-1) = \frac{n}{2}(4n) = 2n^2 = 2S_2 \).
Answer: Proved

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45 Advanced Arithmetic Progression Questions with Solutions

45 More Arithmetic Progression Questions with Solutions

Set 2: Advanced Practice for Class 10 & 12 Board Prep by Omtex Classes.

Part 1: Terms and General Concepts

Question 1

Find the next term of the AP: \( \sqrt{8}, \sqrt{18}, \sqrt{32}, \dots \)

Solution: Simplifying the terms:
\(\sqrt{8} = 2\sqrt{2}\)
\(\sqrt{18} = 3\sqrt{2}\)
\(\sqrt{32} = 4\sqrt{2}\)
The sequence is \( 2\sqrt{2}, 3\sqrt{2}, 4\sqrt{2}, \dots \)
Next term is \( 5\sqrt{2} = \sqrt{25 \times 2} = \sqrt{50} \).
Answer: \(\sqrt{50}\)

Question 2

Find the common difference of the AP where \( a_{18} - a_{14} = 32 \).

Solution: \( a_{18} = a + 17d \) and \( a_{14} = a + 13d \).
\( (a + 17d) - (a + 13d) = 32 \)
\( 4d = 32 \Rightarrow d = 8 \).
Answer: 8

Question 3

Which term of the AP \( 21, 18, 15, \dots \) is -81?

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

Question 4

Is 0 a term of the AP \( 31, 28, 25, \dots \)?

Solution: \( a=31, d=-3 \).
Check if \( 0 = 31 + (n-1)(-3) \).
\( -31 = -3(n-1) \Rightarrow n-1 = 10.33 \).
Since \( n \) is not an integer, 0 is not a term.
Answer: No

Question 5

Find the 4th term from the end of the AP: \( -11, -8, -5, \dots, 49 \).

Solution: Reverse AP: \( 49, \dots, -8, -11 \).
New \( a = 49 \), New \( d = -3 \) (Original was +3).
\( a_4 = 49 + (3)(-3) = 49 - 9 = 40 \).
Answer: 40

Question 6

If the \( n \)th term of an AP is \( 5n - 3 \), find the difference between the 10th and 5th terms.

Solution: \( a_{10} = 5(10) - 3 = 47 \).
\( a_5 = 5(5) - 3 = 22 \).
Difference \( = 47 - 22 = 25 \).
Answer: 25

Question 7

Find the arithmetic mean between 13 and 19.

Solution: AM \( = \frac{a+b}{2} \)
\( = \frac{13+19}{2} = \frac{32}{2} = 16 \).
Answer: 16

Question 8

Find the number of terms in the AP: \( 18, 15\frac{1}{2}, 13, \dots, -47 \).

Solution: \( a=18, d = 15.5 - 18 = -2.5 \).
\( -47 = 18 + (n-1)(-2.5) \)
\( -65 = -2.5(n-1) \Rightarrow n-1 = 26 \Rightarrow n=27 \).
Answer: 27

Question 9

Determine the 10th term of the AP: \( \frac{1}{m}, \frac{1+m}{m}, \frac{1+2m}{m}, \dots \)

Solution: \( a = \frac{1}{m} \).
\( d = \frac{1+m}{m} - \frac{1}{m} = \frac{m}{m} = 1 \).
\( a_{10} = \frac{1}{m} + 9(1) = \frac{1}{m} + 9 = \frac{1+9m}{m} \).
Answer: \(\frac{1+9m}{m}\)

Question 10

Which term of the AP \( 5, 15, 25, \dots \) will be 130 more than its 31st term?

Solution: \( d=10 \).
\( a_n = a_{31} + 130 \)
\( a + (n-1)d = a + 30d + 130 \)
\( (n-1)10 = 30(10) + 130 \)
\( 10n - 10 = 300 + 130 = 430 \)
\( 10n = 440 \Rightarrow n = 44 \).
Answer: 44th term

Part 2: Sum of AP (Sn)

Question 11

Find the sum of the first 22 terms of the AP: \( 8, 3, -2, \dots \)

Solution: \( a=8, d=-5, n=22 \).
\( S_{22} = \frac{22}{2}[2(8) + 21(-5)] \)
\( = 11[16 - 105] = 11[-89] = -979 \).
Answer: -979

Question 12

If \( S_n = 5n^2 + 3n \), find the AP.

Solution: \( a_1 = S_1 = 5(1)^2 + 3(1) = 8 \).
\( S_2 = 5(2)^2 + 3(2) = 20 + 6 = 26 \).
\( a_2 = S_2 - S_1 = 26 - 8 = 18 \).
\( d = a_2 - a_1 = 18 - 8 = 10 \).
Answer: 8, 18, 28, ...

Question 13

How many terms of the AP \( 9, 17, 25, \dots \) must be taken to give a sum of 636?

Solution: \( a=9, d=8, S_n=636 \).
\( 636 = \frac{n}{2}[18 + (n-1)8] = \frac{n}{2}[8n + 10] = 4n^2 + 5n \).
\( 4n^2 + 5n - 636 = 0 \).
Using quadratic formula: \( n = \frac{-5 \pm \sqrt{25 - 4(4)(-636)}}{8} \)
\( n = \frac{-5 \pm 101}{8} \). Taking positive: \( n = 96/8 = 12 \).
Answer: 12

Question 14

Find the sum of all natural numbers between 100 and 200 which are divisible by 4.

Solution: AP: \( 104, 108, \dots, 196 \).
\( 196 = 104 + (n-1)4 \Rightarrow 92 = 4(n-1) \Rightarrow n=24 \).
\( S_{24} = \frac{24}{2}(104 + 196) = 12(300) = 3600 \).
Answer: 3600

Question 15

Find the sum of the first 25 terms of an AP whose nth term is given by \( a_n = 7 - 3n \).

Solution: \( a_1 = 7 - 3(1) = 4 \).
\( a_{25} = 7 - 3(25) = 7 - 75 = -68 \).
\( S_{25} = \frac{25}{2}(4 - 68) = \frac{25}{2}(-64) = 25(-32) = -800 \).
Answer: -800

Question 16

The sum of the first 6 terms of an AP is 36 and the sum of the first 16 terms is 256. Find the sum of the first 10 terms.

Solution: Note the pattern: \( S_n = n^2 \).
\( S_6 = 6^2 = 36 \), \( S_{16} = 16^2 = 256 \).
Therefore, \( S_{10} = 10^2 = 100 \).
Answer: 100

Question 17

Find the sum: \( (-5) + (-8) + (-11) + \dots + (-230) \).

Solution: \( a=-5, d=-3, l=-230 \).
\( -230 = -5 + (n-1)(-3) \Rightarrow -225 = -3(n-1) \Rightarrow n=76 \).
\( S_{76} = \frac{76}{2}(-5 - 230) = 38(-235) = -8930 \).
Answer: -8930

Question 18

If the sum of \( n \) terms of an AP is \( 2n^2 + 5n \), find the common difference.

Solution: Comparing with \( S_n = \frac{d}{2}n^2 + (a - \frac{d}{2})n \), coefficient of \( n^2 \) is \( d/2 \).
\( d/2 = 2 \Rightarrow d = 4 \).
Alternatively: \( S_1 = 7 = a \), \( S_2 = 18 \). \( a_2 = 11 \). \( d = 4 \).
Answer: 4

Part 3: Finding Unknown Variables

Question 19

Find \( x \) if \( 2x, x+10, 3x+2 \) are in AP.

Solution: \( 2(x+10) = 2x + (3x+2) \)
\( 2x + 20 = 5x + 2 \)
\( 18 = 3x \Rightarrow x = 6 \).
Answer: 6

Question 20

If \( k+9, 2k-1, \) and \( 2k+7 \) are in AP, find \( k \).

Solution: \( 2(2k-1) = (k+9) + (2k+7) \)
\( 4k - 2 = 3k + 16 \)
\( k = 18 \).
Answer: 18

Question 21

Determine \( k \) so that \( 4k+8, 2k^2+3k+6, 3k^2+4k+4 \) are in AP.

Solution: \( 2(2k^2+3k+6) = (4k+8) + (3k^2+4k+4) \)
\( 4k^2+6k+12 = 3k^2+8k+12 \)
\( k^2 - 2k = 0 \Rightarrow k(k-2) = 0 \).
Answer: 0 or 2

Question 22

Find the value of \( a \) and \( b \) given that the numbers \( 2, a, 10, b \) are in AP.

Solution: Common difference must be constant.
\( a-2 = 10-a \Rightarrow 2a=12 \Rightarrow a=6 \).
\( d = 6-2=4 \).
\( b = 10+4 = 14 \).
Answer: a=6, b=14

Question 23

If \( \frac{4}{5}, k, 2 \) are in AP, find \( k \).

Solution: \( 2k = \frac{4}{5} + 2 = \frac{14}{5} \).
\( k = \frac{7}{5} \).
Answer: 7/5

Part 4: Real Life Word Problems

Question 24

A man repays a loan of Rs 3250 by paying Rs 20 in the first month and then increasing the payment by Rs 15 every month. How long will it take him to clear the loan?

Solution: \( S_n = 3250, a=20, d=15 \).
\( 3250 = \frac{n}{2}[40 + (n-1)15] \)
\( 6500 = n(40 + 15n - 15) = n(25 + 15n) = 5n(5 + 3n) \).
\( 1300 = 5n + 3n^2 \Rightarrow 3n^2 + 5n - 1300 = 0 \).
Factors of \( 3 \times -1300 = -3900 \) summing to 5 are 65 and -60.
\( n = 20 \) or \( n = -65/3 \).
Answer: 20 months

Question 25

The taxi fare after each km when the fare is Rs 15 for the first km and rises by Rs 8 for each additional km. Write the AP and find the fare for 15 km.

Solution: AP: \( 15, 23, 31, \dots \).
\( a=15, d=8, n=15 \).
\( a_{15} = 15 + 14(8) = 15 + 112 = 127 \).
Answer: Rs 127

Question 26

A manufacturer of TV sets produced 600 sets in the third year and 700 sets in the seventh year. Assuming uniform increase, find production in the 1st year.

Solution: \( a_3 = 600 \Rightarrow a+2d=600 \).
\( a_7 = 700 \Rightarrow a+6d=700 \).
Subtracting: \( 4d = 100 \Rightarrow d = 25 \).
\( a + 50 = 600 \Rightarrow a = 550 \).
Answer: 550 sets

Question 27

In a potato race, a bucket is placed at the starting point, which is 5m from the first potato, and the other potatoes are placed 3m apart in a straight line. There are 10 potatoes. Find the total distance run by a competitor.

Solution: Distances run: \( 2(5), 2(5+3), 2(5+6), \dots \)
AP: \( 10, 16, 22, \dots \). \( n=10 \).
\( S_{10} = \frac{10}{2}[2(10) + 9(6)] = 5[20 + 54] = 5(74) = 370 \).
Answer: 370 m

Question 28

A spiral is made up of successive semicircles, with centers alternately at A and B, starting with center at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, etc. What is the total length of such a spiral made up of 13 consecutive semicircles?

Solution: Perimeter \( \pi r \). \( r_1=0.5, r_2=1.0 \).
AP of lengths: \( 0.5\pi, 1.0\pi, 1.5\pi, \dots \).
\( a=0.5\pi, d=0.5\pi, n=13 \).
\( S_{13} = \frac{13}{2}[2(0.5\pi) + 12(0.5\pi)] = \frac{13}{2}[\pi + 6\pi] = \frac{13}{2}(7\pi) \).
Using \( \pi = 22/7 \): \( \frac{91}{2} \times \frac{22}{7} = 13 \times 11 = 143 \).
Answer: 143 cm

Part 5: Advanced & Properties

Question 29

Divide 32 into four parts which are in AP such that the product of extremes is to the product of means is 7:15.

Solution: Let parts be \( a-3d, a-d, a+d, a+3d \). Sum = 32 \(\Rightarrow 4a=32 \Rightarrow a=8 \).
\( \frac{(8-3d)(8+3d)}{(8-d)(8+d)} = \frac{7}{15} \).
\( \frac{64-9d^2}{64-d^2} = \frac{7}{15} \).
\( 15(64-9d^2) = 7(64-d^2) \Rightarrow 960 - 135d^2 = 448 - 7d^2 \).
\( 512 = 128d^2 \Rightarrow d^2 = 4 \Rightarrow d = 2 \).
Parts: \( 2, 6, 10, 14 \).
Answer: 2, 6, 10, 14

Question 30

If \( p, q, r \) are in AP, prove that \( p^3 + r^3 + 6pqr = 8q^3 \).

Solution: Since in AP, \( p+r = 2q \).
Cube both sides: \( (p+r)^3 = (2q)^3 \).
\( p^3 + r^3 + 3pr(p+r) = 8q^3 \).
Substitute \( p+r=2q \): \( p^3 + r^3 + 3pr(2q) = 8q^3 \).
\( p^3 + r^3 + 6pqr = 8q^3 \).
Answer: Proved

Question 31

Find the sum of all two digit numbers which leave remainder 1 when divided by 3.

Solution: AP: \( 10, 13, 16, \dots, 97 \).
\( 97 = 10 + (n-1)3 \Rightarrow 87 = 3(n-1) \Rightarrow n=30 \).
\( S_{30} = \frac{30}{2}(10+97) = 15(107) = 1605 \).
Answer: 1605

Question 32

If \( a^2, b^2, c^2 \) are in AP, prove that \( \frac{1}{b+c}, \frac{1}{c+a}, \frac{1}{a+b} \) are in AP.

Solution: Given \( 2b^2 = a^2 + c^2 \).
Check AP condition: \( \frac{1}{c+a} - \frac{1}{b+c} = \frac{1}{a+b} - \frac{1}{c+a} \).
LHS = \( \frac{b+c-c-a}{(c+a)(b+c)} = \frac{b-a}{(c+a)(b+c)} \).
RHS = \( \frac{c+a-a-b}{(a+b)(c+a)} = \frac{c-b}{(a+b)(c+a)} \).
Cross multiply \( \frac{b-a}{b+c} = \frac{c-b}{a+b} \Rightarrow b^2-a^2 = c^2-b^2 \Rightarrow 2b^2 = a^2+c^2 \).
Since this is true, the terms are in AP.
Answer: Proved

Question 33

The sum of \( n, 2n, 3n \) terms of an AP are \( S_1, S_2, S_3 \) respectively. Prove that \( S_3 = 3(S_2 - S_1) \).

Solution: \( S_2 - S_1 = \frac{2n}{2}[2a+(2n-1)d] - \frac{n}{2}[2a+(n-1)d] \).
\( = \frac{n}{2} [ 2(2a+2nd-d) - (2a+nd-d) ] \).
\( = \frac{n}{2} [ 4a+4nd-2d -2a-nd+d ] = \frac{n}{2} [ 2a + 3nd - d ] \).
\( 3(S_2-S_1) = \frac{3n}{2} [ 2a + (3n-1)d ] = S_3 \).
Answer: Proved

Question 34

Find the common difference of an AP whose first term is 5 and the sum of the first four terms is half the sum of the next four terms.

Solution: \( S_4 = \frac{1}{2}(S_8 - S_4) \Rightarrow 2S_4 = S_8 - S_4 \Rightarrow 3S_4 = S_8 \).
\( 3 \times \frac{4}{2}[2(5)+3d] = \frac{8}{2}[2(5)+7d] \).
\( 6[10+3d] = 4[10+7d] \)
\( 60 + 18d = 40 + 28d \)
\( 20 = 10d \Rightarrow d = 2 \).
Answer: 2

Part 6: HOTS (Higher Order Thinking Skills)

Question 35

Which term of the AP \( 121, 117, 113, \dots \) is its first negative term?

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

Question 36

If the roots of the cubic equation \( x^3 - 12x^2 + 39x - 28 = 0 \) are in AP, find them.

Solution: Let roots be \( a-d, a, a+d \).
Sum of roots = \( -(-12)/1 = 12 \).
\( (a-d)+a+(a+d) = 12 \Rightarrow 3a = 12 \Rightarrow a = 4 \).
Product of roots = \( -(-28)/1 = 28 \).
\( (4-d)(4)(4+d) = 28 \Rightarrow 16-d^2 = 7 \Rightarrow d^2 = 9 \Rightarrow d=3 \).
Roots: \( 1, 4, 7 \).
Answer: 1, 4, 7

Question 37

The sum of the third and seventh terms of an AP is 6 and their product is 8. Find the sum of the first sixteen 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 \( a \): \( (3-2d)(3+2d) = 8 \).
\( 9 - 4d^2 = 8 \Rightarrow 4d^2 = 1 \Rightarrow d = \pm 1/2 \).
Case 1: \( d=1/2, a=1 \). \( S_{16} = 8[2 + 15(0.5)] = 76 \).
Case 2: \( d=-1/2, a=5 \). \( S_{16} = 8[10 - 7.5] = 20 \).
Answer: 76 or 20

Question 38

If \( S_n \) denotes the sum of first \( n \) terms of an AP, prove that \( S_{12} = 3(S_8 - S_4) \).

Solution: \( S_8 - S_4 = \frac{8}{2}(2a+7d) - \frac{4}{2}(2a+3d) \)
\( = 4(2a+7d) - 2(2a+3d) = 8a+28d-4a-6d = 4a+22d \).
Multiply by 3: \( 12a + 66d \).
\( S_{12} = \frac{12}{2}(2a+11d) = 6(2a+11d) = 12a + 66d \).
Answer: Proved

Question 39

Solve for \( x \): \( -4 + (-1) + 2 + \dots + x = 437 \).

Solution: \( a=-4, d=3 \).
\( 437 = \frac{n}{2}[2(-4) + (n-1)3] \).
\( 874 = n(-8 + 3n - 3) = 3n^2 - 11n \).
\( 3n^2 - 11n - 874 = 0 \). Using quadratic formula, \( n=19 \).
\( x = a_{19} = -4 + 18(3) = 50 \).
Answer: 50

Question 40

Find the AP if the 4th term is 18 and the difference of the 9th and 15th term is 30.

Solution: \( a_{15} - a_9 = 30 \Rightarrow 6d = 30 \Rightarrow d = 5 \).
\( a_4 = 18 \Rightarrow a + 3(5) = 18 \Rightarrow a = 3 \).
Answer: 3, 8, 13, ...

Question 41

Find the sum of all multiples of 7 lying between 500 and 900.

Solution: First multiple: 504. Last multiple: 896.
\( 896 = 504 + (n-1)7 \Rightarrow 392 = 7(n-1) \Rightarrow 56 = n-1 \Rightarrow n=57 \).
\( S_{57} = \frac{57}{2}(504+896) = \frac{57}{2}(1400) = 39900 \).
Answer: 39900

Question 42

If the sum of \( m \) terms of an AP is \( n \) and the sum of \( n \) terms is \( m \), then find the sum of \( (m+n) \) terms.

Solution: \( 2a + (m-1)d = \frac{2n}{m} \) (1)
\( 2a + (n-1)d = \frac{2m}{n} \) (2)
Subtracting: \( d(m-n) = \frac{2n}{m} - \frac{2m}{n} = \frac{2(n^2-m^2)}{mn} \).
\( d = -\frac{2(m+n)}{mn} \).
Substitute to find \( S_{m+n} = -(m+n) \).
Answer: -(m+n)

Question 43

A club consists of members whose ages are in AP, the common difference being 3 months. If the youngest member is 7 years old and the sum of the ages of all members is 250 years, find the number of members.

Solution: Convert all to years. \( d = 3/12 = 0.25 \). \( a=7 \). \( S_n = 250 \).
\( 250 = \frac{n}{2}[14 + (n-1)0.25] \).
\( 500 = n(14 + 0.25n - 0.25) = 13.75n + 0.25n^2 \).
\( 0.25n^2 + 13.75n - 500 = 0 \). Multiply by 4:
\( n^2 + 55n - 2000 = 0 \).
Factors: \( (n+80)(n-25)=0 \).
Answer: 25 members

Question 44

Find the sum of odd integers from 1 to 2001.

Solution: \( a=1, l=2001, d=2 \).
\( 2001 = 1 + (n-1)2 \Rightarrow 2000 = 2(n-1) \Rightarrow n=1001 \).
\( S_n = \frac{1001}{2}(1+2001) = 1001 \times 1001 = 1002001 \).
Answer: 1002001

Question 45

The ratio of the sum of \( m \) and \( n \) terms of an AP is \( m^2 : n^2 \). Show that the ratio of \( m \)th and \( n \)th term is \( 2m-1 : 2n-1 \).

Solution: \( \frac{S_m}{S_n} = \frac{m^2}{n^2} \Rightarrow \frac{\frac{m}{2}[2a+(m-1)d]}{\frac{n}{2}[2a+(n-1)d]} = \frac{m^2}{n^2} \).
\( \frac{2a+(m-1)d}{2a+(n-1)d} = \frac{m}{n} \).
This holds if \( d=2a \).
Ratio of terms: \( \frac{a+(m-1)2a}{a+(n-1)2a} = \frac{a(1+2m-2)}{a(1+2n-2)} = \frac{2m-1}{2n-1} \).
Answer: Proved

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Arithmetic Progression

Definitions and Formulas

A collection of numbers arranged in a definite order according to some definite rule is called a sequence. e.g. 1,5,9,13,.....

If {t_n} is a sequence then we denote the sum of first n terms of this sequence by S_n and formula is S_n = n(n+1)/2.

A special type of sequence in which the relationship between any two consecutive terms is the same is called a progression. e.g. 1,4,9,16,....

A sequence such that for a given first term, each term can be obtained by adding a fixed number to the preceding term is called an Arithmetic Progression. The fixed number is called common difference and is denoted by d. e.g. A.P. = a,a+d,a+2d,a+3d,....

Formula for general term : t_n = a+(n-1)d.

Formula for sum of first n terms of an A.P. whose first term is a and the common difference is d is S_n = n/2 [ 2a+ (n-1)d].

For an A.P. whose first term is a and common difference is d, if any real number k is added to each term of the A.P. then the new sequence is also an A.P. with first term a+k and common difference d.

ALGEBRA WEEKLY TEST

CHAPTERS COVERED: #ARITHMETIC PROGRESSION & #QUADRATIC EQUATION 

MARKS: 30 

1. Find the first five terms of the following sequences, whose nth terms are given.  t_n = 4n – 3

2. Which of the following lists of numbers are Arithmetic Progressions? Justify.  1, 3, 6, 10, ..... 

3. Find the twenty fifth term of the A. P. : 12, 16, 20, 24, ..... 

4. Find four consecutive terms in an A.P. whose sum is 12 and the sum of 3rd and 4th term is 14.

5. What will be her salary after 20 months? [Ans.]

6. Write the following quadratic equations in standard form ax² + bx + c = 0.  7 – 4x –x² = 0  [Ans]

7. Solve the following quadratic equations by  factorization method.  x² – 5x + 6 = 0 [Ans.]

8. Solve the following quadratic equations by using formula. m² – 3m – 10 = 0 [Ans]

9. Determine the nature of the roots of the following equations from their  discriminants : y² – 4y – 1 = 0 [Ans.]

10. Form the quadratic equation if its roots are : 5 and – 7     [Ans.]