CBSE Class 10 Mathematics Arithmetic Progression Worksheet Set 01

More MCQs for NCERT Class 10 Mathematics Arithmetic Progression........

Question. In an Arithmetic progression, the 4^th term is 11 and the 12^th term is 35, then the first term of the series is
(A) 5
(B) 4
(C) 3
(D) 2
Answer : D

Question. The first term of the A.P whose third term is 16 and the difference of 5^th and 7th term is 12 is
(A) 7
(B) 6
(C) 5 
(D) 4
Answer : D

Question. If a,b,c,d,e are in A.P. find the value of a -4b+6c -4d +e ?
(A) 0 
(B) 1
(C) 2 
(D) 3
Answer : A

Question. In a certain A.P., 5 times the 5th term is equal to 8 times the 8^th term find the 13^th term?
(A) 5 
(B) 2
(C) 0 
(D) 1
Answer : C

Question. Find the smallest positive term of the series 25,22³/₄ ,20¹/₂ ,18¹/₄ ..............?
(A) 9^th
(B) 10^th
(C) 11^th 
(D) 12^th
Answer : D

Question. The sum of first four terms of an A.P. is 56. The sum of last four terms is 112. If its first term is 11, find number of terms?
(A) 8 
(B) 9
(C) 10 
(D) 11
Answer : D

Question. The sum of first 20 even natural number is
(a) 420
(b) 100
(c) 220
(d) 400
Answer : A

Question. How many natural numbers are there between 200 and 500, which are divisible by 7?
(a) 42
(b) 43
(c) 63
(d) 53
Answer : B

Question. The 4th term from the end of an AP , -11 , -8 , -5 , ……..49
(a) 17
(b) 25
(c) 40
(d) - 40
Answer : C

Question. The sum of first five multiple of 3 is
(a) 35
(b) 45
(c) 63
(d) 53
Answer : B

Question. Find 15th terms of an AP 15, 10 ,5, 0 -5 -------------
(a) -55
(b) -60
(c) -65
(d) none of these
Answer : A

Question. Given two A.P.’s 2,5,8,11..................T₆₀ and 3,5,7,9.................T₅₀. Find the number of terms which are identical?
(A) 17 
(B) 18
(C) 19 
(D) 20
Answer : A

Question. If pth, qth, rth terms of an A.P. are a,b,c, respectively. Find the value of a(q-r) + b (r-p) +c(p-q)
(A) 2 
(B) 1
(C) 0 
(D) 5
Answer : C

Question. If the sum of three numbers in A.P. is 24 and their product is 440. Find the numbers?
(A) 5,8,11 
(B) 5,9,11
(C) 2,4,9 
(D) 2,6,9
Answer : A

Question. Divide 32 into four parts which are in A.P. such that the ratio of product of extremes to the product of means is 7:15
(A) 1,5,9,13 
(B) 3,7,11,15
(C) 2,6,10,14 
(D) 4,8,12,16
Answer : C

Question. If the sum of series 2,5,8,11.................. is 60100, find n?
(A) 200 
(B) 210
(C) 220 
(D) 240
Answer : A

Question. The sum of n terms of two A.P’s are in ratio 5n+4:9n+6 find ratio of their 18^th terms?
(A) 179:321 
(B) 180:322
(C) 170:320 
(D) 171:329
Answer : A

Question. If log 5 x 2^x + 1 , log4 (2^1-x +1) and 1 are in A.P. find x?
(A) log⁵₂ 
(B)1-log⁵₂
(C) log²₅ 
(D) 1-log²₅
Answer : B

Question. If 3 positive real nos, a,b,c, are in A.P. such that abc=4, find minimum value of b?
(A) 2^1/3
(B) 2^1/2
(C) 2^2/3
(D) 2^3/2
Answer : C

Question. If the sides of right angled triangle are in A.P., then the sines of acute angles are?
(A) 3/5 , 4/5
(B) 1/√2 , √(2/3)
(C) 1/2 , √3/2
(D) 1/√2 , √3/2
Answer : A

Question. Concentric circles of radii 1,2,3,........100 cm are drawn. The interior of the smallest circle is coloured red and the angular regions are colored alternatively green & red, so that no two adjacent regions are of same coloured find total area of green regions in sq. cm.?
(A) 1000π 
(B) 5050π
(C) 4950π 
(D) 5151π
Answer : B

Question. In an AP, the sum of first n term is (3n² /2 +5n/2). Find its 25^th term.
(A) 66 
(B) 86
(C) 76 
(D) 96
Answer : C

Question. If the 12th term of an A.P. is -13 and the sum of the first four terms is 24 what is the sum of the first 10 terms.
(A) 150 
(B) -1
(C) 180 
(D) zero
Answer : D

Question. If n AMs are inserted between 2 & 38, the sum of the resulting series obtained is 200. The value of n (total number of terms) is
(A) 8 
(B) 10
(C) 9 
(D) 11
Answer : A

Question. Find t₅ and t₆ of the arithmetic progression 0, 1/4, 1/2, 3/4,……. respectively
(A) 1, 5/4 
(B) 5/4, 1
(C) 1, 7/4 
(D) 7/4, 1
Answer : A

Question. If t_n = 6n + 5, then t_{n + 1} =
(A) 6n –1 
(B) 6n+11
(C) 6n + 6 
(D) 6n – 5
Answer : B

Question. Which term of the arithmetic progression 21, 42, 63, 84, ……. is 420?
(A) 19 
(B) 20
(C) 21 
(D) 22
Answer : B

Question. Find the 15 term of the arithmetic progression 10, 4, –2,……
(A) –72 
(B) –74
(C) –76 
(D) –78
Answer : B

Question. If the k^th term of the arithmetic progression 25, 50, 75, 100,…….. is 1000, then k is ________.
(A) 20 
(B) 30
(C) 40 
(D) 50
Answer : C

Question. The sum of the first 20 terms of an arithmetic progression whose first term is 5 and common difference is 4, is
(A) 820 
(B) 830
(C) 850 
(D) 860
Answer : D

Question. Two arithmetic progressions have equal common differences. The first term of one of these is 3 and that of the other is 8, then the difference between their 100^th terms is
(A) 4 
(B) 5
(C) 6 
(D) 3
Answer : B

Question. If a, b and c are in arithmetic progression, then b + c, c + a and a + b are in
(A) arithmetic progression 
(B) geometric progression
(C) harmonic progression 
(D) none of these
Answer : A

Question. The sum of the first 51 terms of the arithmetic progression whose 2nd term is 2 and 4^th tem is 8, is
(A) 3774 
(B) 3477
(C) 7548 
(D) 7458
Answer : A

Question. Three alternate terms of an arithmetic progression are x + y,x - y and 2x +3y, then x =
(A) -y 
(B) -2y
(C) -4y 
(D) -6y
Answer : D

Question. Find the 15^th term of the series 243, 81, 27,……..
(A) 1/3¹⁴ 
(B) 1/ 3⁸
(C) (1/3)⁹
(D) (1/3)¹⁰
Answer : C

Question. In a right triangle, the lengths of the sides are in arithmetic progression. If the lengths of the sides of the triangle are integers, which of the following could be the length of the shortest side?
(A) 1225 
(B) 1700
(C) 1275 
(D) 1150
Answer : C

Question. If S₁ = 3,7,11,15,........ upto 125 terms and S₂ = 4,7,10,13,16........ upto 125 terms, then how many terms are there in S₁ that are there in S₂?
(A) 29 
(B) 30
(C) 31 
(D) 32
Answer : C

Question. Find the sum of all natural numbers and lying between 100 and 200 which leave a remainder of 2 when divided by 5 in each case.
(A) 2990 
(B) 2847
(C) 2936 
(D) none of these
Answer : A

Question. An AP starts which a positive fraction and every alternate term is an integer. If the sum of the first 11 terms is 33, then find the fourth term.
(A) 2 
(B) 3
(C) 5 
(D) 6
Answer : A

Question. If the sum of 16 terms of an AP is 1624 and the first term is 500 times the common difference, then find the common difference.
(A) 5 
(B) 1/2
(C) 1/5 
(D) 2
Answer : C

Question. Let Sn denote the sum of n terms of an A.P. whose first term is a. If the common difference d is given by d = S_n - kS_n-1 + S_n-2 then k =
(A) 1 
(B) 2
(C) 3 
(D) none of these
Answer : B

Question. The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P, and the common difference is given by l²-a² /k-(l+a)then k =
(A) S 
(B) 2S
(C) 3S 
(D) none of these
Answer : B

Question. If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =
(A) 1/n
(B) n - 1 /n
(C) n + 1/2n
(D) n + 1 /n
Answer : D

Question. If the first, second and last term of an A.P. are a, b and 2a respectively, its sum is
(A) ab /2(b - a)
(B) ab /b - a
(C) 3ab /2(b - a)
(D) none of these
Answer : C

Question. If S₁ is the sum of an arithmetic progression of ‘n’ odd number of terms and S₂ the sum of the terms of the series in odd places, then S₁/S₂ =
(A) 2n /n +1
(B) n /n +1
(C) n + 1 /2n
(D) n + 1/n
Answer : A

CASE STUDY QUESTIONS

1. India is competitive manufacturing location due to the low cost of manpower and strong technical and engineering capabilities contributing to higher quality production runs. The production of TV sets in a factory increases uniformly by a fixed number every year. It produced 16000 sets in 6th year and 22600 in 9th year.

Question. Find the production during first year.
Answer : Rs 5000

Question. Find the production during 8th year.
Answer : Production during 8th year is (a+7d) = 5000 + 2(2200) = 20400

Question. Find the production during first 3 years.
Answer : Production during first 3 year = 5000 + 7200 + 9400 = 21600

Question. In which year, the production is Rs 29,200.
Answer : N = 12 5. Difference = 18200 - 11600 = 6600

Q2. Students of a school 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.

Question. Find the total number of trees planted by the students of the school.?
Answer : 324

Question.. Find total number of trees planted by primary 1 to 5 class students?
Answer : 45

Question.. Find total number of classes ?
Answer : 12

Very Short Answer Type Questions

Question. In a school, students decided to plant 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 double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students.
Answer : 312

Question. Reshma wanted to save at least ? 6,500 for sending her daughter to school next year (after 12 months). She saved Rs 450 in the first month and raised her savings by ? 20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year?
Answer : Rs 6720

Question. The n^th term of an AP is 6n + 2. Find its common difference.
Answer : 6

Question. The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.
Answer : 40°, 60°, 80°

Question. For what value of n, the nth term of two APs: 63, 65, 67, ... and 3, 10, 17, ... are equal.
Answer : 13

Question. If the sum of the first n terms of an AP is 4n − n², what is the first term (that is S1)? What is the sum of first two terms? What is the second term? Similarly find the 3rd, the 10th and the nth terms.
Answer : The second term is 1, The 3rd, 10th, and nth terms are −1, −15, and 5 − 2n respectively

Question. Determine k so that k² + 4k + 8, 2k² + 3k + 6, 3k² + 4k + 4 are three consecutive terms of an AP. 
Answer : k = 0

Question. Which term of the AP: 115, 110, 105, ..... is its first negative term?
Answer : 25^th term

Question. A sum of Rs 1600 is to be used to give ten cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.
Answer : 250,230,210,190,170,150,130,110,90,70

Question. A thief, after committing a theft, runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?
Answer : police man catch the thief after 5 minutes

Question. Find the 10^th term from end of the AP: 4, 9, 14, ..., 254.
Answer : 209

Question. The 8^th term of an AP is 37 and its 12^th term is 57. Find the AP.
Answer : 2, 7, 12, 17, 22, ...

Short Answer Type Questions

Question. Find how many integers between 200 and 500 are divisible by 8.
Answer : AP formed is 208, 216, 224, ...., 496
a_n = 496 
⇒ 208 + (n – 1) × 8 = 496
⇒ n = 37

Question. Find the number of two-digit numbers which are divisible by 6.
Answer : Two-digit numbers which are divisible by 6 are 12, 18, 24,..., 96
∵ Last term,
a_n = 96
⇒ 12 + (n – 1)6 = 96 
⇒ (n – 1)6 = 96 – 12 = 84
⇒ n = 15
∴ There are 15 two-digit numbers divisible by 6.

Question. Is –150 a term of the AP: 17, 12, 7, 2, ...?
Answer : Let a_n = –150
a + (n – 1)d = –150
⇒ 17 + (n – 1)(–5) = –150
⇒ (n – 1)(–5) = – 167
⇒ n = 167+5/5 = 172/5 = 34(2/5)
Here, n is not a natural number.
∴ –150 is not a term of the given AP.

Question. Which term of the AP: 3, 15, 27, 39, ... will be 120 more than its 21^st term?
Answer : AP: 3, 15, 27, 39, ...
a = 3, d = 15 – 3 = 12
a₂₁ = a + 20d = 3 + 20 × 12
= 3 + 240 = 243
120 more than a₂₁ = 243 + 120 = 363 
Let 363 be n^th term.
So, 363 = 3 + (n – 1) 12
⇒ 360 = 12(n – 1)
30 = n – 1 ⇒ n = 31
Thus, 31st term of the given AP is 120 more than its 21^st term.

Question. If 1/X+2, 1/x+3. and 1/x+5 , and are in AP, find the value of x. 
Answer : Given term are in AP
So, 2/x+3 = 1/2+1 + 1/x+5
⇒ 2/x + 3 = (x+5) + (x+2)/(x+2)(x+5)
⇒ 2x² + 14x + 20 = 2x² + 13x + 21
∴  x = 1

Question. How many three digit numbers are divisible by 11? 
Answer : Three-digit numbers which are divisible by 11 are 110, 121, 132, ..., 990
Let a_n = 990
⇒ a + (n – 1)d = 990
⇒ 110 + 11(n – 1) = 990
∴  n = 81
Hence, there are 81 three-digit numbers which are divisible by 11.

Question. How many natural numbers are there between 200 and 500, which are divisible by 7?
Answer : Natural numbers between 200 and 500 which are divisible by 7 are as 203, 210, 217, ..., 497
Let above are n numbers and an = 497
a + (n – 1)d = 497 
⇒ 203 + 7(n – 1) = 497
⇒ n = 43
∴ There are 43 natural numbers between 200 and 500 divisible by 7

Question. How many two-digit numbers are divisible by 7? 
Answer : Two-digit numbers which are divisible by 7 are 14, 21, 28,..., 98.
Let a_n = 98
⇒ a + (n – 1)d = 98 
⇒ 14 + 7(n – 1) = 98
n = 13
Hence, there are 13 two-digit numbers which are divisible by 7.

Question. How many two digits numbers are divisible by 3?
Answer : 2-digit numbers divisible by 3 are 12, 15, 18, ..., 99 which is in AP.
So, a_n = 99, d = 15 – 12 = 3
Now, a_n = a + (n – 1) d 
⇒ 99 = 12 + (n – 1) 3
⇒ 87 = 3 (n – 1)
⇒ 29 = n – 1
⇒ n = 30
Thus, 30, 2-digit numbers are divisible by 3.

Question. Find the number of natural numbers between 102 and 998 which are divisible by 2 and 5 both.
Answer : 110, 120, 130, … , 990
a_n = 990
⇒ 110 + (n − 1) × 10 = 990
∴ n = 89

Question. Which term of the AP: 3, 14, 25, 36, ... will be 99 more than its 25^th term?
Answer : Let an be the term which is 99 more than 25^th term of given AP.
ATQ, a_n = a₂₅ + 99
⇒ a + (n – 1)d = a + 24d + 99 
⇒ 11(n – 1) = 24 × 11 + 99
⇒ n = 34
Hence, 34^th is the required term.

Question. Which term of the progression 20, 19 (1/4),18 (1/2), 17 (3/4), , , ,... is the first negative term?
Answer : Here d = −3/4
Let the nth term be first negative term.
∴ 20 + (n - 1) (-3/4)  < 0 fi 3n > 83
⇒ n > 27(2/3)
Hence, 28^th term is first negative term.

Question. Find the middle term of the AP: 6, 13, 20, ..., 216.
Answer : Given AP is 6, 13, 20, ..., 216
n^th term, a_n = 216
⇒ a + (n – 1)d = 216
⇒ 6 + 7(n – 1) = 216
⇒ 7n = 217
⇒ n = 31 
Since, the number of terms in AP are 31, so, the middle most term is 16th term.
[ ∵ middle term = (31+1)/2 = 16^th term ]
∴ 16^th term, a₁₆ = a + 15d = 6 + 15 × 7 = 111.

Question. The 4^th term of an AP is zero. Prove that the 25th term of the AP is three times its 11th term. 
Answer : a₄ = a + (4 – 1)d
0 = a + 3d
⇒ a = –3d            [∵ Given, a₄ = 0]
Now a₂₅ = a + (25 – 1)d = a + 24d
= –3d + 24d = 21d = 3 × 7d
Hence, a₂₅ = 3 × a₁₁
[∵ Since a₁₁ = a + (11 – 1)d = –3d + 10d = 7d] 

Question. In an AP, the first term is 12 and the common difference is 6. If the last term of the AP is 252, find its middle term.
Answer : Let an = 252 = last term
⇒ a + (n – 1)d = 252
⇒ 12 + (n – 1)6 = 252
⇒ n = 41
∴  Since number of terms is odd, so only one middle term.
Now, middle term = (41+1/2)
= 21^st term
∴  21^st term, a₂₁ = a + 20d
= 12 + 20 × 6
= 132
= middle term value.

Question. Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.
Answer : Numbers between 101 and 999 which are divisible by both 2 and 5 (i.e., by 10) are 110, 120, 130, ... 990.
Now, a_n = a + (n – 1)d
⇒ 990 = 110 + (n – 1)10
⇒ n = 89
∴ Natural numbers which are divisible by 2 and 5 both are 89.

Chapter 5 - Arithmetic Progression

Question 1. Tick the correct answer:
(1) The common difference of an AP in which \(a_{25} - a_{12} = -52\) is
(a) 4
(b) -4
(c) -3
(d) 3
Answer: (b) -4
We know that the \(n^{\text{th}}\) term of an AP is given by \(a_n = a + (n - 1)d\).
Therefore, \(a_{25} = a + 24d\) and \(a_{12} = a + 11d\).
Given equation: \(a_{25} - a_{12} = -52\)
\((a + 24d) - (a + 11d) = -52\)
\(13d = -52\)
\(d = \frac{-52}{13} = -4\)
In simple words: Using the formula for the \(n^{\text{th}}\) term, we subtract the two terms to find the common difference \(d\).
Exam Tip: Remember the shortcut \(a_m - a_n = (m - n)d\). Here, \((25 - 12)d = -52 \implies 13d = -52 \implies d = -4\).

(2) If the common difference of an AP is -6, then what is \(a_{16} - a_{12}\)?
(a) -24
(b) 24
(c) -30
(d) 30
Answer: (a) -24
Using the formula \(a_n = a + (n - 1)d\):
\(a_{16} = a + 15d\)
\(a_{12} = a + 11d\)
\(a_{16} - a_{12} = (a + 15d) - (a + 11d) = 4d\)
Substitute \(d = -6\):
\(4(-6) = -24\)
In simple words: The difference between the 16th and 12th terms is equal to 4 times the common difference.

Exam Tip: Apply the general property \(a_m - a_n = (m - n)d\) directly to save time in objective questions.

(3) The \(9^{\text{th}}\) term of an AP is 449 and \(449^{\text{th}}\) term is 9. The term which is equal to zero is
(a) \(501^{\text{th}}\)
(b) \(502^{\text{th}}\)
(c) \(508^{\text{th}}\)
(d) None of the options
Answer: (d) None of the options
Given: \(a_9 = 449\) and \(a_{449} = 9\).
\(a + 8d = 449\) ---- (1)
\(a + 448d = 9\) ---- (2)
Subtracting (1) from (2):
\(440d = 9 - 449 = -440\)
\(d = -1\)
Substitute \(d = -1\) in (1):
\(a + 8(-1) = 449 \implies a - 8 = 449 \implies a = 457\)
We need to find \(n\) such that \(a_n = 0\):
\(a + (n - 1)d = 0\)
\(457 + (n - 1)(-1) = 0\)
\(457 - n + 1 = 0\)
\(n = 458\)
Since \(458^{\text{th}}\) term is 0, and it is not listed in (a), (b), or (c), the correct choice is None of the options.
In simple words: Set up simultaneous equations using the given terms to find the first term and common difference, then find which term equals zero.

Exam Tip: Whenever \(a_m = n\) and \(a_n = m\) in an AP, the \((m+n)^{\text{th}}\) term is always zero. Here \(m=9, n=449\), so \(a_{9+449} = a_{458} = 0\).

(4) In an AP if \(a = -15\), \(a_n = 1\) and \(S_n = -63\), then \(n\) is
(a) 9
(b) 10
(c) 11
(d) 12
Answer: (a) 9
Formula for the sum of \(n\) terms: \(S_n = \frac{n}{2}(a + a_n)\)
Substitute the given values:
\(-63 = \frac{n}{2}(-15 + 1)\)
\(-63 = \frac{n}{2}(-14)\)
\(-63 = -7n\)
\(n = \frac{-63}{-7} = 9\)
In simple words: Use the sum formula that includes the first term and the last term to easily find \(n\).

Exam Tip: Always check if the first and last terms are given before using \(S_n = \frac{n}{2}(a + l)\) as it saves considerable calculation time.

(5) The common difference of an AP whose \(n^{\text{th}}\) term is \(6n + 2\) is
(a) 8
(b) 6
(c) 4
(d) 2
Answer: (b) 6
Given \(a_n = 6n + 2\).
First term \(a_1 = 6(1) + 2 = 8\)
Second term \(a_2 = 6(2) + 2 = 14\)
Common difference \(d = a_2 - a_1 = 14 - 8 = 6\)
In simple words: Find the first two terms by substituting \(n = 1\) and \(n = 2\), then find their difference.

Exam Tip: The coefficient of \(n\) in the expression for the \(n^{\text{th}}\) term of an AP is always equal to its common difference.

(6) Which term of the AP \(18, 23, 28, 33, \dots\) is 98?
(a) \(15^{\text{th}}\)
(b) \(16^{\text{th}}\)
(c) \(17^{\text{th}}\)
(d) \(18^{\text{th}}\)
Answer: (b) \(16^{\text{th}}\)
Here, \(a = 18\), \(d = 23 - 18 = 5\), and \(a_n = 98\).
\(a_n = a + (n - 1)d\)
\(98 = 18 + (n - 1)5\)
\(98 - 18 = 5(n - 1)\)
\(80 = 5(n - 1)\)
\(n - 1 = \frac{80}{5} = 16\)
\(n = 16 + 1 = 17\) - wait, let us recalculate.
\(80 / 5 = 16\), so \(n - 1 = 16 \implies n = 17\). Let us check option (c).
Wait, let's re-verify: \(18 + (16-1)\times 5 = 18 + 15 \times 5 = 18 + 75 = 93\).
For \(n = 17\): \(18 + (17-1)\times 5 = 18 + 16 \times 5 = 18 + 80 = 98\).
Therefore, the correct option is (c) \(17^{\text{th}}\).
In simple words: Substitute \(a\), \(d\), and \(a_n\) into the \(n^{\text{th}}\) term formula to solve for \(n\).

Exam Tip: Double-check your arithmetic when dividing by the common difference.

Question 2. For what value of \(k\) are \(2k\), \(k + 10\) and \(3k + 2\) in AP?
Answer:
If three terms \(a, b, c\) are in AP, then the twice of the middle term is equal to the sum of the outer terms: \(2b = a + c\).
Here, \(a = 2k\), \(b = k + 10\), and \(c = 3k + 2\).
\(2(k + 10) = 2k + (3k + 2)\)
\(2k + 20 = 5k + 2\)
\(20 - 2 = 5k - 2k\)
\(18 = 3k\)
\(k = 6\)
In simple words: Use the property that the middle term is the average of the first and third terms in any arithmetic progression of three numbers.

Exam Tip: Alternatively, you can equate common differences: \((k+10) - 2k = (3k+2) - (k+10)\).

Question 3. If the sum of the first \(p\) terms of an AP is \(ap^2 + bp\), find its common difference.
Answer:
Given sum of first \(p\) terms: \(S_p = ap^2 + bp\)
The first term \(a_1 = S_1 = a(1)^2 + b(1) = a + b\)
The sum of first 2 terms \(S_2 = a(2)^2 + b(2) = 4a + 2b\)
The second term \(a_2 = S_2 - S_1 = (4a + 2b) - (a + b) = 3a + b\)
Common difference \(d = a_2 - a_1 = (3a + b) - (a + b) = 3a + b - a - b = 2a\)
In simple words: Find the first term by putting \(p=1\) and the second term by subtracting \(S_1\) from \(S_2\), then find their difference.

Exam Tip: In general, if \(S_n = An^2 + Bn\), the common difference is always equal to twice the coefficient of \(n^2\) (i.e., \(2A\)).

Question 4. Which term of the AP \(115, 110, 105, \dots\) is its first negative term?
Answer:
Here, first term \(a = 115\) and common difference \(d = 110 - 115 = -5\).
We need to find the first negative term, so we set:
\(a_n < 0\)
\(a + (n - 1)d < 0\)
\(115 + (n - 1)(-5) < 0\)
\(115 - 5n + 5 < 0\)
\(120 - 5n < 0\)
\(120 < 5n\)
\(n > \frac{120}{5}\)
\(n > 24\)
Since \(n\) must be an integer strictly greater than 24, the first negative term is the \(25^{\text{th}}\) term.
In simple words: Set the \(n^{\text{th}}\) term formula to be less than zero and solve for \(n\) to find the position of the first negative number.

Exam Tip: Always take the next integer greater than the decimal/fraction value when finding the position of the first negative term.

Question 5. For what value of \(n\), are the \(n^{\text{th}}\) terms of two APs: \(15, 12, 9, \dots\) and \(-15, -13, -11, \dots\) equal?
Answer:
For the first AP: \(a_1 = 15\), \(d_1 = 12 - 15 = -3\).
\(n^{\text{th}}\) term of the first AP: \(A_n = a_1 + (n - 1)d_1 = 15 + (n - 1)(-3) = 15 - 3n + 3 = 18 - 3n\)
For the second AP: \(a_2 = -15\), \(d_2 = -13 - (-15) = 2\).
\(n^{\text{th}}\) term of the second AP: \(B_n = a_2 + (n - 1)d_2 = -15 + (n - 1)(2) = -15 + 2n - 2 = 2n - 17\)
Equating the two \(n^{\text{th}}\) terms:
\(18 - 3n = 2n - 17\)
\(18 + 17 = 2n + 3n\)
\(35 = 5n\)
\(n = 7\)
In simple words: Write expressions for the \(n^{\text{th}}\) term of both APs separately, equate them, and solve for \(n\).

Exam Tip: Ensure you use the correct signs when finding the common difference, especially with negative numbers.

Question 6. Three numbers whose sum is 21 are in AP. If the product of the first and the third numbers exceeds the second number by 6, find the numbers.
Answer:
Let the three numbers in AP be \(a - d\), \(a\), and \(a + d\).
Given that their sum is 21:
\((a - d) + a + (a + d) = 21\)
\(3a = 21 \implies a = 7\)
Given that the product of the first and the third numbers exceeds the second number by 6:
\((a - d)(a + d) = a + 6\)
\(a^2 - d^2 = a + 6\)
Substitute \(a = 7\):
\(7^2 - d^2 = 7 + 6\)
\(49 - d^2 = 13\)
\(d^2 = 49 - 13 = 36\)
\(d = \pm 6\)
Case 1: When \(a = 7\) and \(d = 6\):
Numbers are \(7 - 6 = 1\), \(7\), \(7 + 6 = 13\) (i.e., \(1, 7, 13\)).
Case 2: When \(a = 7\) and \(d = -6\):
Numbers are \(7 - (-6) = 13\), \(7\), \(7 + (-6) = 1\) (i.e., \(13, 7, 1\)).
In simple words: Assume the three numbers as \(a-d\), \(a\), and \(a+d\) to easily eliminate \(d\) when adding, then solve for \(a\) and \(d\).

Exam Tip: Always use \(a-d\), \(a\), and \(a+d\) for three terms in AP, and \(a-3d\), \(a-d\), \(a+d\), \(a+3d\) for four terms in AP.

Question 7. How many terms of the series \(54, 51, 48, \dots\) be taken so that, their sum is 513? Explain the double answer.
Answer:
Here, \(a = 54\), \(d = 51 - 54 = -3\), and \(S_n = 513\).
Using the sum formula: \(S_n = \frac{n}{2}[2a + (n - 1)d]\)
\(513 = \frac{n}{2}[2(54) + (n - 1)(-3)]\)
\(513 \times 2 = n[108 - 3n + 3]\)
\(1026 = n[111 - 3n]\)
\(1026 = 111n - 3n^2\)
Divide the entire equation by 3:
\(342 = 37n - n^2\)
\(n^2 - 37n + 342 = 0\)
Factorizing the quadratic equation by splitting the middle term (-19 and -18):
\(n^2 - 19n - 18n + 342 = 0\)
\(n(n - 19) - 18(n - 19) = 0\)
\((n - 19)(n - 18) = 0\)
\(n = 18\) or \(n = 19\)
Explanation of the double answer: The sum of 18 terms is 513, and the sum of 19 terms is also 513. This happens because the \(19^{\text{th}}\) term of the AP is \(a_{19} = 54 + (19 - 1)(-3) = 54 - 54 = 0\). Adding zero as the \(19^{\text{th}}\) term does not change the sum.
In simple words: Solve the quadratic equation for \(n\). Two positive values appear because the 19th term of the series is zero.

Exam Tip: Whenever you get a quadratic equation for \(n\) yielding two valid positive integer values, check if the terms starting from the larger index sum to zero.

Question 8. Divide 56 into four parts which are in AP such that the ratio of product of extremes to the product of means is \(5:6\).
Answer:
Let the four parts in AP be \(a - 3d\), \(a - d\), \(a + d\), and \(a + 3d\).
Given that their sum is 56:
\((a - 3d) + (a - d) + (a + d) + (a + 3d) = 56\)
\(4a = 56 \implies a = 14\)
The extremes are \((a - 3d)\) and \((a + 3d)\). Their product is \((a - 3d)(a + 3d) = a^2 - 9d^2\).
The means are \((a - d)\) and \((a + d)\). Their product is \((a - d)(a + d) = a^2 - d^2\).
Given ratio of product of extremes to means is \(5:6\):
\(\frac{a^2 - 9d^2}{a^2 - d^2} = \frac{5}{6}\)
Substitute \(a = 14\), so \(a^2 = 196\):
\(\frac{196 - 9d^2}{196 - d^2} = \frac{5}{6}\)
\(6(196 - 9d^2) = 5(196 - d^2)\)
\(1176 - 54d^2 = 980 - 5d^2\)
\(1176 - 980 = 54d^2 - 5d^2\)
\(196 = 49d^2\)
\(d^2 = \frac{196}{49} = 4\)
\(d = \pm 2\)
Taking \(a = 14\) and \(d = 2\), the four numbers are:
\(14 - 3(2) = 8\)
\(14 - 2 = 12\)
\(14 + 2 = 16\)
\(14 + 3(2) = 20\)
The four parts are \(8, 12, 16, 20\).
In simple words: Assume four symmetrical numbers in AP, find \(a\) from their sum, and then use the given product ratio to find \(d\).

Exam Tip: For four numbers in AP, always use \(a-3d, a-d, a+d, a+3d\) with a common difference of \(2d\).

Question 9. If the \(m^{\text{th}}\) term of an AP is \(n\) and the \(n^{\text{th}}\) term is \(m\), prove that its \(k^{\text{th}}\) term is \((m + n - k)\).
Answer:
Let the first term of the AP be \(A\) and the common difference be \(D\).
Given:
\(a_m = A + (m - 1)D = n\) ---- (1)
\(a_n = A + (n - 1)D = m\) ---- (2)
Subtracting equation (2) from equation (1):
\((m - n)D = n - m\)
\(D = \frac{n - m}{m - n} = -1\)
Substitute \(D = -1\) in equation (1):
\(A + (m - 1)(-1) = n\)
\(A - m + 1 = n\)
\(A = n + m - 1\)
Now, find the \(k^{\text{th}}\) term \(a_k\):
\(a_k = A + (k - 1)D\)
\(a_k = (n + m - 1) + (k - 1)(-1)\)
\(a_k = n + m - 1 - k + 1\)
\(a_k = m + n - k\)
Hence proved.
In simple words: Find the first term and common difference in terms of \(m\) and \(n\), then substitute them into the formula for the \(k^{\text{th}}\) term.

Exam Tip: This is a standard and very important textbook proof. Memorize the steps of finding \(D\) first by subtracting the two given equations.

Question 10. Determine \(k\) so that \(k^2 + 4k + 8\), \(2k^2 + 3k + 6\), \(3k^2 + 4k + 4\) are the three consecutive terms of an AP.
Answer:
Let \(a = k^2 + 4k + 8\), \(b = 2k^2 + 3k + 6\), and \(c = 3k^2 + 4k + 4\).
Since they are consecutive terms of an AP, \(2b = a + c\).
\(2(2k^2 + 3k + 6) = (k^2 + 4k + 8) + (3k^2 + 4k + 4)\)
\(4k^2 + 6k + 12 = 4k^2 + 8k + 12\)
Subtracting \(4k^2 + 12\) from both sides:
\(6k = 8k\)
\(8k - 6k = 0\)
\(2k = 0 \implies k = 0\)
In simple words: Use the property that twice the middle term equals the sum of the first and third terms to form an equation in \(k\).

Exam Tip: Be careful while simplifying terms on both sides of the equation; quadratic terms \(4k^2\) cancel out here, making it a simple linear equation.

Question 11. If \(S_n\) denotes the sum of first \(n\) terms of an AP, prove that \(S_{12} = 3(S_8 - S_4)\).
Answer:
Let the first term be \(a\) and common difference be \(d\).
Sum of first \(n\) terms formula: \(S_n = \frac{n}{2}[2a + (n - 1)d]\)
First, find \(S_{12}\):
\(S_{12} = \frac{12}{2}[2a + (12 - 1)d] = 6(2a + 11d) = 12a + 66d\) ---- (1)
Next, find \(S_8\) and \(S_4\):
\(S_8 = \frac{8}{2}[2a + (8 - 1)d] = 4(2a + 7d) = 8a + 28d\)
\(S_4 = \frac{4}{2}[2a + (4 - 1)d] = 2(2a + 3d) = 4a + 6d\)
Now, find \(S_8 - S_4\):
\(S_8 - S_4 = (8a + 28d) - (4a + 6d) = 4a + 22d\)
Multiply by 3:
\(3(S_8 - S_4) = 3(4a + 22d) = 12a + 66d\) ---- (2)
From (1) and (2), \(S_{12} = 3(S_8 - S_4)\).
Hence proved.
In simple words: Expand \(S_{12}\), \(S_8\), and \(S_4\) separately using the sum formula and verify that both sides simplify to the same expression.

Exam Tip: Alternatively, use the property that the sum of terms from \((m+1)^{\text{th}}\) to \(n^{\text{th}}\) is \(S_n - S_m\).

Question 12. The ratio of \(11^{\text{th}}\) term to the \(18^{\text{th}}\) term of an AP is \(2:3\). Find the ratio of the \(5^{\text{th}}\) term to the \(21^{\text{st}}\) term, and also the ratio of the sum of first five terms to the sum of the first 21 terms.
Answer:
Let first term be \(a\) and common difference be \(d\).
Given ratio:
\(\frac{a_{11}}{a_{18}} = \frac{2}{3}\)
\(\frac{a + 10d}{a + 17d} = \frac{2}{3}\)
\(3(a + 10d) = 2(a + 17d)\)
\(3a + 30d = 2a + 34d\)
\(a = 4d\) ---- (1)
Now, find the ratio of the \(5^{\text{th}}\) term to the \(21^{\text{st}}\) term:
\(\frac{a_5}{a_{21}} = \frac{a + 4d}{a + 20d}\)
Substitute \(a = 4d\):
\(\frac{4d + 4d}{4d + 20d} = \frac{8d}{24d} = \frac{1}{3}\) (Ratio is \(1:3\))
Next, find the ratio of the sum of first 5 terms to the sum of first 21 terms:
\(\frac{S_5}{S_{21}} = \frac{\frac{5}{2}[2a + (5 - 1)d]}{\frac{21}{2}[2a + (21 - 1)d]} = \frac{5(2a + 4d)}{21(2a + 20d)} = \frac{5(a + 2d)}{21(a + 10d)}\)
Substitute \(a = 4d\):
\(\frac{5(4d + 2d)}{21(4d + 10d)} = \frac{5(6d)}{21(14d)} = \frac{30}{294} = \frac{5}{49}\) (Ratio is \(5:49\))
In simple words: Express \(a\) in terms of \(d\) using the given ratio of terms, then substitute it into the required ratio formulas.

Exam Tip: Be careful with the factor of 2 when converting sums to simpler bracket terms: \(\frac{n}{2}[2a+(n-1)d]\).

Question 13. Reshma repays her total loan of Rs. 1,75,000 by paying every month starting with the first installment of Rs. 1500. If she increases the installment by Rs. 175 every month, what amount will be paid by her in the \(21^{\text{st}}\) installment? What amount of loan does she still have to pay after the \(21^{\text{st}}\) installment?
Answer:
Here, the installments form an AP where:
First installment \(a = 1500\)
Increase per month (common difference) \(d = 175\)
Total loan = Rs. 1,75,000
Part 1: Amount paid in the \(21^{\text{st}}\) installment (\(a_{21}\)):
\(a_{21} = a + (21 - 1)d = 1500 + 20(175)\)
\(a_{21} = 1500 + 3500 = 5000\)
So, the \(21^{\text{st}}\) installment is Rs. 5000.
Part 2: Total amount paid in 21 installments (\(S_{21}\)):
\(S_{21} = \frac{21}{2}[2a + (21 - 1)d] = \frac{21}{2}[2(1500) + 20(175)]\)
\(S_{21} = \frac{21}{2}[3000 + 3500] = \frac{21}{2}[6500] = 21 \times 3250 = 68250\)
Total amount paid = Rs. 68,250.
Remaining loan amount after 21 installments:
Remaining loan = Total loan - Total amount paid
Remaining loan = \(175000 - 68250 = 106750\)
So, she still has to pay Rs. 1,06,750.
In simple words: Find the 21st term for the last installment and the sum of 21 terms for the total paid amount, then subtract from the total loan.

Exam Tip: Clearly label both parts of your answer so that no sub-question is missed during evaluation.

Question 14. The sum of the first \(n\) terms of an AP is given by \(S_n = 3n^2 - 4n\). Determine the AP and the \(12^{\text{th}}\) term.
Answer:
Given sum of first \(n\) terms: \(S_n = 3n^2 - 4n\)
First term \(a_1 = S_1 = 3(1)^2 - 4(1) = 3 - 4 = -1\)
Sum of first 2 terms \(S_2 = 3(2)^2 - 4(2) = 3(4) - 8 = 12 - 8 = 4\)
Second term \(a_2 = S_2 - S_1 = 4 - (-1) = 4 + 1 = 5\)
Common difference \(d = a_2 - a_1 = 5 - (-1) = 5 + 1 = 6\)
Therefore, the AP is \(-1, 5, 11, 17, \dots\)
Now, find the \(12^{\text{th}}\) term (\(a_{12}\)):
\(a_{12} = a + 11d = -1 + 11(6) = -1 + 66 = 65\)
Alternatively, \(a_{12} = S_{12} - S_{11}\):
\(S_{12} = 3(12)^2 - 4(12) = 3(144) - 48 = 432 - 48 = 384\)
\(S_{11} = 3(11)^2 - 4(11) = 3(121) - 44 = 363 - 44 = 319\)
\(a_{12} = 384 - 319 = 65\)
In simple words: Find the first term as \(S_1\) and the second term as \(S_2 - S_1\), then construct the AP and calculate the 12th term.

Exam Tip: You can verify your AP using the \(n^{\text{th}}\) term formula \(a_n = S_n - S_{n-1}\).

Question 15. If the \(m^{\text{th}}\) term of an AP is \(\frac{1}{n}\) and \(n^{\text{th}}\) term is \(\frac{1}{m}\), then show that its \((mn)^{\text{th}}\) term is 1.
Answer:
Let the first term of the AP be \(A\) and the common difference be \(D\).
Given:
\(a_m = A + (m - 1)D = \frac{1}{n}\) ---- (1)
\(a_n = A + (n - 1)D = \frac{1}{m}\) ---- (2)
Subtracting equation (2) from equation (1):
\((m - n)D = \frac{1}{n} - \frac{1}{m}\)
\((m - n)D = \frac{m - n}{mn}\)
\(D = \frac{1}{mn}\)
Substitute \(D = \frac{1}{mn}\) into equation (1):
\(A + (m - 1)\frac{1}{mn} = \frac{1}{n}\)
\(A + \frac{m}{mn} - \frac{1}{mn} = \frac{1}{n}\)
\(A + \frac{1}{n} - \frac{1}{mn} = \frac{1}{n}\)
\(A = \frac{1}{mn}\)
Now, find the \((mn)^{\text{th}}\) term \(a_{mn}\):
\(a_{mn} = A + (mn - 1)D\)
\(a_{mn} = \frac{1}{mn} + (mn - 1)\left(\frac{1}{mn}\right)\)
\(a_{mn} = \frac{1}{mn} + 1 - \frac{1}{mn} = 1\)
Hence proved.
In simple words: Find \(A\) and \(D\) by solving the simultaneous equations formed by the given terms, then substitute them into the formula for the \((mn)^{\text{th}}\) term.

Exam Tip: This is an important standard proof. Take care with fraction subtractions and common denominators.

Question 16. If the sum of \(m\) terms of an AP is the same as the sum of its \(n\) terms, show that the sum of its \((m + n)\) terms is zero.
Answer:
Let the first term be \(a\) and common difference be \(d\).
Given: \(S_m = S_n\)
\(\frac{m}{2}[2a + (m - 1)d] = \frac{n}{2}[2a + (n - 1)d]\)
Multiply both sides by 2 and expand:
\(m[2a + (m - 1)d] = n[2a + (n - 1)d]\)
\(2am + m(m - 1)d = 2an + n(n - 1)d\)
Bring all terms to one side:
\(2am - 2an + [m(m - 1) - n(n - 1)]d = 0\)
\(2a(m - n) + [(m^2 - m) - (n^2 - n)]d = 0\)
\(2a(m - n) + [(m^2 - n^2) - (m - n)]d = 0\)
Factorize \((m^2 - n^2)\) as \((m - n)(m + n)\):
\(2a(m - n) + [(m - n)(m + n) - (m - n)]d = 0\)
Take \((m - n)\) common out:
\((m - n)[2a + (m + n - 1)d] = 0\)
Since \(m \neq n\), we can divide by \((m - n)\):
\(2a + (m + n - 1)d = 0\) ---- (1)
Now, the sum of \((m + n)\) terms is:
\(S_{m+n} = \frac{m + n}{2}[2a + (m + n - 1)d]\)
Using equation (1), substitute \(2a + (m + n - 1)d = 0\):
\(S_{m+n} = \frac{m + n}{2}(0) = 0\)
Hence proved.
In simple words: Equate \(S_m\) and \(S_n\), simplify the algebraic expression, factor out \((m-n)\), and substitute the resulting relation into the formula for \(S_{m+n}\).

Exam Tip: Grouping terms containing \(2a\) together and terms containing \(d\) together is the key step to factor out \((m-n)\).

Question 17. The ratio of the sums of \(m\) and \(n\) terms of an AP is \(m^2 : n^2\). Show that the ratio of the \(m^{\text{th}}\) and \(n^{\text{th}}\) terms is \((2m - 1) : (2n - 1)\).
Answer:
Let first term be \(a\) and common difference be \(d\).
Given ratio of sums:
\(\frac{S_m}{S_n} = \frac{m^2}{n^2}\)
\(\frac{\frac{m}{2}[2a + (m - 1)d]}{\frac{n}{2}[2a + (n - 1)d]} = \frac{m^2}{n^2}\)
Cancel \(\frac{m}{n}\) from both sides:
\(\frac{2a + (m - 1)d}{2a + (n - 1)d} = \frac{m}{n}\) ---- (1)
We need to find the ratio of the \(m^{\text{th}}\) to the \(n^{\text{th}}\) term:
\(\frac{a_m}{a_n} = \frac{a + (m - 1)d}{a + (n - 1)d}\)
Multiply numerator and denominator of the required ratio by 2:
\(\frac{a_m}{a_n} = \frac{2a + 2(m - 1)d}{2a + 2(n - 1)d} = \frac{2a + (2m - 2)d}{2a + (2n - 2)d}\)
Wait, let's use the property that in any AP, the ratio of sums of \(2k-1\) terms or using the substitution \(m \to 2m-1\):
From equation (1), replace \(m\) by \(2m - 1\) and \(n\) by \(2n - 1\):
\(\frac{2a + (2m - 1 - 1)d}{2a + (2n - 1 - 1)d} = \frac{2m - 1}{2n - 1}\)
\(\frac{2a + (2m - 2)d}{2a + (2n - 2)d} = \frac{2m - 1}{2n - 1}\)
\(\frac{a + (m - 1)d}{a + (n - 1)d} = \frac{2m - 1}{2n - 1}\)
\(\frac{a_m}{a_n} = \frac{2m - 1}{2n - 1}\)
Hence proved.
In simple words: Simplify the ratio of sums to relate \(2a+(m-1)d\) with \(m\), then substitute \(m\) with \(2m-1\) to get the ratio of terms.

Exam Tip: This is a standard property-based question. Remember to multiply the numerator and denominator by 2 inside the sum formula to convert it into terms.

Question 18. The sum of \(n\), \(2n\), \(3n\) terms of an AP are \(S_1\), \(S_2\) and \(S_3\) respectively. Prove that \(S_3 = 3(S_2 - S_1)\).
Answer:
Let the first term of the AP be \(a\) and the common difference be \(d\).
Given:
\(S_1\) is the sum of \(n\) terms:
\(S_1 = \frac{n}{2}[2a + (n - 1)d]\)
\(S_2\) is the sum of \(2n\) terms:
\(S_2 = \frac{2n}{2}[2a + (2n - 1)d] = n[2a + (2n - 1)d]\)
\(S_3\) is the sum of \(3n\) terms:
\(S_3 = \frac{3n}{2}[2a + (3n - 1)d]\)
Now, consider the right-hand side expression \(3(S_2 - S_1)\):
\(S_2 - S_1 = n[2a + (2n - 1)d] - \frac{n}{2}[2a + (n - 1)d]\)
Take \(\frac{n}{2}\) common out:
\(S_2 - S_1 = \frac{n}{2} \left[ 2(2a + (2n - 1)d) - (2a + (n - 1)d) \right]\)
\(S_2 - S_1 = \frac{n}{2} \left[ 4a + 2(2n - 1)d - 2a - (n - 1)d \right]\)
\(S_2 - S_1 = \frac{n}{2} \left[ 2a + (4n - 2 - n + 1)d \right]\)
\(S_2 - S_1 = \frac{n}{2} \left[ 2a + (3n - 1)d \right]\)
Multiply by 3:
\(3(S_2 - S_1) = \frac{3n}{2} [2a + (3n - 1)d]\)
From the above, this is exactly equal to \(S_3\).
Therefore, \(S_3 = 3(S_2 - S_1)\).
Hence proved.
In simple words: Write formulas for \(S_1\), \(S_2\), and \(S_3\), simplify \(S_2 - S_1\) by taking common factors, and multiply by 3 to match \(S_3\).

Exam Tip: Factoring out \(\frac{n}{2}\) makes the subtraction inside the brackets much simpler and less prone to calculation errors.