(a) 4 (b) 5 (c) 3 (d) 2 Answer: (c) 3 We have: a5 = 20 and a7 + a11 = 64. Let a be the first term and d be the common difference of the AP. Then, Thus, the common difference of the AP is...

(a) 50 (b) -50 (c) 30 (d) -30 Answer: (b) -50

(a) (3n + 8) (b) (4n – 7) (c) (15 – 2n) (d) (2n – 15) Answer: (d) (2n – 15)

(a) (6n - 2) (b) (7n – 3) (c) (8n – 2) (d) (8n + 2) Answer: (c) (8n – 2)

(a) ( 5 - 2n) (b) ( 6 – 2n) (c) (2n – 5) (d) (2n – 6) Answer: (b) ( 6 – 2n)

(a) 6                  (b) 9               (c) 15                  (d) -3 Solution: (a) b

(a) 6n+3 (b) 15 (c) 12 (d) 21 Solution: (b) 15

(a) 19 (b) 23 (c) 22 (d) cannot be determined Solution: (c) 22

$ (a)\,\sqrt{84} $ $ (b)\,\sqrt{98} $ $ (c)\,\sqrt{70} $ $ (d)\,\sqrt{112} $ Solution:  $ (d)\,\sqrt{112} $ The AP is $ \sqrt{7},\sqrt{28},\sqrt{63},...... $ $ =\sqrt{7},\sqrt{4\times...

(a)1/3 (b)-1/3 (c) b (d) –b Solution: (d) -b

(a) p (b) –p (c) -1 (d) 1 Solution: (c) -1  

Solution: Because the punishment for each subsequent day is 50 more than the previous day, the total number of penalties is in AP with a common difference of ₹50. The number of days the job has been...

Solution: Let the first installment's value be ₹a. Let us assume that the man increases the value of each installment by ₹d every month because the monthly installments constitute an arithmetic...

Solution: Total savings = ₹ 33000 Total period = 10 months Every month, a man saved $100 more than the month before. Suppose he saves x in the first month, then:

Solution: Total sum = ₹ 700 Number of cash prizes = 7 Each prize in ₹ 20 less than its preceding prize. Let first prize = ₹ x Then second prize = ₹ (x – 20) Third prize = ₹ (x – 40) and so...

Solution: There are a total of 25 trees in this area. In a straight line, the distance between them is 5 metres. A water tank can be found 10 metres from the first tree. These plants are watered...

Solution: Bucket is 5 metres from the first potato in a potato race, and the distance between the two potatoes is 3 metres. There are ten potatoes in total. Pick one potato at a time and place it in...

Solution: According to the question, in the school, there are 12 classes and classes 1 to 12 and each class has two sections. Each class plants double of the class i.e. class 1 plants two plants,...

Solution: Suppose that a is the first term and d is the common difference of an AP, then we have:

Solution: S14 = 1505 Suppose that a is the first term and d is the common difference, then a = 10

Solution: The AP is -12, -9, -6,…, 21 Here, a = -12, d = -9 – (-12) = -9 + 12 = 3, l = 21

Solution:

Solution:

Solution: Suppose that a is the first term and d is the common difference of an AP. Then, we have:

Solution:

Solution: The AP is 5, 12, 19,… to 50 terms Here, a = 5, d = 12 – 5 = 7, n = 50

Solution: Suppose that a is the first term and d is the common difference of an AP, then we have:

Solution: Suppose that a is the first term and d is the common difference, then we have:

Solution: Suppose that a is the first term and d is the common difference, then we have: S10 = -150

Solution: Suppose that a1 and a2 are the first terms of the two APs and d is the common difference, then we have: a1 = 3, a2 = 8

Solution: Suppose that a is the first term and d is the common difference of an AP, then we have:

Solution: Suppose the first term is a and d is the common difference in an AP, then we have:

Solution: Suppose that a is the first term and d is the common difference, then we have:

Solution: Suppose that a is the first term and d is the common difference of an AP, then we have:

Solution: In an AP, let a be the first term and d be the common difference, then

Solution: Suppose that a is the first term, d is the common difference, then we have:  

Solution: In an AP a = 17, d = 9, l = 350 Let number of terms be n, then

Solution: Suppose that a is the first term and d is the common difference, then we have:

Solution: Suppose that a is the first term and d is the common difference, then we have: Here, a = 2, l = 29 and Sn = 155 We know that: l = a + (n-1)d                             (1) (n - 1)d = 29 -...

Solution: Suppose that a is the first term and d is the common difference of the AP, then we have:

Solution:

Solution: In this case, the even natural numbers are 2, 4, 6, 8, 10, … And the even natural numbers which are divisible by 5 will be10, 20, 30, 40, … to 100 terms Here, a = 10, d = 20 – 10 = 10, n =...

Solution: The three digit numbers are 100, 101, …, 999 and the numbers divisible by 13, will be 104, 117, 130, …, 988 Here, a = 104, d = 13, l = 988 Tn (l) = a + (n – 1) d

Solution: The multiples of 9 lying between 300 and 700 are 306, 315, 324, 333, …, 693 Here, a = 306, d = 9, l = 693

Solution: The first forty positive integers are 0, 1, 2, 3, 4, … and the numbers divisible by 6 will be 6, 12, 18, 24, … to 40 terms Here, a = 6, d = 12 – 6 = 6, n = 40. Then we have:

Solution: The numbers between 200 and 400 which are divisible by 7 will be 203, 210, 217,…, 399 Here, a = 203, d = 7, l = 399

Solution: the odd numbers between 0 and 50 are 1, 3, 5, 7, 9, …, 49 Here, a = 1 and the common difference d = 3 – 1 = 2, l = 49 = 25/2 x 50 = 25 x 25 = 625

Solution;

Solution: The AP here is 63, 60, 57, 54,… Here, a = 63 and d = 60 – 63 = -3 Also, the sum = 693 Let number of terms be n, then we can say that: Therefore, the 22nd term is zero. There will be no...

Solution: The given AP is 9, 17, 25,… Here, a = 9 and d = 17 – 9 = 8 We are given that the Sum of terms = 636 Let number of terms be n, then we can write:

Solution: Suppose that a is the first term and d is the common difference of an AP. Since, we have: am = a + (m – 1) d = 1/n …(i)

Solution:

Solution:

Solution: We have, Sn = 3n² – n For n = 1, S1 = 3(1)² – 1 = 3 – 1 = 2 For n = 2, we have: S2 = 3(2)² – 2 = 12 – 2 = 10 T2 = 10 – 2 = 8 and T1 = 2 (i) Tn = a + (n – 1) d = 2 + (n – 1) x 6 = 2 + 6n –...

Solution:

Solution:

Solution;

Solution:

Solution:

Solution: The given AP is -37, -33, -29,…….. Here, a = - 37 and d = -33 - ( -37) = - 33 + 37 = 4

Solution:

Solution: Let a be the first term and d be the common difference of the AP. Then,

Solution:

Solution:

Solution: (2p – 1), 7, 3p are in AP, then ⇒ 7 – (2p – 1) = 3p – 7 ⇒ 7 – 2p + 1 = 3p – 7 ⇒ 7 + 1 + 7 = 3p + 2p ⇒ 5p = 15 ⇒ p = 3 P = 3

Solutions: 2p + 1, 13, 5p – 3 are in AP, then 13 – (2p + 1) = (5p – 3) – 13 ⇒ 13 – 2p – 1 = 5p – 3 – 13 ⇒ 12 – 2p = 5p – 16 ⇒ 5p + 2p = 12 + 16 ⇒ 7p = 28 ⇒ p = 4 P =...

Solution: If 4/5, a and 2 are three consecutive terms of an AP, then we have:

Solution: In an AP First term (a) = p and the common difference (d) = q T10 = a + (n – 1) d T10 = p + (10 – 1) x q T10 = (p + 9q)

Solution: The first n even natural numbers are 2, 4, 6, 8, 10, …., n. Here, a = 2 and d = 4 - 2 = 2 The Sum of n terms of an AP is given by:

Solution: In this case, the first n natural numbers are 1, 2, 3, 4, 5,……..,n Here, a = 1 and d = (2 – 1) = 1 Sum of the n terms of an AP is given by:

Solution: AP is 21, 18, 15, …n Here, a = 21, d = 18 – 21 = -3, l = 0 Tn (l) = a + (n – 1) d ⇒ 0 = 21 + (n – 1) x (-3) ⇒ 0 = 21 – 3n + 3 ⇒ 24 – 3n = 0 ⇒ 3n = 24 ⇒ n = 8 . 0 is the 8th...

Solution:

Solution: AP is √8, √18, √32, ….. ⇒ √(4 x 2) , √(9 x 2) , √(16 x 2), ………

Solution: Tn = 7 – 4n Tn-1 = 7 – 4(n – 1) = 7 – 4n + 4 = 11 – 4n d = Tn – Tn-1  = (7 – 4n) – (11 – 4n) = 7 – 4n – 11 + 4n = -4 d = -4

Solution: Tn = 3n + 5 Tn-1 = 3 (n – 1) + 5 = 3n – 3 + 5 = 3n + 2 d = Tn – Tn-1 = (3n + 5) – (3n + 2) = 3n + 5 – 3n – 2 = 3 Common difference = 3

Solution: AP is 2, 7, 12, 17, … Here, a = 2, d = 7 – 2 = 5 an = a + (n – 1) d = 2 + (n – 1) x 5 = 2 + 5n – 5 = 5n – 3 Now, a30 = 2 + (30 – 1) x 5 = 2 + 29 x 5 = 2 + 145 = 147 and a20 = 2 + (20 – 1)...

Solution: The AP is 2, 7, 12, 17, …… 47 Here, a = 2, d = 7 – 2 = 5, l = 47 nth term from the end = l – (n – 1 )d 5th term from the end = 47 – (5 – 1) x 5 = 47 – 4 x 5 = 47 – 20 =...

Solution: The AP given here is a, 3a, 5a, … Here, a = a, d = 2a

Solution: Sum of first m terms of an AP = 2m² + 3m Sm = 2m² + 3m S1 = 2(1)² + 3 x 1 = 2 + 3 = 5 S2 = 2(2)² + 3 x 2 = 8 + 6=14 S3 = 2(3)² + 3 x 3 = 18 + 9 = 27 Now, T2 = S2 – S1 = 14 – 5 = 9 Second...

Solution: Three digit numbers in this case are 100 to 999 and the numbers which are divisible by 9 will be given as: 108, 117, 126, 135, …, 999 Here, a = 108, d= 9, l = 999 Tn (l) = a + (n – 1) d ⇒...

Solution: Three digit numbers here are 100 to 990 and the numbers which are divisible by 7 will be then given by: 105, 112, 119, 126, …, 994 Here, a = 105, d= 7, l = 994 Tn = (l) = a + (n – 1) d ⇒...

Solution: (2n – 1), (3n + 2) and (6n – 1) are in AP We have: (3n + 2) – (2n – 1) = (6n – 1) – (3n + 2) ⇒ (3n + 2) + (3n + 2) = 6n – 1 + 2n – 1 ⇒ 6n + 4 = 8n – 2 ⇒ 8n – 6n = 4 + 2 ⇒ 2n = 6 ⇒ n = 3...

Solution: a, 9, b, 25 are in AP. We have, 9 – a = b – 9 = 25 – b First case: b – 9 = 25 – b ⇒ b + b = 22 + 9 = 34 ⇒ 2b = 34 ⇒ b= 17 Using, a – b = a – 9 ⇒ 9 + 9 = a + b ⇒ a + b =...

Solution: 18, a, (b – 3) are in AP ⇒ a – 18 = b – 3 – a ⇒ a + a – b = -3 + 18 ⇒ 2a – b = 15

Solution: k, (2k – 1) and (2k + 1) are the three successive terms of an AP. (2k – 1) – k = (2k + 1) – (2k – 1) ⇒ 2 (2k – 1) = 2k + 1 + k ⇒ 4k – 2 = 3k + 1 ⇒ 4k – 3k = 1 + 2 ⇒ k = 3 k =...

Solution: (3y – 1), (3y + 5) and (5y+ 1) are in AP (3y + 5) – (3y – 1) = (5y + 1) – (3y + 5) ⇒ 2 (3y + 5) = (5y + 1) + (3y – 1) ⇒ 6y + 10 = 8y ⇒ 8y – 6y = 10 ⇒ 2y = 10 ⇒ y = 5 y =...

solution; Let the three terms be a – d, a, a + d a – d + a + a + d = 48 ⇒ 3a = 48 ⇒ a = 16 and (a – d) x a = (a + d) + 12 ⇒ a(a – d) = 4 (a + d) + 12 ⇒ 16 (16 – d) = 4(16 + d) + 12 ⇒ 256 – 16d = 64...

Solution: Let the four parts of 32 be a – 3d, a – d, a + d, a + 3d

Solution: Let the four numbers in AP be a – 3d, a – d, a + d, a + 3d, then a – 3d + a – d + a + d + a + 3d = 28

Solution; Sum of angles of a quadrilateral = 360° Let d= 10 The first number be a, then the four numbers will be a, a + 10, a + 20, a + 30 a + a + 10 + a + 20 + a + 30 = 360 4a + 60 = 360 4a = 360 –...

Solution: Let three consecutive in AP be a – d, a, a + d a – d + a + a + d = 21 ⇒ 3a = 21

Solution; Sum of three numbers = 24 Let the three numbers in AP be a – d, a, a + d

Solution: Let the three numbers in AP be a – d, a and a + d

Solution; Let the three numbers in AP be a – d, a, a + d a – d + a + a + d = 15 ⇒ 3a = 15 ⇒ a = 5 and (a – d) x a x (a + d) = 80 a(a² – d²) = 80 ⇒ 5(5² – d²) = 80 ⇒ 25 – d² = 16 ⇒ d² = 25 – 16 = 9 =...

Solution; (a – b)², (a² + b²) and (a + b)² will be in AP. If (a² + b²) – (a – b)² = (a + b)² – (a² + b²) If (a² + b²) – (a² + b² – 2ab) = a² + b² + 2ab – a² – b² 2ab = 2ab which is true. Hence...

Solution: (x + 2), 2x, (2x + 3) are three consecutive terms of an AP. 2x – (x + 2) = (2x + 3) – 2x ⇒ 2x – x – 2 = 2x + 3 – 2x ⇒ x – 2 = 3 ⇒ x = 2 + 3 = 5 x = 5

Solution: (3y – 1), (3y + 5) and (5y + 1) are the three consecutive terms of an AP. (3y + 5) – (3y – 1) – (5y + 1) – (3y + 5) ⇒ 2(3y + 5) = 5y + 1 + 3y – 1 ⇒ 6y + 10 = 8y ⇒ 8y – 6y = 10 ⇒ 2y = 10 ⇒...

Solution: (5x + 2), (4x – 1) and (x + 2) are in AP. (4x – 1) – (5x + 2) = (x + 2) – (4x – 1) ⇒ 2(4x – 1) = (x + 2) + (5x + 2) ⇒ 8x – 2 = 6x + 2 + 2 ⇒ 8x – 2 = 6x + 4 ⇒ 8x – 6x = 4 + 2 ⇒ 2x = 6 x =...

Solution: (3k – 2), (4k – 6) and (k + 2) are three consecutive terms of an AP. (4k – 6) – (3k – 2) = (k + 2) – (4k – 6) ⇒ 2(4k – 6) = (k + 2) + (3k – 2) ⇒ 8k – 12 = 4k + 0 ⇒ 8k – 4k = 0 + 12 ⇒ 4k =...

Solution: Total amount = ₹ 2800 and number of prizes = 4 Let first prize = ₹ a Then second prize = ₹ a – 200 Third prize = a – 200 – 200 = a – 400 and fourth prize = a – 400 – 200 = a – 600 But sum...

Solution: The numbers of rose plants in consecutive rows are 43, 41, 39,..., 11. Difference of rose plants between two consecutive rows = (41 — 43) = (39 - 41) = - 2 [Constant] So, the given...

Solution: Numbers that are divisible by 2 and 5 are likewise divisible by 10. Now, the numbers divisible by 10 integers between 101 and 999 are 110, 120, 130,..., 990. Clearly, these figures are in...

Solution: The three-digit numbers divisible by 9 are 108, 117, 126,...., 999. Clearly, these number are in AP. Here. a= 108 and d= 117 — 108 = 9 Let this AP contains n terms. Then. an = 999 =.108+(n...

Solution; The two-digit numbers divisible by 3 are 12, 15, 18, ..., 99. Clearly, these number are in AP. Here, a = 12 and d =15 —12 = 3 Let this AP contains n terms. Then, an = 99 12 + (n - 1) x...

Solution: The two-digit numbers divisible by 6 are 12, 18, 24,……, 96. Clearly, these numbers are in AP. Here, a = 12 and d = 18 - 12 = 6 Let this AP contains n terms. Then, we have: $ {{a}_{n}}=96 $...

Solution: In the given AP, first term = a and last term = l. Let the common difference be d. Then, nth term from the beginning is given by: Hence, the sum of the nth term from the beginning and the...

Solution; In the given AP, let the first be a and the common difference be d.

Solution; Let a be the first term and d be the common difference of the AP. Then,

Solution: Suppose that a is the first term and d is the common difference of the AP. Then, we have: $ \frac{{{a}_{72}}}{{{a}_{15}}}=\frac{5d+71d}{5d+14d} $ $...

Solution: Suppose that a is the first term and d is the common difference of the AP. Then, we have: $ {{a}_{17}}=2{{a}_{8}}+5 $ $ a+16d=2(a+7d)+5 $ $ a+16d=2a+14d+5 $ $...

Solution; Let the term of the given progressions be tn and Tn respectively. The given AP is 63, 65, 67,... Let the first term be a and the common difference be d. Then a = 63 and d = (65 - 63) = 2...

Solution: Let a be the first term and d be the common difference of the AP. Then, $ {{a}_{2}}+{{a}_{7}}=30 $ $ (a+d)+(a+6d)=30 $ $ 2a+7d=30\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)...

Solution: Suppose that the common difference of the AP is d. We have the first term, a = 5 Now, we have: $...

Solution: Suppose that a is the first term and d is the common difference of the AP. Then, we have: $ a+24d=0 $ $ a+(25-1)d=0 $ $ {{a}_{25}}=0 $ $ Hence,\text{ }the\text{ }25th\text{ }term\text{...

Solution: Suppose that a is the first term and d is the common difference of the AP. Then, we have: $4\times {{a}_{4}}=18\times {{a}_{18}}$ Hence, the 22nd term of the AP is 0.

Solution: Suppose that a is the first term and d is the common difference of the AP. Then, we can say that: $ {{a}_{7}}=-1 $ $ a+(7-1)d=-1 $ $...

Solution: Suppose that a is the first term and d is the common difference of the AP. Then, we have: $ {{a}_{4}}=11 $ $ a+(4-1)d=11 $ $ a+3d=11\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1) $ $...

Solution: Suppose that a is the first term and d is the common difference of the AP. Then, we can write that: Hence, the 38th term of the AP is triple its 18th term.

Solution; In the given AP, suppose that the first term is a and the common difference is d. Then, Tn = a + (n-1)d

Solution:

Solution: $ The\,given\,AP\,is: $ $ 20,19\frac{1}{4},18\frac{1}{2},17\frac{3}{4},.......... $ $ Here,\,a=20\,\,and\,\,d=19\frac{1}{4}-20=-\frac{3}{4} $ $ Let\text{ }the\text{ }nth\text{ }term\text{...

Solution: The given AP is 121, 117, 113, ……. Here, a = 121 and d = 117 - 121 = - 4 Suppose that the nth term of the given AP is the first negative term. Then, we can say that: Hence, it is pretty...

Solution:

The given AP is 3,7,11,15,...... Here, we have a = 3 and d = 7 - 3 = 4 Let the nth term of the given AP be 184. Then, we can write: $ {{a}_{n}}=184 $ $ USING:\,\,{{a}_{n}}=a+(n-1)d $ $ 3+(n-1)\times...

Solution:

Solution:

Solution: The given AP is: 10,7,4,....., -62 So, the first term is a = 10 And the common difference is d = 7 - 10 = -3 Suppose that there are n terms in the given AP. Then, we can write: Thus, we...

Solution: The given AP is 6, 13, 20,...........216 First term, a = 6 and the common difference d = 13 - 6 = 7 Assuming that there are n terms in the AP, we can write: $ {{a}_{n}}=216 $ $...

Solution: In the given AP, let the first term be a and the common difference be d. $ Then,\,{{T}_{n}}=a+(n-1)d $ $ Now.\,we\,have: $ $ {{T}_{10}}=a+(n-1)d=a+9d $ $...

Solution:

Solution: In the given AP, we have a = 3 and d = 8 - 3 = 5 The 20th term is given by: $ {{T}_{20}}=a+(20-1)d=a+19d $ $ {{T}_{20}}=3+19\times 5=98 $ $ \therefore the\,required\,term=98+55=153 $ $...

Solution:

Solution: $ The\,given\,AP\,is: $ $ \frac{5}{6},1,1\frac{1}{6},1\frac{1}{3},...... $ $ where\,\,a=\frac{5}{6}\,\,and\,\,d=\left( 1-\frac{5}{6} \right)=\frac{1}{6} $ $...

Solution: In the given AP, we have: a = 72 and the common difference, d = 68 - 72 = -4

Solution: In the given AP, we have: the first term, a = 3 and the common difference, d = 8 - 3 = 5

Solution: $ The\text{ }given\text{ }AP\text{ }is:\, $ $ 18,15\frac{1}{2},13,.......,-47 $ $ First\,term,\,a=18 $ $ Common\,difference,\,d=15\frac{1}{2}-18=-\frac{5}{2} $ $ Suppose\text{ }there\text{...

Solution:

Solution: In the given Arithmetic Progression. a = 6 and d = 10 - 6 = 4 Let us assume that there are n terms in the given AP. Then, $ {{T}_{n}}=174 $ $ \Rightarrow a+(n-1)d=174 $ $ 6+(n-1)d=174 $ $...

Solution: $ {{T}_{n}}=\left( {{4}_{n}}-10 \right)\,\,\,\,\,\,\,\,\,\,\,\,[Given] $ $ {{T}_{1}}=\left( 4\times 1-10 \right)=-6 $ $ {{T}_{2}}=\left( 4\times 2-10 \right)=-2 $ $ {{T}_{3}}=\left(...

(i) 5, 11, 17, 23 … (ii) 16, 9, 2, -5, ….. Solution: (i) (6n – 1) (ii) (23 – 7n)

Solution: The given AP is $6,7\frac{3}{4},9\frac{1}{2},11\frac{1}{4},............ $ First term,  $ a=6 $ and common difference,  $ d=7\frac{3}{4}-6=\frac{7}{4} $ $ Now,\,{{T}_{37}}=a+(n-1)d=a+36d $...

Solution: $ The\text{ }given\text{ }AP\text{ }is-40,-15,-10,-35,.........\text{ } $ $ First\text{ }term,\text{ }a=-40\text{ } $ $ Common\text{ }difference,d=-15-(-40)=25 $ $...

$\sqrt{2},\text{ }\sqrt{18},\text{ }\sqrt{50},\text{ }\sqrt{98},...........$ (iv) the 9th term of the AP 3/4, 5/4, 7/4, 9/4...................... Solution:  

$ The\text{ }given\text{ }AP\text{ }is\text{ }9,13,17,21,..........\text{ } $ $ First\text{ }term,\text{ }a\text{=}9 $ $ Common\text{ }difference,\text{ }d=13-9=4\text{ } $ $...

$(v)\,\sqrt{20},\text{ }\sqrt{45},\text{ }\sqrt{80},\text{ }\sqrt{125},.........\text{ }$ Solution:  The given progression is $\sqrt{20},\text{ }\sqrt{45},\text{ }\sqrt{80},\text{...

(iii) $ \left( iii \right)\text{ -1,}\frac{-5}{6}\text{,}\frac{-2}{3}\text{,}\frac{-1}{2}\text{,}...........\text{ } $ (iV) $ \left( iv \right)\text{ }\sqrt{2},\text{ }\sqrt{8},\text{...

(i) 9, 15, 21, 27,………… (ii) 11, 6, 1, – 4,…….. Solution:  The given progression is 9, 15, 21, 27,………… We can see that, 15 – 9 = 21 – 15 = 27 – 21 = 6 (Constant) Thus, we can say that each term...