Solution: The distances between the bucket and the potatoes are 5, 8, 11, 14,..., which is in the form of AP. Given that the competitor's journey for gathering these potatoes is two times the...

Solution:  The numbers of logs in each row are in the form of an A.P. as 20, 19, 18,... For the provided A.P., The first term, a = 20 and the common...

Solution: We are aware that, The perimeter of a semi-circle is πr Therefore, P1 = π(0.5) = π/2 cm P2 = π(1) = π cm P3 = π(1.5) = 3π/2 cm We can say that, P1, P2, P3 are...

Solution: It is clear that the number of trees planted by students is in the form of A.P. 1, 2, 3, 4, 5………………..12 The first term, a = 1 The common difference, d = 2−1 = 1...

Solution: Let us assume that Rs. P be the cost of1st prize Then the cost of 2nd prize = Rs. P − 20 And the cost of 3rd prize = Rs. P − 40 We can see...

Solution: The given penalties are in the form of an A.P. having the first term as 200 and the common difference as 50, as can be seen. As a result, a = 200...

Solution: 1, 3, 5, 7, 9 … 49 are the odd numbers between 0 and 50. As a result, we can see that these odd numbers are in the form of A.P. Hence, The first term, a = 1 The common difference,...

Solution: 8, 16, 24, 32… are the multiples of 8 This series creates an A.P., with the first term being 8 and the common difference being 8. As a result, a = 8 d = 8 S15 = ? Using...

Solution: 6, 12, 18, 24 …. are the positive integers that are divisible by 6 This series creates an A.P., with the first term being 6 and the common difference being 6. a = 6 d = 6...

Solution: Provided here that, Sn = 4n−n2 The first term, a = S1 = 4(1) − (1)2 = 4−1 = 3 The sum of first two terms = S2= 4(2)−(2)2 = 8−4 = 4 The second...

(i) an = 3+4n(ii) an = 9−5nAlso find the sum of the first 15 terms in each case. Solutions: (i) an = 3+4n a1 = 3+4(1) = 7 a2 = 3+4(2) = 3+8 = 11 a3 = 3+4(3) = 3+12 = 15...

Solution: Provided here that, S7 = 49 S17 = 289 We are aware that, Sum of n terms; Sn = n/2 [2a + (n – 1)d] As a result, S7= 7/2 [2a +(n -1)d]...

Solution: Provided here that, The second term, a2 = 14 The third term, a3 = 18 The common difference, d = a3−a2 = 18−14 = 4 a2 = a+d 14 = a+4 a = 10 =...

Solution: Provided here,The common difference, d = 7 22nd term, a22 = 149 The sum of the first 22 term, S22 = ? Using the nth term formula, an = a+(n−1)d...

Solution: Provided that, The first term, a = 17 The last term, l = 350 The common difference, d = 9 If the A.P. has n terms, the formula for the last term can be expressed as; l =...

Solution: Provided that, The first term, a = 5 The last term, l = 45 The sum of the AP be, Sn = 400 The sum of the AP formula, as we all know, is; Sn = n/2 (a+l) 400...

Solutions: In the given AP. 9, 17, 25 … let there be n terms For the above A.P., First term, a = 9 The common difference, d = a2−a1 = 17−9 = 8 Say that the total of n terms...

(i) Given a = 3, n = 8, S = 192, find d.(ii) Given l = 28, S = 144 and there are total 9 terms. Find a. Solutions: (i) Provided that,  The first term, a = 3, The number of...

(i) Given a = 8, an = 62, Sn = 210, find n and d.(ii) Given an = 4, d = 2, Sn = − 14, find n and a. Solutions: (i) Provided that, a = 8, an = 62, Sn = 210 In AP,...

Solutions: (i) Provided that, d = 5, S9 = 75 In AP, the sum of n terms is, Sn = n/2 [2a +(n -1)d] As a result, the sum of the first nine terms is;...

(i) Given a12 = 37, d = 3, find a and S12.(i) Given a3 = 15, S10 = 125, find d and a10. Solutions: (i) Provided that, a12 = 37, d = 3 The formula for the nth term in an AP,...

Solutions: (i) Provided that, a = 5, d = 3, an = 50 The formula for the nth term in an AP, as we know, is an = a +(n −1)d, Putting the given...

(i) − 5 + (− 8) + (− 11) + ………… + (− 230) Solution: Given to us that, (−5) + (−8) + (−11) + ………… + (−230) For the above given A.P., The first term, a = −5 The nth term, an= −230 Common...

(i) $7+10 \frac{1}{2}+14+\ldots \ldots \ldots \ldots \ldots . .+84$ (ii) $34+32+30+\ldots \ldots \ldots . .+10$ Solutions: (i) For the given A.P., $7+10 \frac{1}{2}+14+\ldots \ldots \ldots \ldots...

Solution: The first step is $\frac{1}{2}$ m wide, the second step is 1 m wide, and the third step is $\frac{3}{2}$m wide, as seen in the diagram. When the height reaches $\frac{1}{4}$ m, we can see...

Solution: Given, Row houses have numbers ranging from 1,2,3,4,5,......49. As a result, we can observe that the houses in a row are in the form of AP. So, The first term, a = 1 The common difference,...

Solution: Given, The ladder has a 25cm distance between the rungs. The distance between the ladder's top and bottom rungs of ladder is $=2\frac{1}{2}m=2\frac{1}{2}*100cm$= 5/2 ×100cm = 250cm As a...

Solution: We can write from the given statements, a3 + a7 = 6 …………………………….(i) And a3 ×a7 = 8 ……………………………..(ii) By using the nth term formula, an = a+(n−1)d Third...

Solution: Given that the AP series is 121, 117, 113, . . ., The, first term, a = 121 and, The common difference, d = 117-121= -4 According to the nth term formula, an = a+(n −1)d...

Solutions: (i) Provided that, $0.6,1.7,2.8, \ldots$, to 100 terms And for this A.P., The first term, $a=0.6$ And the common difference, $d=a_{2}-a_{1}=1.7-0.6=1.1$ We all know that,for the sum of...

(i) $2,7,12, \ldots .$, to 10 terms. (ii) $-37,-33,-29, \ldots$, to 12 terms Solutions: (i) Provided that, $2,7,12, \ldots$, to 10 terms And for this A.P., The first term, $\mathrm{a}=2$ And the...

Given this, Ramkali saved Rs.5 in the first week and then began saving Rs.1.75 each week after that. As a result, the first term is a = 5 and the common difference is d = 1.75. In addition,...

Solution: As can be observed from the provided question, Subba Rao's earnings increase by Rs.200 every year, forming an AP. As a result, after 1995, the annual salaries are as follows: 5000, 5200,...

Solution: We all know what the AP's nth phrase is; an = a+(n−1)d a4 = a+(4−1)d a4 = a+3d We can write in the same way., a8 = a+7d a6 = a+5d...

Solution: If A.P. is 3, 8, 13,..., 253, then the common difference is d= 5. As a result, we can rewrite the given AP as follows: 253, 248, 243, …, 13, 8, 5 For the new AP, the first term is a = 253...

Solutions: Provided here that, The Third term, a3 = 16 As we all know, a +(3−1)d = 16 a+2d = 16 ………………………………………. (i) It is known that, 7th term exceeds the...

Solution: Two APs are given: 63, 65, 67,... and 3, 10, 17,.... Taking first AP, 63, 65, 67, … The First term, a = 63 The Common difference, d = a2−a1 = 65−63 = 2 We all know that...

Solution: The number 12 is the first multiple of four that is greater than ten. The following multiple will be 16. As a result, the series formed as follows: 12, 16, 20, 24,... All of these are...

Solution: The first three-digit number that is divisible by seven is 105. 105+7 = 112 is the second number. 112+7 = 119 is the third number. As a result, 105, 112, 119,... All three digit integers...

Solution: Let a1 and a2 be the initial terms of two APs, respectively. And let d be the common difference between these APs. For the first A.P.,we know, an = a+(n−1)d Therefore,...

Solution: Given to us that A.P. is 3, 15, 27, 39, … The first term, a = 3 The common difference, d = a2 − a1 = 15 − 3 = 12 We all know that, an = a+(n−1)d...

Solution: For an A.P series we all know that; an = a+(n−1)d a17 = a+(17−1)d a17 = a +16d Similarly, a10 = a+9d As stated in the question,...

Solution: Provided that, a3 = 4 is the 3rd term and, a9 = −8 is the 9th term We all know that, an = a+(n−1)d Therefore, a3 = a+(3−1)d 4...

Solution:  Provided that,  a3 = 12 is the 3rd term  a50 = 106 is the 50th term We all know that, ${{a}_{n}}=a+(n-1)d$ ${{a}_{3}}=a+(3-1)d$ 12...

Solution: Provided that, For the 11th term, a11 = 38 and for the16th term, a16 = 73 We all know that, ${{a}_{n}}=a+(n-1)d$ a11 = a+(11−1)d 38...

Solution: In the above mentioned given series, A.P. 11, 8, 5, 2.. The first term, a = 11 The common difference, d = a2−a1 = 8−11 = −3 Let −150 be the nth term of this...

(i) 7, 13, 19, …, 205 (ii)$18,15\frac{1}{2},13...-47$ Solutions: (i) Given here, 7, 13, 19, …, 205 is the A.P Therefore The first term, a = 7 The common difference,...

Solutions: The A.P. series is given as 3, 8, 13, 18, … The first term, a = 3 The common difference, d = a2 − a1 = 8 − 3 = 5 Let the nth term of the given...

Solution: (i) In the case of given A.P., a2 = 38 a6 = −22 As we all know, for an A.P., an = a+(n −1)d Therefore, putting the values here, a2 = a+(2−1)d 38...

Solutions: (i) In the case of given A.P., a = 5 and a4 = 19/2 As we all know, for an A.P., an = a+(n−1)d Therefore, putting the values here, a4 = a+(4-1)d 19/2...

Solutions: (i) In the case of the given A.P., 2,2 , 26 First and the third term are; a = 2 a3 = 26 As we all know, for an A.P., an = a+(n −1)d Therefore, putting the...

(i) 30th term of the A.P: 10,7, 4, …, is(A) 97 (B) 77 (C) −77 (D) −87 (ii) 11th term of the A.P. -3, -1/2, ,2 …. is(A) 28 (B) 22 (C) – 38 (D)$-48\frac{1}{2}$ Solutions: (i)...

(i) Given here, The first term, a = 3.5 The common difference, d = 0 Number of terms given, n = 105 Nth term, an = ? As we all know, for an A.P.,...

(i) Given here, The first term, a = ? The common difference, d = -3 Number of terms given, n = 18 Nth term, an = -5 As we all know, for an A.P.,...

Solutions:  (i) Given here, The first term, a = 7 The common difference, d = 3 Number of terms given, n = 8 We need to seek out the nth term, ${{a}_{n}}=?$ As we all know, for an A.P.,...

(i) 12, 52, 72, 73 … Solution(i): Given here, ${{a}_{2}}-{{a}_{1}}=25-1=24$ ${{a}_{3}}-{{a}_{2}}=49-25=24$ ${{a}_{4}}-{{a}_{3}}=73-49=24$ Since, an+1 – an or the common...

(i) √3, √6, √9, √12 … (ii) 12, 32, 52, 72 … Solution(i): Given here, ${{a}_{2}}-{{a}_{1}}=\sqrt{6}-\sqrt{3}=\sqrt{3}*\sqrt{2}-\sqrt{3}=\sqrt{3}(\sqrt{2}-1)$...

(i) a, a2, a3, a4 … (ii) √2, √8, √18, √32 … Solution(i): Given here, ${{a}_{2}}-{{a}_{1}}={{a}^{2}}-a=a(a-1)$ ${{a}_{3}}-{{a}_{2}}={{a}^{3}}-{{a}^{2}}={{a}^{2}}(a-1)$...

(i) 1, 3, 9, 27 … (ii) a, 2a, 3a, 4a … Solution(i):Given here, ${{a}_{2}}-{{a}_{1}}=3-1=2$ ${{a}_{3}}-{{a}_{12}}=9-3=6$ ${{a}_{4}}-{{a}_{3}}=27-9=18$ Since, an+1 – an or the...

(i) 0, -4, -8, -12 … (ii) -1/2, -1/2, -1/2, -1/2 …. Solution(i):Given here, ${{a}_{2}}-a=(-4)-0=-{{4}_{{}}}$  ${{a}_{3}}-{{a}_{2}}=(-8)-(-4)=-{{4}_{{}}}$...

(i) Given, 3, 3+√2, 3+2√2, 3+3√2(ii) 0.2, 0.22, 0.222, 0.2222 …. Solution(i):Given here, ${{a}_{2}}-a=3=\sqrt{2}-3={{\sqrt{2}}_{{}}}$...

(i) Given, -1.2, – 3.2, -5.2, -7.2 … (ii) Given, -10, – 6, – 2, 2 … Solution(i): Given here, ${{a}_{2}}-a=(-3.2)-(1.2)=-{{2}_{{}}}$ ${{a}_{3}}={{a}_{2}}=(-5.2)-(-3.2)=-2$...

(i) 2, 4, 8, 16 … (ii) 2, 5/2, 3, 7/2 …. Solution(i): It was given to us that, 2, 4, 8, 16 … The main difference here is; ${{a}_{2}}-{{a}_{1}}=4-2=2$ ${{a}_{3}}-{{a}_{2}}=8-4=4$...

(i) 1/3, 5/3, 9/3, 13/3 …. (ii) 0.6, 1.7, 2.8, 3.9 … Solution(i): The first term is, a = 1/3 And, the Common difference, d = Second term – First term Therefore, ⇒ 5/3 – 1/3...

(i) 3, 1, – 1, – 3 … (ii) -5, – 1, 3, 7 … Solution(i): 3, 1, – 1, – 3 … The first term is, a = 3 And, the Common difference, d = Second term – First term Therefore, ⇒  1 – 3...

(i) a = – 1.25, d = – 0.25 Solution: Assume that the Arithmetic Progression series is a1, a2, a3, a4, a5 … a1 = a = – 1.25...

(i) a = 4, d = – 3  (ii) a = -1 d = 1/2 Solution(i): Assume that the Arithmetic Progression series is  a1, a2, a3, a4, a5 …...

(i) a = 10, d = 10 (ii) a = -2, d = 0    Solution(i): Assume that the Arithmetic Progression series is a1, a2, a3, a4, a5 … a1 = a =...

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km. (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air...

(i) The cost of digging a well after every metre of digging, when it costs Rs 150 for the first metre and rises by Rs 50 for each subsequent metre. (ii) The amount of money in the account every...

(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km. (ii) The amount of air present in a cylinder when a vacuum pump removes 1/4 of the air...