Kaliyuga Ekalavya

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 20 Question: In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato, and the other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line. A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? Given: First potato distance (a) = 5 m Common difference (d) = 3 m Number of potatoes (n) = 10 To Find: Total distance the competitor has to run. Formula: Sum of n terms of an AP: S n = n 2 × [2a + (n − 1)d] Solution: The potato are kept at a distance such that it forms an arithmetic progression.  Here,  a = 5,  d = 3,  n = 10 Total distance covere...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 19 Question: 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 (see Fig. 5.5). In how many rows are the 200 logs placed and how many logs are in the top row? Given: Total number of logs = 200 Number of logs in the bottom row = 20 Number of logs in each subsequent row decreases by 1. To Find: Number of rows and number of logs in the top row. Formula: The sum of an arithmetic series is given by S n = n 2 (2a + (n-1)d).  where  n is the number of rows,  a is the number of logs in the top row, and  d is the common difference. Solution: In this case,  first term a = 20,  common difference d = -1, and  the total number of logs is 200.  Substituting the values into the formula: n 2 (2a + (n-1)d). 200 = n 2  (2 (20) + (n-1)(-1)). ⇒ 400 = n(40 - n + 1)  ⇒ 400 = 41n -...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 18 Question: A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A, of radii 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm,... What is the total length of such a spiral made up of thirteen consecutive 22 semicircles? (Take pi = 22/7) Given: Spiral is made of semicircles. Total number of semicircles = 13.  Radii of semicircles are 0.5 cm, 1.0 cm, 1.5 cm, 2.0 cm, ...  π = 22/7 To Find: Total length of the spiral. Formula: Length of a semicircle = πr The nth term of an arithmetic progression is given by a n = a + (n-1)d.   The sum of an arithmetic progression with n terms is given by S n = n 2 (2a + (n-1)d). Solution: The radii of the semicircles are in an arithmetic progression with  first term a = 0.5 and  common difference d = 0.5. The nth term of an arithmetic progression is given by a n = a + (n-1)d.  Ther...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 17 Question: 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 by the students? Given: Number of classes = 12 Number of sections per class = 3 Trees planted by each section of class n = n To Find: Total number of trees planted by the students. Formula: Sum of n terms of AP: S n = \( \frac{n}{2} \) × (a + l) Solution: Let us find the number of trees planted by the 1st section of the students. Class 1, section 1 = 1  Class 2, section 1 = 2 Class 12, section 1 = 12 This forms an AP 1, 2, 3, ....., 12 Here, a = 1, l ...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 16 Question: A sum of Rs 700 is to be used to give seven 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. Given: Total sum = Rs 700 Number of prizes = 7 Each prize is Rs 20 less than its preceding prize. To Find: The value of each of the seven prizes. Formula: Arithmetic Progression sum formula: S = n 2 (a + l), where  S is the sum,  n is the number of terms,  a is the first term, and  l is the last term. Solution: Let the first prize be 'a'.  Then the 7 prizes are a, a-20, a-40, a-60, a-80, a-100, a-120. Sum of the A.P = S = n 2  (a + l) = 700 ⇒ 7 2 (a + a - 120) = 700 ⇒ 7 2 (2a - 120) = 700 ⇒ 7(2a - 120) = 1400 ⇒ 2a - 120 = 200 ⇒ 2a = 320 ⇒ a = 160 Therefore, the prizes are: 160, 140, 120, 100, 80, 60, 40 Result: The values o...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 15 Question: A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows: Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days? Given: Penalty for the first day = Rs 200 Penalty increases by Rs 50 each day. Delay = 30 days To Find: Total penalty for 30 days delay. Formula: The sequence of penalties forms an arithmetic progression.    The sum of an arithmetic series is given by: S n = n 2 (2a + (n-1)d) Where: S n = sum of the series n = number of terms a = first term d = common difference Solution: Here, a = 200, d = 50, n = 30 Substituting the values in the formula: S n = n 2 (2a + (n-1)d) S 30 = 30 2 (2(200) + (30-1...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 14 Question: Find the sum of the odd numbers between 0 and 50. Given: The range of numbers is between 0 and 50. To Find: The sum of odd numbers in the given range. Formula: The sum of an arithmetic series is given by: S n = n 2 (a 1 + a n ), where  n is the number of terms,  a 1 is the first term, and  a n is the last term. Solution: Odd numbers between 0 and 50 are 1, 3, 5, ..., 49. To find the number of terms (n), we can use the formula for the nth term of an arithmetic sequence: a n = a 1 + (n-1)d, where  a n = 49,  a 1 = 1, and  d = 2. ⇒ 49 = 1 + (n-1)2  ⇒ 48 = 2(n-1)  ⇒ 24 = n-1  ⇒ n = 25 Therefore, there are 25 odd numbers between 0 and 50. Using the formula, sum of an arithmetic series given by: S n = n 2 (a 1 + a n ),   ⇒ S 25 = 25 2 ( 1 + 49 )  ⇒ S 25 = 25 x 50 2   ...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 12 Question: Find the sum of the first 40 positive integers divisible by 6. Given: We need to find the sum of the first 40 positive integers divisible by 6. To Find: The sum of the first 40 positive integers divisible by 6. Formula: The sum of an arithmetic series is given by: S n = n 2 (a 1 + a n ), where  n is the number of terms,  a 1 is the first term, and  a n is the last term. Solution: The first 40 positive integers divisible by 6 are 6, 12, 18, ..., 240. This is an arithmetic series with  first term a 1 = 6,  common difference d = 6, and  number of terms n = 40. The last term is a 40 ⇒ a 40 = a 1 + (n-1)d  ⇒ a 40 = 6 + (40-1)6  ⇒ a 40 = 6 + 39(6)  ⇒ a 40 = 240. Using the formula for the sum of an arithmetic series: S n = n 2 (a 1 + a n ), ⇒ S 40 = 40 2 (6 + 240)  ⇒ S 40 = 20(246)  ⇒...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 11 Question: If the sum of the first n terms of an AP is 4 n – n 2 , what is the first term (that is S 1 )? What is the sum of first two terms? What is the second term? Similarly, find the 3rd, the 10th and the n th terms. Given: The sum of the first n terms of an AP is given by S n = 4n – n 2 To Find: First term (S 1 ),  sum of first two terms (S 2 ),  second term (a 2 ),  third term (a 3 ),  tenth term (a 10 ), and  nth term (a n ). Formula: S n = n 2 (2a + (n-1)d) ,  a n = a + (n-1)d Solution: a 1 = 4(1) – (1) 2 = 3 ⇒ First term = 3 S 2 = 4(2) – (2) 2 = 4 ⇒ Sum of first two terms = 4 a 2 = S 2 – S 1 = 4 – 3 = 1 ⇒ Second term = 1 Common difference (d) = a 2 – a 1 = 1 – 3 = -2 a 3 = a 1 + 2d = 3 + 2(-2) = -1 ⇒ Third term = -1 a 10 = a 1 + 9d = 3 + 9(-2) = -15 ⇒ Tenth term = -15 a n = a 1 + (n-1)d = 3 + (n-1)(-2) = 5 – 2n...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 13 Question: Find the sum of the first 15 multiples of 8. Given: First 15 multiples of 8 are required. To Find: The sum of the first 15 multiples of 8. Formula: The sum of an arithmetic series is given by: S n = n 2 (a 1 + a n ), where  n is the number of terms,  a 1 is the first term, and  a n is the last term. Solution: First 15 multiples of 8 are required are 8, 16, 24, .... , 120 This forms an AP, with   first term a 1 = 8,  common difference d = 8 and  last term a n = 120 The sum of an arithmetic series till 15 is given by: S 15 = 15 2 ( 8 + 120 ), where ⇒ S 15 = 15 × 128 2 ⇒ S 15 = 15 × 64 ⇒ S 15 = 960 Result: The sum of the first 15 multiples of 8 is 960. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 14 Explore more in Arithmetic Prog...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 10(ii) Question: Show that a 1 , a 2 , . . ., a n , . . . form an AP where a n is defined as below : a n = 9 – 5n. Also find the sum of the first 15 terms in each case. Given: a n = 9 – 5n To Find: Whether the sequence forms an AP and the sum of the first 15 terms. Formula: The common difference of an AP is given by d = a n+1 - a n .  The sum of first n terms of an AP is given by S n = n 2 (2a + (n-1)d) Solution: a n = 9 – 5n a n+1 = 9 – 5(n+1)  ⇒ a n+1 = 9 – 5n – 5  ⇒ a n+1 = 4 – 5n d = a n+1 – a n   ⇒ d = (4 – 5n) – (9 – 5n)  ⇒ d = 4 – 5n – 9 + 5n  ⇒ d = -5 Since the common difference is constant (-5), the sequence forms an AP. a 1 = 9 – 5(1) = 4 n = 15 d = -5 ⇒ S 15 = 15 2  (2(4) + (15-1)(-5))  ⇒ S 15 = 15 2  (8 + 14(-5))  ⇒ S 15 = 15 2  (8 -70)  ⇒ S 15 = 15 2  (-62)  ⇒ S 15 = 15 (-31) ...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 10(i) Question: Show that a 1 , a 2 , . . ., a n , . . . form an AP where a n is defined as below : a n = 3 + 4n. Also find the sum of the first 15 terms in each case. Given: a n = 3 + 4n To Find: 1. Show that the sequence forms an AP. 2. Find the sum of the first 15 terms. Formula: The sum of an arithmetic series is given by: S n = n 2 (a 1 + a n ) , where  n is the number of terms,  a 1 is the first term, and  a n is the last term. Solution: To show it's an AP, let's find the difference between consecutive terms: a n+1 - a n = [3 + 4(n+1)] - (3 + 4n)  ⇒ a n+1 - a n = 3 + 4n + 4 - 3 - 4n  ⇒ a n+1 - a n = 4 Since the difference is constant (4), the sequence is an arithmetic progression. Now, let's find the sum of the first 15 terms (n=15): ⇒ a 1 = 3 + 4(1) = 7 ⇒ a 15 = 3 + 4(15) = 63 ⇒ S 15 = 15 2  (7 + 63)  ⇒ S 15 = 15 2 (70) ...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 9 Question: If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms. Given: Sum of first 7 terms (S 7 ) = 49 Sum of first 17 terms (S 17 ) = 289 To Find: Sum of first n terms (S n ) Formula: Sum of n terms of an AP: S n = n 2 [2a + (n-1)d] Solution: S 7 = 7 2 [2a + 6d] = 49 ⇒ 2a + 6d = 14 (1) S 17 = 17 2 [2a + 16d] = 289 ⇒ 2a + 16d = 34 (2) Subtracting (1) from (2) : 10d = 20  ⇒ d = 2 Substituting d = 2 in (1):  2a + 6(2) = 14  ⇒ 2a = 2  ⇒ a = 1 Substitute a = 1 in S n   ⇒ S n = n 2  [2(1) + (n-1)2]  ⇒ S n = n 2  [2 + 2n -2]  ⇒ S n = n 2  [2n] = n 2 ⇒ S n = n x n  ⇒ S n = n 2 Result: The sum of first n terms is n 2 Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 10(i) Explore more in Arithmetic Prog...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 8 Question: Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. Question: Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively. Given: Second term (a 2 ) = 14 Third term (a 3 ) = 18 To Find: Sum of first 51 terms (S 51 ) Formula: a n = a + (n-1)d S n = n 2 (2a + (n-1)d) Solution: Common difference (d) = a 3 - a 2 = 18 - 14 = 4 a 2 = a + d  ⇒ 14 = a + 4  ⇒ a = 10 S 51 = 51 2 (2(10) + (51-1)4) ⇒ S 51 = 51 2 (20 + 200)  ⇒ S 51 = 51 2 (220)  ⇒ S 51 = 51 × 110  ⇒ S 51 = 5610 Result: The sum of the first 51 terms is 5610. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 9 Explore more in Arithmetic Progressions chapter: Click this link to explore more NCERT Class X C...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 7 Question: Find the sum of first 22 terms of an AP in which d = 7 and 22nd term is 149. Given: Common difference (d) = 7 22 nd term (a 22 ) = 149 To Find: Sum of first 22 terms (S 22 ) Formula: a n = a + (n-1)d S n = n 2 (2a + (n-1)d) Solution: We have a 22 = 149 and d = 7. Using the formula a n = a + (n-1)d, we get: 149 = a + (22-1)7  ⇒ 149 = a + 147  ⇒ a = 149 - 147  ⇒ a = 2 Now, we use the formula for the sum of an AP: S n = n 2 (2a + (n-1)d) Substituting n = 22, a = 2, and d = 7, we get: ⇒ S 22 = 22 2 (2(2) + (22-1)7)  ⇒ S 22 = 11(4 + 147)  ⇒ S 22 = 11(151)  ⇒ S 22 = 1661 Result: The sum of the first 22 terms is 1661. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 8 Explore more in Arithmetic Progressions c...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 6 Question: The first and the last terms of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum? Given: First term (a) = 17 Last term (l) = 350 Common difference (d) = 9 To Find: Number of terms (n) and sum of terms (S n ) Formula: l = a + (n - 1)d S n = n(a + l) 2 Solution: Using the formula l = a + (n - 1)d: 350 = 17 + (n - 1)9  ⇒ 350 - 17 = (n - 1)9  ⇒ 333 = (n - 1)9  ⇒ 333 9 = n - 1  ⇒ 37 = n - 1  ⇒ n = 38 Using the formula S n = n(a + l) 2 : ⇒ S 38 = 38(17 + 350) 2   ⇒ S 38 = 38(367) 2   ⇒ S 38 = 19(367)  ⇒ S 38 = 6973 Result: Number of terms (n) = 38 Sum of terms (S n ) = 6973 Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 7 Explore more in Arithmetic Progressions chapter...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 5 Question: The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference. Given: First term (a) = 5 Last term (l) = 45 Sum of the AP (S n ) = 400 To Find: Number of terms (n) and common difference (d) Formula: Sum of an AP: S n = n 2 (a + l) Last term of an AP: l = a + (n - 1)d Solution: Using the sum formula: 400 = n 2 (5 + 45)  ⇒ 400 = 25n 1   ⇒ n = 16 Using the last term formula: 45 = 5 + (16 - 1)d  ⇒ 40 = 15d  ⇒ d = 40 15   ⇒ d = 8 3   ⇒ d = 2.5 Result: Number of terms (n) = 16 Common difference (d) = 2.5 Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 6 Explore more in Arithmetic Progressions chapter: Click this link to explore more NCERT Class X Chapter 5 Arithmetic Progressions solutio...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.3 Question 4 Question: How many terms of the AP : 9, 17, 25, . . . must be taken to give a sum of 636? Given: Arithmetic Progression (AP): 9, 17, 25, ... Sum of terms (S n ) = 636 To Find: Number of terms (n) to be taken to get a sum of 636. Formula: The sum of n terms of an AP is given by: S n = n 2 [2a + (n - 1)d] where  a is the first term,  d is the common difference, and  n is the number of terms. Solution: Here,  a = 9,  d = 17 - 9 = 8, and  S n = 636. Substituting the values in the formula, S n = n 2 [2a + (n - 1)d] we get: ⇒ 636 = n 2 [2(9) + (n - 1)8] ⇒ 1272 = n[18 + 8n - 8] ⇒ 1272 = n(8n + 10) ⇒ 1272 = 8n 2 + 10n ⇒ 8n 2 + 10n - 1272 = 0 ⇒ 4n 2 + 5n - 636 = 0 For a quadratic equation ax 2 + bx + c = 0, the roots are given by the quadratic formula:                   ...