Kaliyuga Ekalavya

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 19 Question: 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? Given: Starting salary in 1995 = Rs 5000 Annual increment = Rs 200 To Find: The year in which his income reached Rs 7000. Formula: n th term of an AP = a + (n-1)d, where  'a' is the first term and  'd' is the common difference. Solution: Let a = 5000 d = 200 and  The salary forms an AP: 5000, 5200, 5400, .......  an = 7000 Substituting an= 7000, a = 5000 and d = 200 in nth term of an AP formula we have 7000 = 5000 + (n - 1) 200 ⇒ n - 1 = 7000 - 5000 200   ⇒ n - 1 = 7000 - 5000 200   ⇒ n - 1 = 2000 200 ⇒ n - 1 =10 ⇒ n = 10 + 1 ⇒ n = 11 Therefore, his income reached Rs 7000 in 11th years. Result: Subba Rao's income reached Rs 7000 in the 11th year. ...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 18 Question: 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 of the AP. Given: Sum of 4th and 8th terms = 24 Sum of 6th and 10th terms = 44 To Find: First three terms of the AP Formula: n th term of an AP = a + (n-1)d, where  'a' is the first term and  'd' is the common difference. Solution: Let the first term be 'a' and the common difference be 'd'. 4 th term = a + 3d 8 th term = a + 7d Sum = (a + 3d) + (a + 7d) = 2a + 10d = 24  ⇒ a + 5d = 12 6 th term = a + 5d 10 th term = a + 9d Sum = (a + 5d) + (a + 9d) = 2a + 14d = 44  ⇒ a + 7d = 22 Subtracting the first equation from the second:  (a + 7d) - (a + 5d) = 22 - 12  ⇒ 2d = 10  ⇒ d = 5 Substituting d = 5 in a + 5d = 12 ⇒ a + 5(5) = 12  ⇒ a = 12 - 25 = -13 First term (a 1 ) = a = -13 Second term (a 2 ) = a + d = -1...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 17 Question: Find the 20th term from the last term of the AP : 3, 8, 13, . . ., 253. Given: An Arithmetic Progression (AP): 3, 8, 13, ..., 253 To Find: The 20th term from the last term of the AP. Formula: The nth term of an AP is given by a n = a + (n-1)d, where  a is the first term and  d is the common difference.   Also, nth term from the last term (l) = (l - n - 1 )th term from the first term Solution: The common difference (d): d = 8 - 3 = 5 The nth term is given by a n = a + (n-1)d. We have, 253 = 3 + (n-1)5 ⇒ 250 = (n-1)5 ⇒ n-1 = 50 ⇒ n = 51 We want the 20th term from the last term, which is equal to the (51 - 20 + 1)th term from the first term i.e 32nd term. Now we find the 32nd term:  ⇒ a 32 = 3 + (32-1)5  ⇒ a 32 = 3 + 31(5)  ⇒ a 32 = 3 + 155  ⇒ a 32 = 158 Result: The 20th term from the last term is 158. Next question sol...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 16 Question: Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12. Given: Third term (a 3 ) = 16 Seventh term (a 7 ) exceeds fifth term (a 5 ) by 12. i.e., a 7 =  a 5  + 12 To Find: The arithmetic progression (AP). Formula: The nth term of an AP is given by a n = a + (n-1)d, where  'a' is the first term and  'd' is the common difference. Solution: We are given that a 3 = 16.  Using the formula,  a 3 = a + 2d = 16  ⇒ a + 2d = 16 (Equation 1) Also, a 7 = a 5 + 12.  Using the formula,  (a + 6d) - (a + 4d) = 12  ⇒ 2d = 12  ⇒ d = 6 Substituting d = 6 in Equation 1:  a + 2(6) = 16  ⇒ a + 12 = 16  ⇒ a = 4 Therefore, the AP is a, a+d, a+2d, a+3d... which is 4, 8, 12, 16, 20, 24, 28... Result: The arithmetic progression is 4, 8, 12, 16, 20, 24, 28... Next question soluti...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 15 Question: For what value of n, are the nth terms of two APs: 63, 65, 67, . . . and 3, 10, 17, . . . equal? Given: Two Arithmetic Progressions (APs): AP1: 63, 65, 67, ... AP2: 3, 10, 17, ... To Find: The value of n for which the nth terms of AP1 and AP2 are equal. Formula: The nth term of an AP is given by a n = a + (n-1)d, where  'a' is the first term and  'd' is the common difference. Solution: For AP1:  a = 63,  d = 65 - 63 = 2 ⇒ a n = 63 + (n-1)2  ⇒ a n = 63 + 2n - 2  ⇒ a n = 61 + 2n For AP2:  a = 3,  d = 10 - 3 = 7 ⇒ a n = 3 + (n-1)7  ⇒ a n = 3 + 7n - 7 ⇒ a n = 7n - 4 Equating the nth terms of both APs since they are equal 61 + 2n = 7n - 4 ⇒ 61 + 4 = 7n - 2n ⇒ 65 = 5n ⇒ n = 65 5 = 13 Result: The 13th terms of both APs are equal. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 14 Question: How many multiples of 4 lie between 10 and 250? Given: The range of numbers is between 10 and 250. We need to find the number of multiples of 4 within this range. To Find: The number of multiples of 4 between 10 and 250. Formula: The number of multiples of i between a and b is given by \( \frac{m}{i} - \frac{n}{i} + 1\) where m ≤ b and n ≥ a and are divisible by i. Solution: The smallest multiple of 4 greater than 10 is 12.  The largest multiple of 4 less than 250 is 248. The number of multiples of 4 between 10 and 12 can be calculated using the formula as:   ⇒  The number of multiples of 4 = \( \frac{248}{4} - \frac{12}{4} + 1\)   ⇒  The number of multiples of 4 = 62 - 3 + 1   ⇒  The number of multiples of 4 = 60 Result: Therefore, there are 60 multiples of 4 between 10 and 250. Next question solution: NCERT C...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 13 Question: How many three-digit numbers are divisible by 7? Given: Three-digit numbers. To Find: The number of three-digit numbers divisible by 7. Formula: The number of multiples of i between a and b is given by \( \frac{m}{i} - \frac{n}{i} + 1\) where m ≤ b and n ≥ a and are divisible by i. Solution: The smallest three-digit number is 100.  The largest three-digit number is 999. Therefore a = 100 and b = 999. Since we have to find numbers divisible by 7, i = 7 We need to find the number of multiples of 7 between 100 and 999 (inclusive).   999 is not divisible by 7. 994 is the next smallest number divisible by 7.  100 is not divisible by 7. 105 is the next greatest number divisible by 7. Using the formula,  ⇒ The number of three-digit numbers = \( \frac{994}{7} - \frac{105}{7}+ 1 \)  ⇒ The number of three-digit numbers = 142 - 15 + 1  ...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 20 Question: Ramkali saved Rs 5 in the first week of a year and then increased her weekly savings by Rs 1.75. If in the nth week, her weekly savings become Rs 20.75, find n. Given: First week savings (a) = Rs 5 Common difference (d) = Rs 1.75 nth week savings = Rs 20.75 To Find: The value of n (number of weeks). Formula: n th term of an AP = a + (n-1)d, where  'a' is the first term and  'd' is the common difference. Solution: We have  a = 5,  d = 1.75, and  a n = 20.75.  Substituting these values into the formula: a n = a + (n-1)d ⇒ 20.75 = 5 + (n-1)1.75  ⇒ 15.75 = (n-1)1.75 ⇒ 15.75 1.75 = n-1 9 = n - 1  ⇒ n = 9 + 1  ⇒ n = 10 Let's verify:  a 10 = 5 + (10-1)1.75  ⇒ a 10 = 5 + 9(1.75)  ⇒ a 10 = 5 + 15.75  ⇒ a 10 = 20.75.  This is correct. Result: Therefore, n = 10 Next question solu...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 12 Question: Two APs have the same common difference. The difference between their 100th terms is 100, what is the difference between their 1000th terms?. Given: Let the two arithmetic progressions be {a n } and {b n }.  They have the same common difference, say 'd'. a 100 - b 100 = 100 To Find: The difference between their 1000th terms, i.e., a 1000 - b 1000 . S Formula: The nth term of an AP is given by a n = a 1 + (n-1)d, where  a 1 is the first term and d is the common difference. Solution: a 100 = a 1 + 99d b 100 = b 1 + 99d a 100 - b 100 = (a 1 + 99d) - (b 1 + 99d) = a 1 - b 1 = 100 a 1000 = a 1 + 999d b 1000 = b 1 + 999d a 1000 - b 1000 = (a 1 + 999d) - (b 1 + 999d) = a 1 - b 1 Since a 1 - b 1 = 100,  a 1000 - b 1000 = 100 Result: The difference between their 1000th terms is 100. Next question solution: NCERT ...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 11 Question: Which term of the AP : 3, 15, 27, 39, . . . will be 132 more than its 54th term? Given: An arithmetic progression (AP): 3, 15, 27, 39, ... To Find: The term which is 132 more than the 54th term. Formula: The nth term of an AP is given by a n = a + (n-1)d, where a is the first term and d is the common difference. Solution: First, find the common difference (d): d = 15 - 3 = 12 Find the 54th term (a 54 ): a 54 = 3 + (54-1)12 = 3 + 53(12) = 639 Let the required term be a n . Then a n = a 54 + 132 ⇒ a n = 639 + 132 = 771 Now, find n: 771 = 3 + (n-1)12 ⇒ 768 = (n-1)12 ⇒ n - 1 = 768 12 ⇒ n - 1 = 64 ⇒ n = 65 Therefore, the 65th term is 132 more than the 54th term. Let's verify: a 65 = 3 + (65-1)12 = 771. a 54 = 639. 771 - 639 = 132 Result: The 65th term of the AP is 132 more than its 54th term. Next question solution: NCERT Class X Chapter 5:...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 10 Question: The 17th term of an AP exceeds its 10th term by 7. Find the common difference. Given: Let the arithmetic progression be denoted by {a n }. The 17th term (a 17 ) exceeds the 10th term (a 10 ) by 7. Therefore, a 17 - a 10 = 7 To Find: The common difference (d) of the arithmetic progression. Formula: The nth term of an AP is given by a n = a + (n-1)d, where 'a' is the first term and 'd' is the common difference. Solution: We are given that a 17 - a 10 = 7. Using the formula for the nth term, we have: a 17 = a + (17-1)d = a + 16d a 10 = a + (10-1)d = a + 9d Substituting these into the given equation: (a + 16d) - (a + 9d) = 7 ⇒ a + 16d - a - 9d = 7 ⇒ 7d = 7 Solving for d: d = 7 7 = 1 Result: The common difference is 1. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 11 ...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 9 Question: If the 3rd and the 9th terms of an AP are 4 and – 8 respectively, which term of this AP is zero? Given: 3rd term (a 3 ) = 4 9th term (a 9 ) = -8 To Find: The term of the AP which is zero. Formula: a n = a + (n-1)d, where a n is the nth term, a is the first term, n is the number of terms, and d is the common difference. Solution: a 3 = a + 2d = 4 ⇒ a = 4 - 2d a 9 = a + 8d = -8 Substitute a = 4 - 2d into a 9 : (4 - 2d) + 8d = -8 ⇒ 6d = -12 ⇒ d = -2 Substitute d = -2 into a = 4 - 2d: a = 4 - 2(-2) = 8 Let a n = 0 0 = 8 + (n-1)(-2) 2(n-1) = 8 n - 1 = 4 n = 5 Result: The 5th term of this AP is zero. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 10 Explore more in Arithmetic Progressions chapter: Click this link to explore more NCERT Class X Chapter 5 Arithmetic Progressions solutions ...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 8 Question: An AP consists of 50 terms of which 3rd term is 12 and the last term is 106. Find the 29th term. Given: Number of terms (n) = 50 3rd term (a 3 ) = 12 Last term (a 50 ) = 106 To Find: 29th term (a 29 ) Formula: a n = a + (n-1)d where a is the first term, d is the common difference, and n is the number of terms. Solution: a 3 = a + 2d = 12 ⇒ a = 12 - 2d a 50 = a + 49d = 106 Substitute a = 12 - 2d into a 50 equation: (12 - 2d) + 49d = 106 ⇒ 47d = 94 ⇒ d = 2 Substitute d = 2 into a = 12 - 2d: a = 12 - 2(2) = 8 Now find a 29 : a 29 = a + (29-1)d = 8 + 28(2) = 8 + 56 = 64 a 29 = a + 28d = 8 + 28(2) = 64 Result: The 29th term is 64. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 9 Explore more in Arithmetic Progressions chapter: Click this link to explore more NCERT Class X Chapter 5 Arithm...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 7 Question: Find the 31st term of an AP whose 11th term is 38 and the 16th term is 73. Given: 11 th term (a 11 ) = 38 16 th term (a 16 ) = 73 To Find: 31 st term (a 31 ) Formula: a n = a + (n-1)d where  a is the first term and  d is the common difference. Solution: a 11 = a + 10d = 38 ......(1) a 16 = a + 15d = 73 ......(2) Subtracting (1) from (2): (a + 15d) - (a + 10d) = 73 - 38  ⇒ a + 15d - a - 10d = 73 - 38  ⇒ 5d = 35  ⇒ d = 7 Substituting d = 7 in (1): a + 10(7) = 38  ⇒ a + 70 = 38  ⇒ a = 38 - 70  ⇒ a = -32 Now,  a 31 = a + 30d  ⇒ a 31 = -32 + 30(7)  ⇒ a 31 = -32 + 210  ⇒ a 31 = 178 Result: The 31 st term of the AP is 178. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 8 Explore more in Arithmetic Progressions chapter: Click this li...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 6 Question: Check whether – 150 is a term of the AP : 11, 8, 5, 2 . . . Given: Arithmetic Progression (AP): 11, 8, 5, 2, ... To Find: Whether –150 is a term of the given AP. Formula: The nth term of an AP is given by: a n = a + (n-1)d, where  a is the first term,  d is the common difference, and  n is the number of terms. Solution: First term (a) = 11 Common difference (d) = 8 - 11 = -3 Let –150 be the nth term of the AP. Then, a n = a + (n-1)d –150 = 11 + (n-1)(-3) –150 = 11 -3n + 3 –150 = 14 -3n 3n = 14 + 150 3n = 164 n = 164 3 n = 54.666... Since n is not an integer, –150 is not a term of the given AP. Result: –150 is not a term of the AP. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 7 Explore more in Arithmetic Progressions chapter: Click this link to explore more NCERT Cla...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 5(ii) Question: Find the number of terms in each of the following APs : 18, 15 1/2, 13, ... , -47 Given: The given AP is 18, 15 1 2 , 13, ... , -47 To Find: The number of terms in the given AP. Formula: The nth term of an AP is given by a n = a + (n-1)d, where a is the first term, d is the common difference, and n is the number of terms. Solution: Here,  a = 18,  d = 15 1 2 - 18 = - 5 2 , and  a n = -47. Using the formula a n = a + (n-1)d, we have: -47 = 18 + (n-1)(- 5 2 ) ⇒ -47 - 18 = (n-1)(- 5 2 ) ⇒ -65 = (n-1)(- 5 2 ) ⇒ n - 1 = 65 5/2 = 26 ⇒ n = 26 + 1 ⇒ n = 27 Therefore, the number of terms in the AP is 27 Result: The number of terms in the given AP is 27. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 6 Explore more in Arithmetic Progressions chapter: Click this link to explore ...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 5(i) Question: Find the number of terms in each of the following APs : 7, 13, 19, . . . , 205 Given: Arithmetic Progression (AP): 7, 13, 19, ..., 205 To Find: The number of terms (n) in the given AP. Formula: The general formula for the nth term of an AP is given by: a n = a + (n - 1)d, where a n is the nth term, a is the first term, n is the number of terms, and d is the common difference. Solution: Here,  the first term is a = 7,  the common difference is d = 13 - 7 = 6, and  the last term is a n = 205. Using the formula a n = a + (n - 1)d, we have: 205 = 7 + (n - 1)6 ⇒ 205 - 7 = (n - 1)6 ⇒ 198 = (n - 1)6 ⇒ 198 6 = n - 1 ⇒ 33 = n - 1 ⇒ n = 33 + 1 ⇒ n = 34 Result: Therefore, the number of terms in the given AP is 34. Next question solution: ...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 4 Question: Which term of the AP : 3, 8, 13, 18, . . . ,is 78? Given: Arithmetic Progression (AP): 3, 8, 13, 18, ... Last term (a n ) = 78 To Find: The term number (n) of the AP. Formula: The general term of an AP is given by: a n = a + (n - 1)d where: a n = nth term a = first term n = number of terms d = common difference Solution: Here, a = 3, d = 8 - 3 = 5, and a n = 78. Using the formula a n = a + (n - 1)d, we have: 78 = 3 + (n - 1)5 ⇒ 78 - 3 = (n - 1)5 ⇒ 75 = (n - 1)5 ⇒ 75 5 = n - 1 ⇒ 15 = n - 1 ⇒ n = 15 + 1 = 16 Result: Therefore, the 16th term of the AP is 78. Next question solution: NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 5(i) Explore more in Arithmetic Progressions chapter: Click this link to explore more NCERT Class X Chapter 5 Arithmetic Progressions solutions Explore more: Click th...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 3(v) Question: In the following APs, find the missing terms in the boxes : ?,38,?,?,?,-22 Given: An arithmetic progression (AP) with the known term 38 and -22. To Find: The missing terms in the given AP. Formula: The nth term of an AP is given by a n = a + (n-1)d, where  'a' is the first term and  'd' is the common difference. Solution: Let the AP be a 1 , a 2 , a 3 , a 4 , a 5 , a 6 .  We are given that a 2 = 38 and a 6 = -22. Using the formula a n = a + (n-1)d, we have: a 2 = a + d = 38  ⇒ a = 38 - d a 6 = a + 5d = -22 Substituting a = 38 - d into a 6 = a + 5d = -22, we get: (38 - d) + 5d = -22  ⇒ 4d = -60  ⇒ d = -15 Now, we find 'a':  a = 38 - d  ⇒ a = 38 - (-15)  ⇒ a = 38 + 15 ⇒ a = 53 Therefore, the AP is: 53, 38, 23, 8, -7, -22 Result: The missing terms are 53, 23, 8, -7 Next question solution: N...

NCERT Class X Chapter 5: Arithmetic Progression Exercise 5.2 Question 3(iv) Question: In the following APs, find the missing terms in the boxes : -4,?,?,?,?,6 Given: An Arithmetic Progression (AP) with first term a = -4 and last term l = 6. The number of terms n = 6. To Find: The missing terms in the AP. Formula: The nth term of an AP is given by a n = a + (n-1)d, where a is the first term, n is the number of terms, and d is the common difference. The sum of an AP is given by S n = n 2 (a + l), where a is the first term, l is the last term, and n is the number of terms. Solution: We have a = -4 and l = 6, and n = 6. Using the sum formula: S 6 = 6 2 (-4 + 6) = 6 Now, let's find the common difference (d): a n = a + (n-1)d ⇒ 6 = -4 + (6-1)d ⇒ 10 = 5d ⇒ d = 2 Therefore, the missing terms are: a 2 = -4 + 2 = -2 a 3 = -2 + 2 = 0 a 4 = 0 + 2 = 2 a 5 = 2 + 2 = 4 Result: The missing terms are -2, 0, 2, 4. ...