The number of terms in the in the AP 84, 112, 119. The last term less than 500, which is divisible by 7 is 497. Solution: Here 1st term = a = 84 ( which is the 1 st term greater than 80 that is divisible by 7.) = (50/2) – (50/2) Įxample-18: Find the sum of all number divisible by 7 in between 80 to 500.
BASIC ARITHMETIC PROBLEMS WITH SOLUTIONS SERIES
to 50 terms)įor 1st series a=1, d =1 & n = 50, For 2nd series a=5, d =1 & n = 50 Solution: The given series is 1 – 5 + 2 – 6 +3 -7 +4 -8 +. Solution: Here 7 th terms is 6 and 21 st term is -22 Soįrom equations (i) and (ii) a = 18 and d = -2Įxample-17: Find the value of the expression 1 – 5 + 2 – 6 +3 -7 +4 -8 +. ( i) in above thenĮxample- 16: In an A.P 7 th and 21 st terms are 6 and -22 respectively. So a n = a + ( n-1)d = p + q – 1 + (n – 1) (-1) = p + q – nĮxample- 15: In an A.P 31 st term is 40, then the sum of 61 terms of that A.P Įxample- 14: In an A.P a p = q an a q = p then a n = ? Now substitute these values in above equation then -6, 0, 6.Įxample- 13: Find the Arithmetic progression if a 5 + a 9 = 72 and a 7 + a 12 = 97. Solution: n th term = n 3 – 6n 2 + 11n – 6. Then find the sum of the first three terms of that sequence. So in the given sequence 65 number of terms requiredĮxample- 12: The n th term of sequence of number is a n = n 3 – 6n 2 + 11n – 6. Solution: Here the sequence is 96, 93, 90. The number of terms needed to get Sn = 0 in the A.P of 96, 93, 90.
SoĪ 18 = 3 + (18-1)6 = 105 Arithmetic Progression Hard QuestionsĮxample – 11. Now 35 th terms is equal to 18 th term of the 1 st sequence. Solution: The above sequence can be written as two sequences Solution: The maximum sum of the above sequence is “2” So Solution: Here the sequence is 3, 6, 9, 12, 15. S 13 = ( 13/2) = ( 13/2) 2 = 13 x 30 = 390Įxample – 8: The sum of first 50 positive integers divisible by 3 Here a 7 = 30 so 7th term is a + 6d = 30 then S 13 = ? Solution: General form of A.P is a, a+d, a+2d, a+3d. Įxample – 7 : In an A.P a 7 = 30, then find sum of first 13 terms of that A.P Įxample – 5: Find the values of 20 th terms and sum of first 20 terms of the following series 1, 9, 17, 25. Solution: General form of A.M is a, a+d, a+2d, a+3d.
So charge for borewell work ( i.e 160 feet digging) = 1000 + ( 161-1)250 = ₹41,000 Arithmetic Progression Basic ProblemsĮxample – 4: Find the A.P with a = -1.5 and d = -0.5 Solution: Here a = 1000 and d = 110 and nth term is 160 Find the charge when good water found after digging borewell about 161 feet. Solution: Here a = 30 and d = 12 and nth term is 50Įxample – 3: The cost of borewell drilling cost per feet is ₹1000 for first feet and rises by ₹250 for each subsequent feet. what will be the charge after traveling of 50km Įxample – 2: Cab/Taxi Rental Rates after each km when the fare is ₹ 30 for the first km and raise by 12 for each additional km.
General form of A.P is a, a+d, a+2d, a+3d. Find the amount of money in the kiddy bank on her on his 1st, 2nd, 3rd, 4th. Please go through the below link for basic concepts of Sequence and series, fundamental concepts with formulas and properties for arithmetic progressionĬlick Here Arithmetic Progression real life problemsĮxample – 1: Jhon put ₹ 800 into his son’s kiddy bank when he was one year old and increased the amount by 1000 every year. Or, 2 × 3.14r = 3.14.Arithmetic Progression Examples with Solutions for class 10 What is the radius of a circle that has a circumference of 3.14 meters? Solution: Rice weighing 3 3/ 4 pounds was divided equally and placed in 4 containers.
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