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Ex 5.2, 13 - Chapter 5 Class 10 Arithmetic Progressions
Last updated at Dec. 13, 2024 by Teachoo
Divisible/ multiples
Chapter 5 Class 10 Arithmetic Progressions
Concept wise
- Checking if AP or not and finding a, d
- Finding nth term
- Finding n
- Finding AP
- Finding nth term from end
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Divisible/ multiples
- Statement questions - nth term
- Finding sum of n terms
- Finding number of terms given s
- Given nth term finding s
- Statement questions - Sum of n terms
- Determining AP and finding sum
Transcript
Ex 5.2, 13 How many three digit numbers are divisible by 7 Numbers divisible by 7 are 7, 14, 21, 28, …….. Lowest 3 digit number 100/7 = 142/7 101/7 = 143/7 102/7 = 144/7 105/7 = 15 So lowest number is 105 Highest 3 digit number, 999/7 = 1425/7 998/7 = 1424/7 997/7 = 1423/7 994/7 = 142 Highest number is 994 Thus, our series will be 105, 112, 119, 126 …. 994 Since difference is same, it is an AP Last term of this series is 994 We need to find number of terms, i.e., n Here an = 994 , a = 105 , d= 7 Putting values in formula an = a + (n – 1) d 994 = 105 + (n – 1) × 7 994 = 105 + n (7) – 1 × (7) 994 = 105 + 7n – 7 994 – 105 + 7 = 7n 896 = 7n 896/7 = n 128 = n n = 128 Therefore, 128 3-digit numbers are divisible by 7
