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Ex 5.2, 12 - Two APs have the same common difference. - Finding AP

  1. Chapter 5 Class 10 Arithmetic Progressions
  2. Serial order wise
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Ex 5.2 ,12 Two APs have the same common difference. The difference between their 100th term is 100, what is the difference between their 1000th terms? We know that an = a + (n – 1) d So, 100th term will be a100 = a + (100 – 1) d a100 = a + 99d So, 100th term of 1st AP = a + 99d Let second AP be denoted as bn = b + (n – 1) d b100 = b + (100 – 1) d b100 = b + 99d So, 100th term of 2nd AP = b + 99d As per question , Difference between their 100th terms = 100 a100 – b100 = 100 (a + 99d) – (b + 99d) = 100 a + 99d – b – 99d = 100 a – b + 99d – 99d = 100 a – b = 100 Now we need to find Difference between 1000th terms = a1000 – b1000 = [ a + (1000 – 1)d ] – [ b + (1000 – 1)d] = [ a + 999d ] – [ b + 999d] = a + 999d – b – 999d = a – b + 999d – 999d = a – b Putting value from (1) = 100 So, difference between 1000th terms is 100

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