Ex 5.3, 1 (iv) - Chapter 5 Class 10 Arithmetic Progressions

Last updated at Dec. 16, 2024 by

Ex 5.3, 1 - Chapter 5 Class 10 Arithmetic Progressions - Part 7

Ex 5.3, 1 - Chapter 5 Class 10 Arithmetic Progressions - Part 8
Ex 5.3, 1 - Chapter 5 Class 10 Arithmetic Progressions - Part 9

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Ex 5.3, 1 Find the sum of the following APs. (iv) 1/15, 1/12, 1/10 ,………, to 11 terms 1/15, 1/12, 1/10, ….. to 11 terms We know that Sum = 𝑛/2 (2a + (n – 1) d) Here n = 11 , a = 1/15 & d = 1/12 – 1/15 = (15 − 12)/(12 × 15) = 3/180 = 1/60 Putting values in formula Sum = 𝒏/𝟐 (2a + (n – 1) d) = 11/2 ("2 " ×" " 1/15 " + (11 – 1) " ×" " 1/60) = 11/2 (2/15 " + 10 " × 1/60) = 11/2 (2/15 " + " 10/60) = 11/2 ((2 × 4 + 10)/60) = 11/2 ((8 + 10)/60) = 11/2 × 18/60 = 11 × 9/60 = 11 × 3/20 = 𝟑𝟑/𝟐𝟎

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo