Example 15 - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at Aug. 3, 2021 by Teachoo
Transcript
Example 15 Find the sum of first 24 terms of the list of numbers whose nth term is given by an = 3 + 2n Given an = 3 + 2n Now, 1st term = a1 = 3 + 2 (1) a = 3 + 2 a = 5 2nd term = a2 = 3 + 2(2) = 3 + 4 = 7 3rd term = a3 = 3 + 2 (3) = 3 + 6 = 9 4th term = a4 = 3 + 2 (4) = 3 + 8 = 11 So, the series is 5, 7, 9, 11 ……… Since Difference between consecutive terms is same, So, it is an AP We need to find sum of first 24 terms So, n = 24, a = 5 d = 7 – 5 = 2 Putting these values in formula Sum = 𝑛/2[2𝑎+(𝑛−1)𝑑] = 24/2[2 × 5+(24−1)2] = 12 [10+(23)(2)] = 12 [10+46] = 12 × 56 = 672 Hence, Sum of 24th terms is 672
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Example 7
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Example 9
Example 10
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Example 13 Important
Example 14 (i)
Example 14 (ii)
Example 15 Important You are here
Example 16 Important Deleted for CBSE Board 2022 Exams
Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Serial order wise
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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 10 years. He provides courses for Maths and Science at Teachoo.