Example 13 - How many terms of AP: 24, 21, 18 ... must - Examples

Example 13 - Chapter 5 Class 10 Arithmetic Progressions - Part 2
Example 13 - Chapter 5 Class 10 Arithmetic Progressions - Part 3

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Example 13 How many terms of the AP : 24, 21, 18, . . . must be taken so that their sum is 78? Given AP 24, 21, 18,……… Here, a = 24 d = 21 – 24 = –3 Also, given Sum = 78 Sn = 78 We have to find value of n Putting these values in equation Sum = 𝒏/𝟐[𝟐𝒂+(𝒏−𝟏)𝒅] 78 = 𝑛/2[2×24+(𝑛−1)(−3)] 78 ×2=𝑛[48+(𝑛−1)(−3)] 156 = n [ 48 – 3n + 3] 156 = n [ 51 – 3n] 156 = 51n – 3n2 3n2 – 51n + 156 = 0 Dividing the equation by 3 3𝑛2/3−51𝑛/3+156/3=0 n2 – 17n + 52 = 0 We factorize by splitting the middle term n2 – 13n – 4n + 52 = 0 n (n – 13) – 4 (n – 13) = 0 (n – 13) (n – 4) = 0 Splitting the middle term method We need to find two numbers whose Sum = – 17 Product = 52 × 1 = 52 So, n = 13, n = 4 Since, both values are positive natural number. So either of them can be taken .

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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