Example 13 - Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Last updated at Aug. 3, 2021 by Teachoo
Examples
Example 1
Example 2 Important
Example 3
Example 4 Important
Example 5
Example 6 Important
Example 7
Example 8 Important
Example 9
Example 10
Example 11
Example 12
Example 13 Important You are here
Example 14 (i)
Example 14 (ii)
Example 15 Important
Example 16 Important Deleted for CBSE Board 2022 Exams
Chapter 5 Class 10 Arithmetic Progressions (Term 2)
Serial order wise
Transcript
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 .
