Learn all Concepts of Chapter 5 Class 10 (with VIDEOS). Check - Arithmetic Progressions - Class 10    1. Chapter 5 Class 10 Arithmetic Progressions
2. Serial order wise
3. Ex 5.3

Transcript

Ex 5.3 ,4 How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636? AP is 9,17,25, ….. We know that Sum = 𝑛/2 (2a + (n – 1) d) Here, a = 9 d = 17 – 9 = 8 & Sum = Sn = 636 We need to find n Putting values in equation Sum = 𝑛/2 (2a + (n – 1) d) 636 = 𝑛/(2 ) (2 × 9 + (n – 1) × 8) 636 × 2 = n (18 + (n – 1) × 8) 1272 = n (18 + 8n – 8) 1272 = n (10 + 8n) 1272 = 10n + 8n2 0 = 10n + 8n2 – 1272 10n + 8n2 – 1252 = 0 2(5n + 4n2 – 636)= 0 5n + 4n2 – 636 = 0 4n2 + 5n – 636 = 0 We solve by quadratic formula Comparing equation with ax2 + bx + c = 0 Here, a = 4, b = 5, c = – 636 We know that, D = b2 – 4ac D = (5)2 – 4 × 4 × (–636) D = 25 + 10176 D = 10201 Hence, roots to equation are given by n = (− 𝑏 ± √𝐷)/2𝑎 Putting values n = (− 5 ± √10201)/(2 × 4) n = (− 5 ± 101)/8 Solving So, n = 12 & n = (−53)/4 Since n cannot be negative, n = 12

Ex 5.3 