Finding number of terms given s

Chapter 5 Class 10 Arithmetic Progressions
Concept wise

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Ex 5.3, 4 How many terms of the AP. 9, 17, 25 ā¦ must be taken to give a sum of 636? Given AP is 9, 17, 25, ā¦.. Here, a = 9 d = 17 ā 9 = 8 & Sum = Sn = 636 We need to find n We know that Sum = š/2 (2a + (n ā 1) d) Putting a = 9, d = 8, Sn = 636 636 = š/(2 ) (2 Ć 9 + (n ā 1) Ć 8) 636 Ć 2 = n (18 + 8n ā 8) 1272 = n (10 + 8n) 1272 = 10n + 8n2 0 = 10n + 8n2 ā 1272 10n + 8n2 ā 1272 = 0 2(5n + 4n2 ā 636)= 0 5n + 4n2 ā 636 = 0 4n2 + 5n ā 636 = 0 We solve by quadratic formula Comparing equation with ax2 + bx + c = 0 Here, a = 4, b = 5, c = ā 636 We know that, D = b2 ā 4ac D = (5)2 ā 4 Ć 4 Ć (ā636) D = 25 + 10176 D = 10201 Hence, roots to equation are given by n = (ā š Ā± āš«)/šš Putting values n = (ā 5 Ā± āššššš)/(2 Ć 4) n = (ā š Ā± ššš)/š Finding root of 10201 10201 = 101 Ć 101 10201 = 1012 So, ā("10201" ) = 101 Solving So, n = 12 & n = (ā53)/4 Since n cannot be negative, ā“ n = 12 n = (āš + ššš)/š n = 96/8 n = 12 n = (āš ā ššš)/š n = (ā106)/8 n = (āšš)/š