Last updated at March 22, 2021 by Teachoo
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Ex 5.3, 4 How many terms of the AP. 9, 17, 25 โฆ must be taken to give a sum of 636? Given AP is 9, 17, 25, โฆ.. Here, a = 9 d = 17 โ 9 = 8 & Sum = Sn = 636 We need to find n We know that Sum = ๐/2 (2a + (n โ 1) d) Putting a = 9, d = 8, Sn = 636 636 = ๐/(2 ) (2 ร 9 + (n โ 1) ร 8) 636 ร 2 = n (18 + 8n โ 8) 1272 = n (10 + 8n) 1272 = 10n + 8n2 0 = 10n + 8n2 โ 1272 10n + 8n2 โ 1272 = 0 2(5n + 4n2 โ 636)= 0 5n + 4n2 โ 636 = 0 4n2 + 5n โ 636 = 0 We solve by quadratic formula Comparing equation with ax2 + bx + c = 0 Here, a = 4, b = 5, c = โ 636 We know that, D = b2 โ 4ac D = (5)2 โ 4 ร 4 ร (โ636) D = 25 + 10176 D = 10201 Hence, roots to equation are given by n = (โ ๐ ยฑ โ๐ซ)/๐๐ Putting values n = (โ 5 ยฑ โ๐๐๐๐๐)/(2 ร 4) n = (โ ๐ ยฑ ๐๐๐)/๐ Finding root of 10201 10201 = 101 ร 101 10201 = 1012 So, โ("10201" ) = 101 Solving So, n = 12 & n = (โ53)/4 Since n cannot be negative, โด n = 12 n = (โ๐ + ๐๐๐)/๐ n = 96/8 n = 12 n = (โ๐ โ ๐๐๐)/๐ n = (โ106)/8 n = (โ๐๐)/๐
Ex 5.3
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