Ex 5.3, 4 - How many terms of AP 9, 17, 25 … must be - Finding number of terms given s

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  1. Chapter 5 Class 10 Arithmetic Progressions
  2. Serial order wise
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Ex 5.3 ,4 How many terms of the AP. 9, 17, 25 โ€ฆ must be taken to give a sum of 636? AP is 9,17,25, โ€ฆ.. We know that Sum = ๐‘›/2 (2a + (n โ€“ 1) d) Here, a = 9 d = 17 โ€“ 9 = 8 & Sum = Sn = 636 We need to find n Putting values in equation Sum = ๐‘›/2 (2a + (n โ€“ 1) d) 636 = ๐‘›/(2 ) (2 ร— 9 + (n โ€“ 1) ร— 8) 636 ร— 2 = n (18 + (n โ€“ 1) ร— 8) 1272 = n (18 + 8n โ€“ 8) 1272 = n (10 + 8n) 1272 = 10n + 8n2 0 = 10n + 8n2 โ€“ 1272 10n + 8n2 โ€“ 1252 = 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— (โ€“636) D = 25 + 10176 D = 10201 Hence, roots to equation are given by n = (โˆ’ ๐‘ ยฑ โˆš๐ท)/2๐‘Ž Putting values n = (โˆ’ 5 ยฑ โˆš10201)/(2 ร— 4) n = (โˆ’ 5 ยฑ 101)/8 Solving So, n = 12 & n = (โˆ’53)/4 Since n cannot be negative, n = 12

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