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4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Grade:12th pass

1 Answers

Pawan Prajapati
askIITians Faculty 8740 Points
4 months ago
Let there be n terms of the AP. 9, 17, 25 … For this A.P., First term, a = 9 Common difference, d = a2−a1 = 17−9 = 8 As, the sum of n terms, is; Sn = n/2 [2a+(n -1)d] 636 = n/2 [2×a+(8-1)×8] 636 = n/2 [18+(n-1)×8] 636 = n [9 +4n −4] 636 = n (4n +5) 4n2 +5n −636 = 0 4n2 +53n −48n −636 = 0 n (4n + 53)−12 (4n + 53) = 0 (4n +53)(n −12) = 0 Either 4n+53 = 0 or n−12 = 0 n = (-53/4) or n = 12 n cannot be negative or fraction, therefore, n = 12 only.

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