What is the best solution for solving this tough question of arithmetic progressions in exercise 5.3 of class 10th, what is the easy way In an AP(vi) Given a = 2, d = 8, Sn = 90, find n and an.
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Given that, a = 2, d = 8, Sn = 90
As, sum of n terms in an AP is,
Sn = n/2 [2a +(n -1)d]
90 = n/2 [2a +(n -1)d]
⇒ 180 = n(4+8n -8) = n(8n-4) = 8n2-4n
⇒ 8n2-4n –180 = 0
⇒ 2n2–n-45 = 0
⇒ 2n2-10n+9n-45 = 0
⇒ 2n(n -5)+9(n -5) = 0
⇒ (n-5)(2n+9) = 0
So, n = 5 (as n only be a positive integer)
∴ a5 = 8+5×4 = 34