in an ap, nth term is 20 and sum of 1st n th term is 399 . find n
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Step-by-step explanation:
(c) Sn = n/2 [2a + (n-1)d]
399 = n/2 [2 × 1 + (n-1)d]
798 = 2n + n(n-1)d ...(i)
and an = 20
⇒ a + (n-1)d = 20 [∵ an = a + (n-1)d]
⇒ 1 + (n-1)d = 20 ⇒ (n-1)d = 19 ...(ii)
Using Eq.(ii) in Eq (i), we get
798 = 2n + 19n
⇒ 798 = 21n
∴ n = 798 / 21 = 38
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