Question

inanA.P.ifa=1, tn=20andsumofallthetermsisequalto399then n=?​

Answer 1

Step-by-step explanation:

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