THE SUM OF 1st 19 TERMS IS 399.FIND ITS 10th TERM
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(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
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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