sum of the first 30 terms of an AP is 0. Then first term is- 29, then find the sum of 28th, 29th and 30th terms of AP.
Answers
et the sum of the first 30 term be S30, first term be a, fourth term be a4 and the sum of first six terms be S6.
the sum of the first six terms be S6.
Given that S6 = 0 and the fourth term a4 = 2
⇒ a + 3d = 2..........(i)
S6 = 0
n/2(2a + 5d) = 0
⇒ 2a + 5d = 0 .........(ii)
(i) ×2,
2a + 6d = 4...........(iii)
∴ (iii) - (ii)
∴ d = 4
Substituting the value of d = 4 in (i),
a + 3 ×(4) = 2
⇒ a = 2 - 12 = -10Let the sum of the first 30 term be S30, first term be a, fourth term be a4 and the sum of first six terms be S6.
the sum of the first six terms be S6.
Given that S6 = 0 and the fourth term a4 = 2
⇒ a + 3d = 2..........(i)
S6 = 0
n/2(2a + 5d) = 0
⇒ 2a + 5d = 0 .........(ii)
(i) ×2,
2a + 6d = 4...........(iii)
∴ (iii) - (ii)
∴ d = 4
Substituting the value of d = 4 in (i),
a + 3 ×(4) = 2
⇒ a = 2 - 12 = -10
∴ a30 = a + 29d
= -10 + 29 × (4)
= -10 + 116
= 106
∴ Sum to first 30 terms = S30 = n/2(a+l)
= 30/2(-10 + 106)
= 15 × 96
= 1140.
∴ a30 = a + 29d
= -10 + 29 × (4)
= -10 + 116
= 106
∴ Sum to first 30 terms = S30 = n/2(a+l)
= 30/2(-10 + 106)
= 15 × 96
= 1140.