Question

The sum of first six terms of an ap is zero and forth term is 2. Find the sum of its first 30 terms

Answer 1

Step-by-step explanation:

Let 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. 

Answer 2

Answer:

your answer is here

Step-by-step explanation:

Let 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.