Answer :

Given: T _{ 4 } = 2

S _{ 6 } = T _{ 1 } + T _{ 2 } + T _{ 3 } + T _{ 4 } + T _{ 5 } + T _{ 6 } = 0

Let the sum of first 30 terms be S_{30 }

And, the sum of first six terms be S_{6}.

T _{ 4 } = 2

⇒ a + (4 – 1)d = 2

⇒ a + 3d = 2 …………..(i)

We know that,

Also,

S _{ 6 } = 0

⇒ 2a + 5d = 0 …………..(ii)

(i)× 2,

a + 3d = 2

2a + 6d = 4 …………………..(iii)

(iii) – (ii),

(2a + 6d) – (2a + 5d) = 4 – 0

⇒ 2a + 6d – 2a – 5d = 4

⇒ **d = 4**

Substituting the value of d = 4 in (i),

a + 3 × (4) = 2

⇒ a + 12 = 2

⇒ a = 2 - 12 = - 10

∴ T _{ 30 } = a + 29d

= - 10 + 29 × (4)

= – 10 + 116

= 106

∴ Sum to first 30 terms,

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