Question

pls answer 34th question​

Answer 1

Solution :

Let us consider a₁ be the first term and d be the common difference.

  7th term (a₇) = a₁ + 6d

  2nd term (a₂) = a₁ + d

  12th term (a₁₂) = a₁ + 11d

  4th term (a₄) = a₁ + 3d

By the given conditions,

    a₇ = 4a₂

⇒ a₁ + 6d = 4 (a₁ + d)

⇒ a₁ + 6d = 4a₁ + 4d

⇒ 3a₁ = 2d ...(i)

    a₁₂ = 3a₄ + 2

⇒ a₁ + 11d = 3 (a₁ + 3d) + 2

⇒ a₁ + 11d = 3a₁ + 9d + 2

⇒ 2d = 2a₁ +2

⇒ d = a₁ + 1 ...(ii)

Putting d = a₁ + 1 in (i), we get

3a₁ = 2 (a₁ + 1)

⇒ 3a₁ = 2a₁ + 2

⇒ 3a₁ - 2a₁ = 2

⇒ a₁ = 2

From (ii), we get

d = 2 + 1

⇒ d = 3

Therefore, the required arithmetic progression is

a₁ , a₁ + d , a₁ + 2d , a₁ + 3d , ...

⇒ 2 , 2 + 3 , 2 + 6 , 2 + 9 , ...

⇒ 2 , 5 , 8 , 11 , ...

Answer 2

Hope it helps...................