Question

The fourth term of an AP is 0. Prove that it's 25th term is triple its 11th term .

plz solve this for me it is very urgent .

Answer 1

Solution :-

Let the first term of an AP be 'a'and the common difference be 'd'

Fourth term of an AP = 0

⇒ a₄ = a + (4 - 1)d = 0

⇒ a + 3d = 0

⇒ a = - 3d

Now, find the 11th term and 25th term

⇒ a₁₁ = a + (11 - 1)d

⇒ a₁₁ = a + 10d

Substituting a = - 3d in the above equation

⇒ a₁₁ = - 3d + 10d

⇒ a₁₁ = 7d ---- [ Equation 1 ]

⇒ a₂₅ = a + (25 - 1)d

⇒ a₂₅ = a + 24d

Substituting a = - 3d in the above equation

⇒ a₂₅ = - 3d + 24d

⇒ a₂₅ = 21d ---- [ Equation 2 ]

Dividing eq(2) by eq(1)

⇒ a₂₅ / a₁₁ = 21d / 7d

⇒ a₂₅ / a₁₁ = 3

⇒ a₂₅ = 3 * a₁₁

Hence proved.