# The 4th term of an AP is zero. Prove that its 25th term is triple its 11th term.

Let a be the first term and d be the common difference.

Given: a4 = 0

To prove: a25 = 3 × a11

Now, Consider a4 = 0

a + (4 - 1)d = 0

a + 3d = 0

a = - 3d ………. (1)

Consider a25 = a + (25 - 1)d

a25 = - 3d + 24d (from equation (1))

a25 = 21d ………. (2)

Now, consider a11 = a + (11 - 1)d

a11 = - 3d + 10d (from equation (1))

a11 = 7d ……….(3)

From equation (2) and (3), we get,

a25 = 3 × a11

Hence, proved.

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