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

Given: a4 = 0

To Prove: a25 = 3 × a11

Now, a4 = 0

a + 3d = 0

a = –3d

We know that,

an = a + (n – 1)d

a11 = –3d + (11 – 1)d [from (i)]

a11 = –3d + 10d

a11 = 7d …(ii)

Now,

a25 = a + (25 – 1)d

a25 = –3d + 24d [from(i)]

a25 = 21d

a25 = 3 × 7d

a25 = 3 × a11 [from(ii)]

Hence Proved

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