#### Question

The 4^{th} term of an A.P. is zero. Prove that the 25^{th} term of the A.P. is three times its 11^{th} term

#### Solution

4th term of an A.P.= a_{4} = 0

∴ a + (4 – 1)d = 0

∴ a + 3d = 0

∴ a = –3d ….(1)

25^{th} term of an A.P. = a_{25}

= a + (25 – 1)d

= –3d + 24d ….[From (1)]

= 21d.

3 times 11^{th} term of an A.P. = 3a_{11}

= 3[a + (11 – 1)d]

= 3[a + 10d]

= 3[–3d + 10d]

= 3 × 7d

= 21d

∴ a_{25} = 3a_{11}

i.e., the 25^{th} term of the A.P. is three times its 11^{th} term.

Is there an error in this question or solution?

Solution The 4th term of an A.P. is zero. Prove that the 25th term of the A.P. is three times its 11th term Concept: Arithmetic Progression.