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

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#### 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.

Concept: Arithmetic Progression

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