Question

If the sixth term of an AP is zero then show that its 33rd term is three times its 15th term.

Answer 1

Sixth term of an AP is zero

that is a₆ = 0

a + 5d = 0

a = -5 d

Now,

\(a_{15}\) = a + (15 - 1)d

= a + 14d

= -5d + 14d

= 9d

and \(a_{33}\) = a + (33 - 1)d

= a + 32d

= -5d + 32d

= 27d

= 3 × 9d

= 3\(a_{15}\) ( \(\because a_{15}\) = 9d)

Hence, \(a_{33}\) = 3\(a_{15}\)

Hence proved.