Question

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

Answer 1

We know that nth term of an AP is an = a + (n - 1) * d.

(i)

Given that 6th term of an AP is 0.

⇒ a6 = a + (6 - 1) * d

⇒ 0 = a + 5d

⇒ a = -5d.

(ii)

⇒ 15th term a15 = a + (15 - 1) * d

                           = a + 14d

                           = -5d + 14d

                           = 9d.

(iii)

⇒ 33rd term = a + (33 - 1) * d

                     = a + 32d

                     = -5d + 32d

                     = 27d

                     = 3 * a15

Therefore, 33rd term is three times its 15th term.

Hope this helps!

Answer 2

Given,
a6 = 0
a + 5d = 0 ----(1)

a33 = a+32d
a+32d = 3(a+14d)

from (1)
a = -5d

a + 32d
-5d + 32d = 27d (L.H.S)

3(a+14d)
= 3 (-5d+14d)
= 3×9d = 27 d (R.H.S)

L.H.S = R.H.S
Hence, proved