Question

31f 6th term of an AP is zero then show that its 33rd term is three times its 15th​

Answer 1

see explanation

Explanation:

the nth term of an AP is

`a_(n)` = a₁ + (n - 1)d

given a₆ = 0 , then

a₁ + 5d = 0 ( subtract 5d from both sides )

a₁ = - 5d

Then

a₃₃ = a₁ + 32d = - 5d + 32d = 27d

a₁₅ = a₁ + 14d = - 5d + 14d = 9d

27d = 3 × 9d

Thus a₃₃ = 3 × a₁₅