Question

The 25^th term of an A.P. exceeds its 9^th term by 16. Find its common difference.

Answer 1

n^th term of an A.P. is given by t_n= a + (n – 1) d.

`\implies` t₂₅  = a + (25 – 1)d = a + 24d

And t₉ = a + (9 – 1)d = a + 8d

According to the condition in the question, we get

t₂₅ = t₉ + 16

`\implies` a + 24d = a + 8d + 16

`\implies` 16d = 16

`\implies` d = 1