Question

If (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.

Answer 1

It is given that (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP.

∴ (3y + 5) – (3y – 1) = (5y + 1) – (3y + 5)

⇒ 3y + 5 – 3y + 1 = 5y + 1 – 3y – 5

⇒ 6 = 2y – 4

⇒ 2y = 6 + 4 = 10

⇒ y = 5 

Hence, the value of y is 5.