Solution:
(3y – 1), (3y + 5) and (5y + 1) are the three consecutive terms of an AP.
(3y + 5) – (3y – 1) – (5y + 1) – (3y + 5)
⇒ 2(3y + 5) = 5y + 1 + 3y – 1
⇒ 6y + 10 = 8y
⇒ 8y – 6y = 10
⇒ 2y = 10
⇒ y = 5
y = 5
If (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.
If (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.