Solution:
(3y – 1), (3y + 5) and (5y+ 1) are in 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
The first three terms of an AP are respectively (3y – 1), (3y + 5) and (5y + 1), find the value of y.
The first three terms of an AP are respectively (3y – 1), (3y + 5) and (5y + 1), find the value of y.