The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and

1.	The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1. Then, y equals A. –3 B. 4 C. 5 D. 2
Answer» A.P here is 3y – 1, 3y + 5 and 5y + 1. So d= common difference = a₂– a₁= a₃–a₂ 3y + 5 – (3y–1) = (5y + 1) – (3y + 5) 6 = 2y – 4 10 / 2 = y y = 5

The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1. Then, y equals A. –3 B. 4 C. 5 D. 2

Answer»

A.P here is 3y – 1, 3y + 5 and 5y + 1.

So d= common difference = a₂– a₁= a₃–a₂

3y + 5 – (3y–1)

= (5y + 1) – (3y + 5) 6

= 2y – 4 10 / 2 = y

y = 5