Q9 of 84 Page 5

The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1. Then, y equals (CBSE 2014)

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

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