Question

the first three terms of an AP are respectively (3y - 1), (3y +5) and (5y + 1), find hte value of y.

Answer 1

In given AP, $a_2-a_1=a_3-a_2$

$3y+5-3y+1=5y+1-3y-5$

=> $6=2y-4$

=> $y=5$

Then the given AP will be 14, 20, 26, .....

Here common difference is 20 - 14 = 6

Thereforem the fourth term is 26 + 6 = 32