Question

The first $3$ terms of an $A.P$ are $3y-1,3y+5$ & $5y+1.$ Find $y.$

Answer 1

As terms are in AP

$(3y+5)-(3y-1)=d$ (1)

$(5y+1)-(3y+5)=d$ (2)

Equating (1) and (2)

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

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

$6=2y-4$

$10=2y$

$y=5$