# If (3y − 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.

Question:

If (3y − 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.

Solution:

It is given that (3y − 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP.

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

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

$\Rightarrow 6=2 y-4$

$\Rightarrow 2 y=6+4=10$

$\Rightarrow y=5$

Hence, the value of is 5.