Question

If (3y − 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y. [CBSE 2014]

Answer 1

It is given that (3y − 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP.
$$\begin{gathered} \therefore \left(3y+5\right)-\left(3y-1\right)=\left(5y+1\right)-\left(3y+5\right) \\ \Rightarrow 3y+5-3y+1=5y+1-3y-5 \\ \Rightarrow 6=2y-4 \\ \Rightarrow 2y=6+4=10 \end{gathered}$$
$\Rightarrow y=5$

Hence, the value of y is 5.