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 y is 5.
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