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in Arithmetic Progression by (29.9k points)
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If (3y – 1), (3y + 5) and (5y + 1) are three consecutive terms of an AP then find the value of y.

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

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

\(\Rightarrow\) 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5

\(\Rightarrow\) 6 = 2y - 4

\(\Rightarrow\) 2y = 6 + 4 = 10

\(\Rightarrow\) y = 5

Hence, the value of y is 5.

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