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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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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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Given: (3y – 1), (3y + 5) and (5y + 1) are 3 consecutive terms of an AP.

So, (3y + 5) – (3y – 1) – (5y + 1) – (3y + 5)

2(3y + 5) = 5y + 1 + 3y – 1

6y + 10 = 8y

8y – 6y = 10

2y = 10

Or y = 5

The value of y is 5.

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