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Question
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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Solution
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)
⇒ 3y +5-3y +1 = 5y +1 -3y -5
⇒ 6 = 2y - 4
⇒ 2y = 6+4 =10
⇒ y = 5
Hence, the value of y is 5.
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