First 3 terms of an AP are 3y-1,3y+5,5y+1. Find y
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3y-1, 3y+5, 5y+1
Since the number are in AP
Their common difference is constant
T3 - T2 = T2 - T1
=> 5y+1 - 3y-5 = 3y+5 - 3y +1
=> 2y - 4 = 6
=> 2y = 10
=> y = 5
Since the number are in AP
Their common difference is constant
T3 - T2 = T2 - T1
=> 5y+1 - 3y-5 = 3y+5 - 3y +1
=> 2y - 4 = 6
=> 2y = 10
=> y = 5
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