the first three terms of an AP are respectively (3y-1) (3y+1) and (5y+1) find the value of y.
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as these terms in a.p.
t2=t1+d
3y+1=3y-1+d
d=1+1=2
also t3=t2+d
5y+1=3y+1+2
2y=2
y=1
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