The first three tremendous an AP …

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Sia ? 6 years, 3 months ago
Since {tex}(3y-1), (3y+5)\ and\ (5y+1){/tex} are in AP, we have
{tex}(3y+5)-(3y-1)=(5y+1)-(3y+5){/tex}
{tex} \Rightarrow {/tex} {tex}3y+5-3y+1=5y+1-3y-5{/tex}
{tex} \Rightarrow {/tex} {tex}6=2y-4{/tex}
{tex} \Rightarrow {/tex} 2y=10
{tex} \Rightarrow {/tex} y=5
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