Determine k so that k+2,k-6,3k-2 are …
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Posted by Vinit Chahar 6 years, 5 months ago
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Sia ? 6 years, 5 months ago
Since {tex}(3k-2), (4k-6)\ and \ (k+2){/tex} are in AP, we have
{tex}(4k-6)-(3k-2)=(k+2)-(4k-6){/tex}
{tex}\Rightarrow{/tex} {tex}4k - 6 - 3k + 2 = k + 2 - 4k + 6{/tex}
{tex}\Rightarrow{/tex} {tex}k - 4 = -3k + 8{/tex}
{tex}\Rightarrow{/tex} 4k = 12
{tex}\Rightarrow{/tex} k = 3.
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