For what value of k will $k+9,\ 2k-1$ and $2k+7$ are consecutive terms of A.P.
Given: $k+9,\ 2k-1$ and $2k+7$ are consecutive terms of A.P.
To do: To find the value of k.
Solution:
Let us assume,
$k+9=a$
$2k-1=b$
$2k+7=c$
To be in AP,
$a+b=2b$
$\Rightarrow k+9+2k+7=2( 2k-1)$
$\Rightarrow 3k+16=4k-2$
$\Rightarrow 3k-4k=-2-16$
$\Rightarrow -k=-18$
$\therefore k=18$
For $k=18$, the terms $k+9,\ 2k-1,\ 2k+7$ are in A.P
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