Question

For what value of $k$ will $k+9,2k-1$ and $2k+7$ are the consecutive terms of an A.P?

Answer 1

Let

$k+9=a$

$2k-1=b$

$2k+7=c$

To be in AP,

$a+c=2b$

$k+9+2k+7=2(2k-1)$

$\Rightarrow$ $3k+16=4k-2$

$\Rightarrow$ $3k-4k=-2-16$

$\Rightarrow$ $-k=-18$

$\therefore k=18$

For $k=18$, the terms $k+9,2k-1,2k+7$ are in A.P