Question

For what value of ${}^{\prime }{k}^{\prime }$, $k-3,2k+1$ and $4k+3$ are in A.P?

Answer 1

$\Rightarrow$ First term $\left(a_1\right)=k-3$, Second term $\left(a_2\right)=2k+1$ and third term $\left(a_3\right)=4k+3$

We know,

$a_2-a_1=a_3-a_2$

$\Rightarrow$ $2k+1-(k-3)=(4k+3)-(2k+1)$

$\Rightarrow$ $2k+1-k+3=4k+3-2k-1$

$\Rightarrow$ $k+4=2k+2$

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

$\therefore$ $k=2$