Find a, b and c such that the

Solution:

Since a, 7, b, 23 and c are in AR

$\therefore \quad 7-a=b-7=23-b=c-23=$ Common difference

Taking second and third terms, we get

$b-7=23-b$

$\Rightarrow \quad 2 b=30$

$\therefore \quad b=15$

Taking first and second terms, we get

$7-a=b-7$

$\Rightarrow \quad 7-a=15-7 \quad[\because b=15]$

$\Rightarrow \quad 7-a=8$

$\therefore \quad a=-1$

Taking third and fourth terms, we get

$23-b=c-23$

$\Rightarrow \quad 23-15=c-23 \quad[\because b=15]$

$\Rightarrow \quad 8=c-23$

$\Rightarrow \quad 8+23=c \Rightarrow c=31$

Hence, $a=-1, b=15, c=31$