Question:
Find a, b and c such that the following numbers are in AP, a, 7, b, 23 and c.
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, $a=-1, b=15, c=31$