24. Find the arithmetic progression whose third term is $16$ and the seventh term exceeds its fifth term by $12.$

An arithmetic progression or arithmetic sequence is a number’s sequence such that the difference between the consecutive terms is constant.

Solution:

Given, in an A.P

${{a}_{3}}=16$ and ${{a}_{7}}={{a}_{5}}+12$

We know that ${{a}_{n}}=a+\left( n-1 \right)d$

⇒ $a+2d=16$…… $\left( i \right)$

And,

$a+6d=a+4d+12$

$2d=12$

⇒ $d=6$

Using $d$ in $\left( i \right),$ we have

$a+2\left( 6 \right)=16$

$a=16-12=4$

Hence, the A.P is $4,10,16,22,$ …….