Question

Find the A.P whose $3^{rd}$ term is $16$ and the $7^{th}$ term exceeds its $5^{th}$ term by $12$

Answer 1

Third term$=a+2d=16$

Seventh term$-$fifth term$=12$

$\Rightarrow a+6d-a-4d=12$

$\Rightarrow 2d=12$

$\Rightarrow d=\frac{12}{2}=6$

We have $a+2d=16\Rightarrow a+2\times 6=16$

$\Rightarrow a=16-12=4$

Hence the A.P is $4,4+6,4+2\times 6,...$

or $4,10,16,...$