Question

Determine the A.P. whose third term is $16$ and the $7^{th}$ term exceeds the $5^{th}$ term by $12.$

Answer 1

Given: Third term, $a_3=16$

$a+(3-1)d=16$ [$\because n^{th}$ term of an AP is given by $a_n=a+(n-1)d$]

$\Rightarrow a+2d=16\dots \dots \left(i\right)$

And, it is given that $7^{th}$ term exceeds $5^{th}$ term by $12.$

$\Rightarrow a_7-a_5=12$

$[a+(7-1\left)d\right]-[a+(5-1\left)d\right]=12$

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

$\Rightarrow 2d=12$

$\Rightarrow d=6$

From equation $\left(i\right),$ we get

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

$\Rightarrow a+12=16$

$\Rightarrow a=4$

Therefore, required A.P is $a,(a+d),(a+2d),(a+3d),.......$

$4,10,16,22,.\dots$