The sum of four numbers in A.P. is $48,$ and the product of the extremes is to the product of the two middle terms as $27$ to $35.$ The largest term of the A.P. is

The correct option is D: $18$

Let the A.P. be $a–3d,a–d,a+d,a+3d.$

The sum of the terms $=48$

$\Rightarrow a-3d+a-d+a+d+a+3d=48$

$\Rightarrow 4a=48$

$\Rightarrow a=12$

It is given that ratio of product of extremes to the product of two middle terms $=27:35$

$\Rightarrow \frac{(12-3d)(12+3d)}{(12-d)(12+d)}=\frac{27}{35}$

$\Rightarrow \frac{9[4^2-(d{)}^2]}{[{12}^2-d^2]}=\frac{27}{35}$ [$\because (a-b)(a+b)=(a^2-b^2)$]

$\Rightarrow \frac{[16-d^2]}{144-d^2}=\frac{3}{35}$

$\Rightarrow 35(16-d^2)=3(144-d^2)$

$\Rightarrow 560-35d^2=432-3d^2$

$\Rightarrow 560-432=-3d^2+35d^2$

$\Rightarrow 128=32d^2$

$\Rightarrow \frac{128}{32}=d^2$

$\Rightarrow d^2=4$

$\Rightarrow d=\pm 2$

When $d=+2,$ The numbers are

$a+3d=12-3\times 2=6$

$a-d=12-2=10$

$a+d=12+2=14$

$a+3d=12+3\times 2=18$

When $d=-2$, the numbers are

$a+3d=12+3\times -2=6$

$a-d=12-(-2)=14$

$a+d=12+(-2)=10$

$a+3d=12+3\times (-2)=6$

Hence, the largest term is $18.$