Question

The sum of four numbers in an arithmetic progression is $48$ and the ratio of the product of the extremes to the product of the two middle terms is $27:35$ , then the largest term of this arithmetic progression is

• A. $10$
• B. $12$
• C. $14$
• D. $18$

Answer 1

Answer: (D)

Let the A. P. be $a-3d,a-d,a+d,a+3d$
The sum of the terms $=48=4a\Rightarrow a=12$
Given, $\frac{\left(12-3d\right)\left(12+3d\right)}{\left(12-d\right)\left(12+d\right)}=\frac{27}{35}$
$\Rightarrow \frac{9\left(4-d\right)\left(4+d\right)}{\left(12-d\right)\left(12+d\right)}=\frac{27}{35}$
$\Rightarrow \left(16-d^2\right)35=\left(144-d^2\right)3$
$\Rightarrow 35d^2-3d^2=16\times 35-144\times 3$
$\Rightarrow 32d^2=16\left(35-27\right)=16\times 8$
$\Rightarrow d^2=4\Rightarrow d=\pm 2$
So the numbers are: $6,10,14,18$