Question

If the sum of four number in $A.P.$ be $48$ and that the product of the extremes is to the product of the means is $27$ to $35$ then the number are-

• A. $3,9,15,21$
• B. $9,5,7,3$
• C. $6,10,14,18$
• D. None of these

Answer 1

Answer: (C)

$6,10,14,18$

Let the terms of AP be $a-3d,a-d,a+d,a+3d$

so, $a-3d+a-d+a+d+a+3d=48$

$\Rightarrow 4a=48$

$\Rightarrow a=12$

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

$\Rightarrow \frac{a^2-9d^2}{a^2-d^2}=\frac{27}{35}$

$\Rightarrow 35a^2-315d^2=27a^2-27d^2$

$\Rightarrow 8a^2-288d^2=0$

$\Rightarrow 8(12{)}^2-288d^2=0$

$\Rightarrow d^2=\frac{8\times 12\times 12}{288}=4$

$d=\pm 2$

Case $1$ $\Rightarrow d=2;a=12$

Terms$=6,10,14,18$

Case $2$ $\Rightarrow d=-2,a=12$

Terms$=18,14,10,6$

Option $\left[C\right]$ is correct.