Question

Divide 64 into 4 parts such that all the 4 parts are in A.P and the ratio of the product of the extremes to the product of the means is 7:15. Wich of the following could be the common difference of the A.P?

• A. 4
• B. 8
• C. 2
• D. 6

Answer 1

The correct option is B 8

Let the four terms be a-3d, a-d, a+d, a+3d in which the first term and common difference be 'a-3d' and '2d' respectively.
Sum $=4a=64\to a=16$
Product of extremes $=(a-3d)(a+3d)=a^2-9d^2$
Product of means $=(a-d)(a+d)=a^2-d^2$
Ratio $=\frac{a^2-9d^2}{a^2-d^2}=\frac{7}{15}\to 15a^2-135d^2=7a^2-7d^2\to 8a^2=128d^2\to a^2=16d^2\to a=\pm 4d$
$\to 16=\pm 4d$
$\to d=\pm 4$

Common difference, $=a-d-(a-3d)=2d$
$=2\times \pm 4$
$=\pm 8$

Only 8 is there in the options.
So the coomon difference = 8