There are two sets A and B each of which consists of three numbers in A.P.whose sum is 15 and where D and d are the common differences such that `D-d=1. If p/q=7/8`, where p and q are the product of the numbers ,respectively, and ` d gt 0` in the two sets .
The sum of the products of the numbers is set A taken two at at time is

Options

A. 162
B. 64
C. 32
D. 2

Answer

A, C

Explanation

Given `a_(1)=2,(a_(n))/(a_(n-1))=(a_(n-1))/(a_(n-2))`
Hence,`a_(1),a_(2),a_(3),a_(4),a_(5),..` in G.P.
Let `a_(2)=x`. Then for n=3, we have
`(a_(3))/(a_(2))=(a_(2))/(a_(1))`
`rArra_(2)^(2)=a_(1)a_(3)`
`rArra_(3)=x^(2)/2`
i.e., `2,x,x^(2)/2,x^(3)/4,x^(4)/8`,... with common ratio `r=x/2`
Given `x^(4)/8` are integers.
So if x is even, then only `x^(4)/8` will be an integer.
Hence, possible value of x is 4 and 6. (`xne2` as terms are distinct)
Hence, possible value of `a_(5)=x^(4)/8` is `4^(4)/8,6^(4)/8`