There are two sets A and B each of which consists of three numbers in AP whose sum is 15 and where D and d are the common differences such that `D=1+d,dgt0.` If `p=7(q-p)`, where p and q are the product of the nymbers respectively in the two series.
The value of `7D+8d` is

Options

A. 37
B. 22
C. 67
D. 52

Answer

B

Explanation

Let `A={A-D,A,A+D},B={a-d,a,a+d}`
according to the question,
`A-D+A+A+D=15`
`implies 3A=15`
`implies A=15" " "…….(i)"`
and `a-d+a+a+d=15`
`implies a=5" "………(ii)"`
and `D=1+d" " "……..(iii)"`
`p=(A-D)A(A+D)`
`p=A(A^(2)-D^(2))" " ......(iv)"`
`p=5(25-D^(2))" " "........(v)"`
Similarly, `q=5(25-d^(2))`
Given that, `p=7(q-p)`
`8p=7q`
From Eqs. (iv) and (v), we get
`8xx5(25-D^(2))=7xx5(25-d^(2))`
`200-8D^(2)=175-7d^(2)`
`25=8D^(2)-7d^(2)`
`25=8(1+d)^(2)-7d^(2)" " [" from Eq. (iii) "]`
`25=8+8d^(2)+16d-7d^(2)`
`17-d^(2)-16d=0`
`d^(2)+16d-17=0`
`(d+17)(d-1)=0`
`d=-17` or `d=1`
`implies d=1 " " [:.dgt0]`
`impliesD=2`
`7D+8d=14+8=22`