Question

There are 2 sets A and B of three numbers in AP whose sum is 15 and where D and d are the common differences such that D-d=1. Find the numbers if $\dfrac{P}{p} = \dfrac{7}{8}$ where P and p are the product of the numbers in the 2 sets respectively.$D,d > 0$

Answer 1

Assume the three numbers in AP in 2 sets A and B to be $a - D,a,a + D$ and $b - d,b,b + d$ respectively and then proceed as asked in the question.

Complete step-by-step answer:
Let the three numbers in AP in the 2 sets A and B be
$A = \\{ a - D,a,a + D\\} \\\ B = \\{ b - d,b,b + d\\} \\\$
Where a and b are the first terms and D and d are the common differences of the two APs respectively.
Since the sum of the three terms of the two APs=15 as given in the question
$\Rightarrow$ $a - D + a + a + D = 15$ and $b - d + b + b + d = 15$
$\Rightarrow$ $3a = 15$ And $3b = 15$
$\Rightarrow$ $a = 5$ And $b = 5$
As given in question,
$\dfrac{P}{p} = \dfrac{7}{8}$ Where P and p are the product of three numbers in 2 sets A and B respectively.
$\Rightarrow$ $\dfrac{{(a - D)a(a + D)}}{{(b - d)b(b + d)}} = \dfrac{7}{8}$
$\Rightarrow$ $\dfrac{{(5 - D)5(5 + D)}}{{(5 - d)5(5 + d)}} = \dfrac{7}{8}$
$\Rightarrow$ $\dfrac{{(25 - {D^2})}}{{(25 - {d^2})}} = \dfrac{7}{8}$
$\Rightarrow$ $8(25 - {D^2}) = 7(25 - {d^2})$
$\Rightarrow$ $200 - 8{D^2} = 175 - 7{d^2}$
$\Rightarrow$ $7{d^2} - 8{D^2} + 25 = 0$
It is given that $D - d = 1$
$\Rightarrow$ $7{d^2} - 8{(1 + d)^2} + 25 = 0$
$\Rightarrow$ $7{d^2} - 8(1 + {d^2} + 2d) + 25 = 0$
$\Rightarrow$ $7{d^2} - 8 - 8{d^2} - 16d + 25 = 0$
$\Rightarrow$ ${d^2} + 16d - 17 = 0$
On solving the quadratic equation we get,
$d = \dfrac{{ - 16 \pm \sqrt {{{16}^2} - 4(1)( - 17)} }}{{2(1)}}$
$\Rightarrow$ $d = \dfrac{{ - 16 \pm \sqrt {324} }}{2}$
$\Rightarrow$ $d = \dfrac{{ - 16 \pm 18}}{2}$
$\Rightarrow$ $d = 1$ And $d = - 17$
Since d>0 $\Rightarrow$ d=1
$\Rightarrow$ $D = 1 + d$
$\Rightarrow$ $D = 2$
Hence the numbers in AP are
$A = \\{ 3,5,7\\} \\\ B = \\{ 4,5,6\\} \\\$

Note: Always prefer to assume three numbers in AP as $a - D,a,a + D$ rather than $a,a + D,a + 2D$ as it will simplify your calculations and you will be able to solve any problem easily.