Answer
Verified
398.7k+ views
Hint: 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 $$D = 2$
Hence the numbers in AP are
$
A = \{ 3,5,7\} \\
B = \{ 4,5,6\} \\
$
Note: Always prefer to assume three numbers in AP as $a - D,a,a + D$ rather than $a,a + D,a + 2D$ as it will simplify your calculations and you will be able to solve any problem easily.
Complete step-by-step answer:
Let the three numbers in AP in the 2 sets A and B be
$
A = \{ a - D,a,a + D\} \\
B = \{ b - d,b,b + d\} \\
$
Where a and b are the first terms and D and d are the common differences of the two APs respectively.
Since the sum of the three terms of the two APs=15 as given in the question
$ \Rightarrow $ $a - D + a + a + D = 15$ and $b - d + b + b + d = 15$
$ \Rightarrow $$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.
Recently Updated Pages
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
How do you arrange NH4 + BF3 H2O C2H2 in increasing class 11 chemistry CBSE
Is H mCT and q mCT the same thing If so which is more class 11 chemistry CBSE
What are the possible quantum number for the last outermost class 11 chemistry CBSE
Is C2 paramagnetic or diamagnetic class 11 chemistry CBSE
Trending doubts
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Difference Between Plant Cell and Animal Cell
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Change the following sentences into negative and interrogative class 10 english CBSE
Give 10 examples for herbs , shrubs , climbers , creepers
What organs are located on the left side of your body class 11 biology CBSE
Write an application to the principal requesting five class 10 english CBSE
What is the type of food and mode of feeding of the class 11 biology CBSE
Name 10 Living and Non living things class 9 biology CBSE