Question

There are two sets of numbers each consisting of 3 terms in A.P. and the sum of each set of numbers is 12. The common difference of the first set is greater by 1 than the common difference of the second set. The product of the first set of numbers and the product of the first set of numbers is in the ratio 5:4. Find the numbers in both the sets.

Answer 1

Hint: Given numbers are in A.P. So we will consider the numbers of the first set be in a form and common difference in D form and for second set numbers in a form and common difference in d form. Then applying the conditions of sum and product given we will find the common differences. This at the end will give the numbers in both the sets.

Complete step-by-step answer:
Given that there are two sets of numbers each consisting of three terms and those are in A.P.
Let the numbers in the first series be a-D, a, a+D.
And of second series be a-d, a, a+d
From the condition sum of each set of numbers is 12 we can write

Set 1	Set 2
\[ \Rightarrow a - D + a + a + D = 12\]Cancelling D we get,\[ \Rightarrow 3a = 12\]Dividing 12 by 3 we get,\[ \Rightarrow a = \dfrac{{12}}{3}\]\[ \Rightarrow a = 4\]	\[ \Rightarrow a - d + a + a + d = 12\]Cancelling d we get,\[ \Rightarrow 3a = 12\]Dividing 12 by 3 we get,\[ \Rightarrow a = \dfrac{{12}}{3}\]\[ \Rightarrow a = 4\]

Here we found the value of a.
Now we have one more condition given that, the product of first set of numbers and the product of first set of numbers is in the ratio 5:4
Thus we can write,
\[ \Rightarrow \dfrac{{\left( {a - D} \right)\left( a \right)\left( {a + D} \right)}}{{\left( {a - d} \right)\left( a \right)\left( {a + d} \right)}} = \dfrac{5}{4}\]
Cancelling a we get,
\[ \Rightarrow \dfrac{{\left( {a - D} \right)\left( {a + D} \right)}}{{\left( {a - d} \right)\left( {a + d} \right)}} = \dfrac{5}{4}\]
This we can observe as \[\left( {a - D} \right)\left( {a + D} \right) = {a^2} - {D^2}\] and \[\left( {a - d} \right)\left( {a + d} \right) = {a^2} - {d^2}\]. So by putting this value we get,
\[ \Rightarrow \dfrac{{{a^2} - {D^2}}}{{{a^2} - {d^2}}} = \dfrac{5}{4}\]
On cross multiplying we get,
\[ \Rightarrow 4({a^2} - {D^2}) = 5({a^2} - {d^2})\]
Multiplying the terms we get,
\[ \Rightarrow 4{a^2} - 4{D^2} = 5{a^2} - 5{d^2}\]
Separating the terms,
\[ \Rightarrow 5{d^2} - 4{D^2} = 5{a^2} - 4{a^2}\]
Then we get,
\[ \Rightarrow 5{d^2} - 4{D^2} = {a^2}\]
But one more condition is that, the common difference of the first set is greater by 1 than the common difference of the second set
So we can write, D-d=1 or D=d+1
So putting this value in the equation formed above
\[ \Rightarrow 5{d^2} - 4{(d + 1)^2} = {a^2}\]
On solving we get,
\[ \Rightarrow 5{d^2} - 4({d^2} + 2d + 1) = {a^2}\]
\[ \Rightarrow 5{d^2} - 4{d^2} - 8d - 4 = {a^2}\]
\[ \Rightarrow {d^2} - 8d - 4 = {a^2}\]
Now putting the value of a
\[ \Rightarrow {d^2} - 8d - 4 = {4^2}\]
\[ \Rightarrow {d^2} - 8d - 4 = 16\]
\[ \Rightarrow {d^2} - 8d - 4 - 16 = 0\]
\[ \Rightarrow {d^2} - 8d - 20 = 0\]
On solving this quadratic equation,
\[ \Rightarrow {d^2} - 10d + 2d - 20 = 0\]
Taking d common from first two terms and 2 common from last two terms,
\[ \Rightarrow d(d - 10) + 2(d - 10) = 0\]
The two brackets will be,
\[ \Rightarrow (d - 10)(d + 2) = 0\]
Thus values of d can be d=10 or d=-2.
Now let’s tabulate the data and find the series.

Value of d	Value of D	SET1	SET2
d=10	D=d+1=11	a-D=4-11=-7a=4a+D=4+11=15set is→-7,4,15	a-d=4-10=-6a=4a+d=4+10=14set is →-6,4,14
d=-2	D=d+1=-1	a-D=4-(-1)=5a=4a+D=4+(-1)=3set is →5,4,3	a-d=4-(-2)=6a=4a+d=4+(-2)=2set is →6,4,2

Thus we found the two sets of numbers.
Note: Remember students that if we have taken numbers as a, a+d and a+2d it will make the problem more time consuming and complicated. That is the only reason we have taken numbers as a-d, a and a+d.
We can also take $d_1$ and $d_2$ as the common differences.
We actually get four different sets because we have 2 different values of common differences.