Question

Find the three number of A.P. whose sum is 6 and their product is 6

Answer 1

the a.p is 0,2,4

Let the sum of three numbers in a. p be a, a+d, a-d

then,

3a=6

a=2-------eq1

similarly

Let the product of three numbers in a. p be

a(a+d) (a-d) =6

substitute a=2 (from equation 1)

8-4d+4d-2d^2

-2d+8

d=±2

then the three numbers in ap are as follows:

0,2,4 and 2,2,4

as the second one I. e 2,2,4 doest not follow the sequences, therefore 0,2,4 is the correct answer.

Answer 2

To Find :

The first three number of the AP.

Given :

• Sum of three number of AP = 6

• Product of three numbers of AP = 6

We know :-

The three numbers of an AP is taken as :-

$\boxed{\underline{\bf{(a - d) , a , (a + d)}}}$

Where :-

• a = First Term of the AP .

• d = Common Difference

Concept :-

According to the Question , sum of three numbers are given , and then it is given that the product of the three numbers of the AP is 6.

So by the first information that sum of numbers , we can find the first term.

Then by putting the value of a in the the second condition i.e, product of no. is 6 , we can find the common difference and then by putting the value in the three numbers we can find the required value .

Solution :-

First term of the AP (a) =

Given :-

• First Term = (a - d)

• Second Term = a

• Third Term = (a + d)

A/c,

$\underline{:\implies \bf{(a - d) + a + (a + d) = 6}}$

On solving the above equation , we get :-

$:\implies \bf{(a - d) + a + (a + d) = 6} \\ \\ \\ :\implies \bf{a - d + a + a + d = 6} \\ \\ \\ :\implies \bf{a - \not{d} + a + a + \not{d} = 6} \\ \\ \\ :\implies \bf{a + a + a = 6} \\ \\ \\ :\implies \bf{3a = 6} \\ \\ \\ :\implies \bf{a = \dfrac{6}{3}} \\ \\ \\ :\implies \bf{a = \dfrac{\not{6}}{\not{3}}} \\ \\ \\ :\implies \bf{a = 2} \\ \\ \\ \therefore \purple{\bf{a = 2}}$

Hence, the first of the AP is 2.

Common difference of the AP (d) :-

Given :-

• First Term = (a - d)

• Second Term = a

• Third Term = (a + d)

A/c,

$\underline{:\implies \bf{(a - d) \times a \times (a + d) = 6}}$

On solving the above equation , we get :-

$:\implies \bf{(a - d) \times a \times (a + d) = 6} \\ \\ \\ :\implies \bf{(a - d)(a + d) \times a = 6} \\ \\ \\ :\implies \bf{(a{^2} - ad + ad - d^{2}) \times a = 6} \\ \\ \\ :\implies \bf{(a^{2} - d^{2}) \times a = 6} \\ \\ \\ :\implies \bf{(a^{3} - ad^{2}) = 6}$

Putting the value of a i.e, First Term in the equation , we get :-

$:\implies \bf{(2^{3} - 2 \times d^{2}) = 6} \\ \\ \\ :\implies \bf{(8 - 2d^{2}) = 6} \\ \\ \\ :\implies \bf{(-2d^{2}) = 6 - 8} \\ \\ \\ :\implies \bf{- 2d^{2} = -2} \\ \\ \\ :\implies \bf{\not{-} 2d^{2} = \not{-} 2} \\ \\ \\ :\implies \bf{d^{2} = \dfrac{2}{2}} \\ \\ \\ :\implies \bf{d^{2} = \dfrac{\not{2}}{\not{2}}} \\ \\ \\ :\implies \bf{d^{2} = 1} \\ \\ \\ :\implies \bf{d = \sqrt{1}} \\ \\ \\ :\implies \bf{d = 1} \\ \\ \purple{\bf{d = \pm1}}$

Hence, the common difference of the AP is -1 and 1.

Three terms of the AP :-

We know :-

• First term = (a - d)
• Second term = a
• Third term = (a + d)

Values found :-

• First term (a) = 2
• Common difference (d) = 1

Using the three numbers and substituting the values in it ,we get :-

• First Number :-

$:\implies \bf{a_{1} = (a - d)} \\ \\ :\implies \bf{a_{1} = 2 - 1} :\implies \bf{a_{1} = 1} \\ \\ \therefore \purple{\bf{a_{1} = 1}}$

Hence, the first number is 1.

• Second Number :-

$:\implies \bf{a_{2} = a} \\ \\ :\implies \bf{a_{2} = 2} \\ \\ \therefore \purple{\bf{a_{2} = 2}}$

Hence, the second number is 2.

• Third Number :-

$:\implies \bf{a_{3} = (a + d)} \\ \\ :\implies \bf{a_{3} = 2 + 1} \\ \\ :\implies \bf{a_{1} = 3} \\ \\ \therefore \purple{\bf{a_{3} = 3}}$

Hence, the third number is 3.

The three numbers are 1 , 2 and 3.

or/

We know :-

• First term = (a - d)
• Second term = a
• Third term = (a + d)

Values found :-

• First term (a) = 2
• Common difference (d) = - 1

Using the three numbers and substituting the values in it ,we get :-

• First Number :-

$:\implies \bf{a_{1} = (a - d)} \\ \\ :\implies \bf{a_{1} = 2 - (-1)} :\implies \bf{a_{1} = 2 + 1} \\ \\ :\implies \bf{a_{1} = 3} \\ \\ \therefore \purple{\bf{a_{1} = 3}}$

Hence, the first number is 3.

• Second Number :-

$:\implies \bf{a_{2} = a} \\ \\ :\implies \bf{a_{2} = 2} \\ \\ \therefore \purple{\bf{a_{2} = 2}}$

Hence, the second number is 2.

• Third Number :-

$:\implies \bf{a_{3} = (a + d)} \\ \\ :\implies \bf{a_{3} = 2 + (-1)} \\ \\ :\implies \bf{a_{3} = 2 - 1} \\ \\ :\implies \bf{a_{3} = 1}\\ \\ \therefore \purple{\bf{a_{3} = 1}}$

Hence, the third number is 1.

The three numbers are 3 , 2 and 1.

----------------------------------------------------------------