Question

Divide 72 in four parts in A.P, such that the ratio of the product of their extremes (1st and 4th ) to the product of means (2nd and 3rd) is 27/35 Find the four parts.​

Answer 1

Answer:

Let the four parts be (a−3d),(a−d),(a+d) and (a+3d).

Then, Sum of the numbers =32

⟹(a−3d)+(a−d)+(a+d)+(a+3d)=32⟹4a=32⟹a=8

It is given that

(a−d)(a+d)

(a−3d)(a+3d)

​

=

15

7

​

⟹

a

2

−d

2

a

2

−9d

2

​

=

15

7

​

⟹

64−d

2

64−9d

2

​

=

15

7

​

⟹128d

2

=512⟹d

2

=4⟹d=±2

Thus, the four parts are a−3d,a−d,a+d and 3d, i.e. 2,6,10,14.

Step-by-step explanation:

Answer 2

Answer:

The four parts  given AP are 9,15,21,27

Step-by-step explanation:

• Arithmetic progression is defined as progression in which every term after the first is obtained by adding a constant value called the common difference.
• For four terms, the AP is $a-3d,a-d,a+d,a+3d$

Step 1 of 1:

• Sum of all parts of AP =72  

        $a-3d,a-d,a+d,a+3d=72\\4a=72\\a=18$

• The ratio of the product of their extremes (1st and 4th ) to the product of means (2nd and 3rd) is 27/35.
• This can be represented as      

       $\frac{(a-3d)(a+3d)}{(a-d)(a+d)} =\frac{27}{35} \\\frac{a^2-9d^2}{a^2-d^2} =\frac{27}{35} \\\\35a^2-(35*9)d^2=27a^2-27d^2\\35a^2-315d^2=27a^2-27d^2\\\\8a^2=288d^2\\288d^2=8*18^2\\288d^2=2592\\d^2=9\\d=-3,3$

• Putting values of a and d, the AP is given as  9,15,21,27 and 27,21,15,9