Question

Divide 32 into four parts which are in AP such that product of extremes to the product of means is 7:15 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−3d)(a+3d)     =  15/  7

(a−d)(a+d)  

⟹  a2  −9d2

a 2  −9d2  =  15/7

⟹  64−d2   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:

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)

= 7/15

⟹  

a2−d2  

a2−9d2

= 7/15

⟹  64−d2

64−9d2

=  7/15

⟹128d2

=512⟹d2

=4⟹d=±2

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