Question

divide 16 into 4 parts which are in A.P. such that the product of extremes is one less than sum of means.

Answer 1

Hope it is helpful.....

Answer 2

Solution :-

Let the four part be (a - 3d), (a + d), (a - d) and (a + 3d)

Sum of these four parts is 16

⇒ (a - 3d) + (a + d) + (a - d) + (a + 3d) = 16

⇒ - 3d and + 3d are cancelled and also - d and + d are cancelled.

⇒ 4a = 16

⇒ a = 16/4

⇒ a = 4

Now,

Sum of the means (middle part) = (a + d) + (a - d)

⇒ + d and - d are cancelled.

⇒ 2a

= 2*4

= 8

Sum of the means is 8

Product of the extreme (both ends) = (a + 3d)*(a - 3d) 

⇒ (4 + 3d)*(4 - 3d)

= 16 - 9d² 

Now, according to the question.

⇒ 16 - 9d² = 8 - 1

⇒ - 9d² = 7 - 16

⇒ - 9d² = - 9

⇒ 9d² = 9

⇒ d² = 9/9

⇒ d² = 1

⇒ d = 1

Now, substituting the values of a and d.

⇒ a + 3d 

⇒ 4 + 3*1

= 7

a + d

⇒ 4 + 1

= 5

a - d

⇒ 4 - 1

= 3

a - 3d

⇒ 4 - 3*1

⇒ 4 - 3

= 1

So, four parts are 7, 5, 3, and 1 respectively, which are in an AP.

Answer.