Question

12. Divide 32 into four parts which are the four terms of an AP such that the
product of the first and the fourth terms is to the product of the second
and the third terms as 7:15.
ICBSE 2014​

Answer 1

In case of 4 terms, we usually take 2d as the common difference and the average of the two middle terms should be a.

Therefore, Let a-3d, a-d, a+d and a+3d be the 4 terms.

Now, By Question, we have

(a-3d) + (a-d) + (a+d) + (a+3d) = 32

=> 4a = 32

=> a = 8

Again, By Question, we have

(a-3d)*(a+3d) : (a-d)*(a+d) = 7 : 15

=> a^2 - 9d^2 : a^2 - d^2. = 7 : 15

=>15(a^2 - 9d^2) = 7(a^2 - d^2)

=>15a^2 - 135d^2. = 7a^2 - 7d^2

=>15a^2 - 7a^2. = 135d^2 - 7d^2

=>8a^2. = 128d^2

=>8(8)^2. = 128d^2 (Substituting the value of a)

=>512. = 128d^2

=>4. = d^2

=>d =(+)(-)2

Therefore Common Difference = 2d = (+)(-)4

Therefore, The required AP is -

a-3d = 8 - 3(2) = 2. Or. a-3d = 8-3(-2) = 14

a-d. = 8 - 2. = 6. Or. a-d. = 8-(-2) = 10

a+d. = 8+2. = 10. Or. a+d. = 8+(-2) = 6

a+3d = 8+3(2) = 14. Or. a+3d = 8+3(-2) = 2

Note: In one case the Common Difference is 4 where as in the 2nd case the Common Difference is -4.