Question

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 .

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)

=>$512.=128d^2$
=>$4.=d^2$
=>$d=+2,-2$

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.