Question

Divide 56 in four parts in ap such that the ratio of the product of their extermes to the product of means is 5:6

Answer 1

≡QUESTION≡

Divide 56 in four parts in AP such that the ratio of the product of their extremes to the product of means is 5:6.

                                                         

║⊕ANSWER⊕║

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

Then, sum = 56

$(a-3d) + (a-d) + (a + d) + (a + 3d) = 56\\ 4a = 56\\ a = 14$

It is given that,

$\frac{(a+3d)(a-3d)}{(a+d)(a-d)} = \frac{5}{6}$

$6(a^{2} -9d^{2} ) = 5(a^{2} -d^{2} )$

$a^{2} =49d^{2}$

d = ±2

Thus, the four parts are a – 3d, a – d, a + d, a + 3d i.e., 8, 12, 16 and 20.