Divide 32 into 4 parts which are in AP such that the ratio of the product of the extremes to the product of the means is 7:5 .

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- Mar 8th 2010, 10:24 PMprantik007Arithmetic progression and ratio of terms.
Divide 32 into 4 parts which are in AP such that the ratio of the product of the extremes to the product of the means is 7:5 .

- Mar 9th 2010, 09:26 AMqmech
If I'm interpreting your problem correctly, something is wrong in your problem.

Consider the sequence:

a-3d,a-d,a+d,a+3d

This is an arithmetic progression with constant difference 2d.

The product of the 2 outer terms is a^2 - 9d^2. The product of the 2 inner terms is a^2 - d^2.

Your requested ratio is:

$\displaystyle \frac{a^2-9d^2}{a^2-d^2} = \frac {7}{5}$

which leads to

$\displaystyle -38d^2 = 2a^2$

which is impossible. - Mar 9th 2010, 07:23 PMprantik007
But I got this problem in my exam

- Mar 10th 2010, 11:19 AMSoroban
Hello, prantik007!

Please check the problem for typos.

As stated, the situation is patently impossible.

Quote:

Divide 32 into 4 parts which are in AP such that the ratio of the product

of the extremes to the product of the means is 7:5 .??

The four parts are: .$\displaystyle a,\;a+d,\;a+2d,\;a+3d$

The product of the extremes is: .$\displaystyle P_E \:=\:a(a+3d) \:=\:a^2 + 3ad$

The product of the means is: .$\displaystyle P_M \:=\:(a+d)(a+2d) \:=\:a^2 + 3ad + 2d^2$

. . It isthat: .$\displaystyle P_E \:<\:P_M$*obvious*

So how can . $\displaystyle P_E:P_M \:=\:{\color{red}7:5}$ ?