Question

Divide 32 into four parts which are in A. P such that the product of extremes is to the product of means is 7:15.​

Answer 1

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Explanation:

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Answer 2

Given Question :-

Divide 32 into four parts which are in A. P such that the product of extremes is to the product of means is 7:15.

Solution :-

Let 32 be divided into four parts such as,

(a-3d),(a-d),(a+d),(a+d) and (a+3d).

As the sum of the numbers is 32,

Therefore:-

By adding all the numbers we have,

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

4a = 32

$a \: = \: \frac{32}{4} = 8$

Now According to the question,

$\frac{product \: of \: extremes}{product \: of \: means} = \frac{7}{15}$

So here the extremes are (a-3d)×(a+3d).

Means are (a-d) ×(a+d).

Therefore,

$\frac{(a - 3d)(a + 3d)}{(a - d)(a + d)} = \frac{7}{15}$

$\frac{ {a}^{2} - 9 {d}^{2} }{ {a}^{2} - {d}^{2} } = \frac{7}{15}$

By putting the value of a =8,we get

$\frac{64 - 9 {d}^{2} }{64 - {d}^{2} } = \frac{7}{15}$

$128 {d}^{2} = 512$

${d}^{2} = \frac{512}{128} = 4$

$d \: = \: 2$

Therefore:-

The four parts are 2,6,10 and 14 respectively.