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

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Solution

Let the four parts be $a - 3 d, a - d, a + d, a + 3 d$
Sum, $4 a = 56$
$a = 14$
$\frac{(a - 3 d) (a + 3 d)}{(a - d) (a + d)} = \frac{5}{6}$
$6 (a^{2} - 9 d^{2}) = 5 (a^{2} - d^{2})$
$a^{2} = 49 d^{2}$
$d = \pm 2$
For d=2,
Numbers are $8, 12, 16, 20$ .

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Divide $56$ into $4$ parts which are in AP such that the ratio of products of extremes to the products of means is $5 : 6.$

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