Question

Divide 56 in four parts in A.P such that the ratio of the product of their extremes (1st and 4th) to the product of means (2nd and 3rd) is 5:6.

Answer 1

done...................

Answer 2

$\large\underline{\sf{\red{Given}}}$

✭ 56 is divided into four parts which form an A.P

✭ Ratio of the product of extremes of given A.P to the product of their means is 5 : 6

$\large\underline{\sf{\blue{To\:Find}}}$

✭ All the parts of 56?

$\large\underline{\sf{\gray{Solution}}}$

Things to know before solving this question,

$\underline{\boxed{\sf{a^2 - b^2 = (a + b)(a - b)}}}$

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Now,

Let the four parts be (a - 3d), (a - d), (a + d), (a + 3d) such that they are in A.P.

$⇝\sf Sum \:of\: all \:parts = 56 \\ ⇝\sf \: \small\sf (a - 3d) + (a - d) + (a + d) + (a + 3d) = 56 \\⇝ \sf a - 3d + a - d + a + d + a + 3d = 56 \\ ⇝\sf \: \sf a + a + a + a - d + d - 3d + 3d = 56 \\⇝ \sf \: \sf 4a = 56 \\ ⇝ \sf \: \sf a = \dfrac{56}{4}\sf a = {\cancel{\dfrac{56}{4}}}\:(Cancelling) \\ ⇝ \sf \: \sf \pink{ a = 14}$

Now,

$\underline{\sf{\bigstar\:According\:to\:the\:given\:Question\::}}$

$\sf \dfrac{Product\:of\:extremes\:of\:A.P}{Product\:of\:means\:of\:A.P} = \dfrac{5}{6}$

i.e,

$\sf \dfrac{Product\:of\:1^{st}\:and\:4^{th}\:terms\:of\:A.P}{Product\:of\:2^{nd}\:and\:3^{rd}\:terms\:of\:A.P} = \dfrac{5}{6}\sf \dfrac{(a - 3d)(a + 3d)}{(a - d)(a + d)} \\ \\ \sf= \dfrac{5}{6}\sf \dfrac{a^2 - (3d)^2}{a^2 - d^2} = \dfrac{5}{6}\\ \\ \sf\sf \dfrac{a^2 - 9d^2}{a^2 - d^2} = \dfrac{5}{6}$

By cross multiplication,

$\sf 6(a^2 - 9d^2) = 5(a^2 - d^2) \\ \sf \: \sf 6a^2 - 54d^2 = 5a^2 - 5d^2)\\ \sf \: \sf 6a^2 - 5a^2 = - 5d^2 + 54d^2\\ \sf \: \sf a^2 = 49d^2$

Putting the value of 'a' in above eqⁿ,

$→\sf {14}^{2} = 49d^2 \\ \\ →\sf 196 = 49d^2 \\ \\→ \sf \dfrac{196}{49} = d^2 \\ \\→ \sf \sqrt{\dfrac{196}{49}} = d \\ \\ →\sf \sqrt{\dfrac{14\:\times\:14}{7\:\times\:7}} = d \\ \\ →\sf \dfrac{14}{7} = d \\ \\→ \sf d = {\cancel{\dfrac{14}{7}}}\:(Cancelling) \\ \\→ \sf\red{d = \pm 2}$

Now,

Case - 1, when d = 2

⇉$\small\sf 1^{st}\: Part = a - 3d = 14 - 3(2) = 14 - 6 = \bf{8}$

⇉$\small\sf 2^{nd}\:Part = a - d = 14 - 2 = \bf{12}$

⇉$\small\sf 3^{rd}\:Part = a + d = 14 + 2 = \bf{16}$

⇉$\small\sf 4^{th}\: Part = a + 3d = 14 + 3(2) = 14 + 6 = \bf{20}$

Case - 2, when d = -2

⇉$\small\sf 1^{st}\: Part = a - 3d = 14 - 3(2) = 14 - 6 = \bf{8}$

⇉$\small\sf 2^{nd}\:Part = a - d = 14 - 2 = \bf{12}$

⇉$\small\sf 3^{rd}\:Part = a + d = 14 + 2 = \bf{16}$

⇉$\small\sf 4^{th}\: Part = a + 3d = 14 + 3(2) = 14 + 6 = \bf{20}$

Case - 2, when d = -2

⇉$\small\sf 1^{st}\:Part = a - 3d = 14 - 3(-2) = 14 + 6 =\bf{20}$

⇉$\small\sf 2^{nd}\:Part = a - d = 14 - (-2) = 14 + 2 = \bf{16}$

⇉$\small\sf 3^{rd}\:Part = a + d = 14 + (-2) = 14 - 2 = \bf{12}$

⇉$\small\sf 4^{th}\: Part = a + 3d = 14 + 3(-2) = 14 - 6 = \bf{8}$

$\therefore\:{\underline{\sf{All\:the\:parts\:of\:56\:=\:\bf{8,\:12,\:16,\:20}}}}$

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