Math, asked by harnoor4662, 5 hours ago

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

Answers

Answered by amansharma264
303

EXPLANATION.

56 into four parts.

The ratio of the products of their extremes to the products of means = 5 : 6.

As we know that,

Four terms of an A.P.

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

⇒ a - 3d + a - d + a + d + a + 3d = 56.

⇒ 4a = 56.

⇒ a = 14.

⇒ (a - 3d)(a + 3d)/(a - d)(a + d) = 5/6.

As we know that,

Formula of :

⇒ (x² - y²) = (x - y)(x + y).

Using this formula in the equation, we get.

⇒ (a² - 9d²)/(a² - d²) = 5/6.

⇒ 6(a² - 9d²) = 5(a² - d²).

⇒ 6[(14)² - 9d²] = 5[(14)² - d²].

⇒ 6[196 - 9d²] = 5[196 - d²].

⇒ 1176 - 54d² = 980 - 5d².

⇒ 1176 - 980 = - 5d² + 54d².

⇒ 196 = 49d².

⇒ d² = 4.

⇒ d = √4.

⇒ d = ± 2.

Case = 1.

When a = 14 and d = 2.

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

⇒ [14 - 3(2)], [14 - 2], [14 + 2], [14 + 3(2)].

⇒ (14 - 6), (14 - 2), (14 + 2), (14 + 6).

⇒ (8), (12), (16), (20).

Case = 2.

When a = 14 and d = - 2.

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

⇒ [14 - 3(-2)], [14 - (-2)], [14 + (-2)], [14 + 3(-2)].

⇒ [14 + 6], [14 + 2], [14 - 2], [14 - 6].

⇒ (20), (16), (12), (8).

                                                                                                                       

MORE INFORMATION.

Supposition of an A.P.

(1) = Three terms as : a - d, a, a + d.

(2) = Four terms as : a - 3d, a - d, a + d, a + 3d.

(3) = Five terms as : a - 2d, a - d, a, a + d, a + 2d.

Answered by MяMαgıcıαη
467

\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

\small\sf (a - 3d) + (a - d) + (a + d) + (a + 3d) = 56

\sf a - 3d + a - d + a + d + a + 3d = 56

\sf a + a + a + a - d + d - 3d + 3d = 56

\sf 4a = 56

\sf a = \dfrac{56}{4}

\sf a = {\cancel{\dfrac{56}{4}}}\:(Cancelling)

\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)} = \dfrac{5}{6}

\sf \dfrac{a^2 - (3d)^2}{a^2 - d^2} = \dfrac{5}{6}

\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 6a^2 - 54d^2 = 5a^2 - 5d^2)

\sf 6a^2 - 5a^2 = - 5d^2 + 54d^2

\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\pink{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{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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