Question

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.
​

Answer 1

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.

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.

➝ $\sf Sum \:of\: all \:parts = 56$

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

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

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

➝ $\sf 4a = 56$

➝ $\sf a = \dfrac{56}{4}$

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

➝ $\sf \pink{ a = 14}$

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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