Question

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5:6 .

Answer properly step by step .
pls it urgent .

Answer 1

Hey ❗

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↪Let the AP be a - d, a + d, a + 3d, a - 3d

↪According to the question

↪a - d + a + d + a + 3d + a - 3d = 56

↪4a = 56

↪a = 56 / 4

↪a = 14

↪a - 3d ( a + 3d ) / a - d ( a + b ) = 5 / 6

↪a ^2 - 9d^2 / a^2 - d^2 = 5 / 6

↪Put value of a

↪14 ^2 - 9d^2 / 14^2 - d^2 = 5 / 6

↪1176 - 54d^2 = 980 - 5d^2

↪1176 - 980 = - 5d^2 + 54d^2

↪196 = 49d^2

↪196 / 49 = d^2

↪4 = d^2

↪2 = d

↪Now a = 14 , d = 2 but here d can be positive also and negative also.

↪So, If d = + 2

↪a + d = 14 + 2 = 16

↪a - d = 14 - ( + 2 ) = 12

↪a - 3d = 14 - 3 ( 2 ) = 14 - 6 = 8

↪a + 3d = 14 + 3 ( 2 ) = 14 + 6 = 20

↪And we will get same AP when d = - 2

↪Thus the AP is 8 , 12 , 16 and 20

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Answer 2

Let the terms of the AP be ( a - 3d ), ( a - d ), ( a + d ), ( a + 3d )

Ratio of Product of Extremes to Product of means = 5 : 6

Product of Extremes = ( a - 3d ) ( a + 3d ) ( Form = ( a + b ) ( a - b ) )

Product = a² - 9d² ( Since ( a + b ) ( a - b ) = a² - b²

Similarly Product of means = ( a - d ) ( a + d ) = a² - d²

Ratio = 5 : 6

=> $\frac{ a^2 - 9d^2}{ a^2 - d^2} = \frac{5}{6}$

Cross multiplying we get,

= 6 ( a² - 9d² ) = 5 ( a² - d² )

= 6a² - 54d² = 5a² - 5d²

= 6a² - 5a² = 54d² - 5d²

=> a² = 49d²

Taking square root we get,

= a = 7d

Hence according to the question

( a + 3d ) + ( a + d ) + ( a - d ) + ( a - 3d ) = 56

a + a + a + a + 3d + d - 3d - d = 56

4a + 4d - 4d = 56

=> 4a = 56

=> a = 56 / 4 = 14

We know 7d = a

=> d = a / 7 = 14 / 7 = 2

Hence the four parts are :

( 7 - 6 ) , ( 7 - 2 ) , ( 7 + 2 ) , ( 7 + 6 )

1, 5, 9, 13