Question

Divide 20 into 4 parts which are in A. P. and such that the product of first and fourth is to
product of the second and third in the ratio 2 : 3.
IAns.2,4,6,8या8,6,4,2)
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Answer 1

Step-by-step explanation:

Let the terms are a - 3d, a - d, a + d, a + 3d. Here, common differemce is 2d.

Sum of parts = 20

=> (a - 3d) + (a - d) + (a + d) + (a + 3d) = 20

=> 4a = 20

=> a = 5

Ratio of product of 1st & 4th to 2nd & 3rd is 2:3.

=> (a - 3d)(a + 3d) : (a - d)(a + d) = 2:3

=> (a² - (3d)²) : (a² - d²) = 2:3

=> (5² - 9d²) / (5² - d²) = 2/3

=> 3(25 - 9d²) = 2(25 - d²)

=> 75 - 27d² = 50 - 2d²

=> 75 - 50 = 27d² - 2d²

=> 25 = 25d²

=> ± 1 = d

Hence terms are,

a - 3d = 5 - 3(±1) = 5-3 = 2 or 5+3 = 8

a - d = 5 - (±1) = 5-1 = 4 or 5+1 = 6

a + d = 5 + (±1) = 5+1=6 or 5-1 = 4

a + 3d= 5+ 3(±1) = 5+3=8 or 5-3 = 2

Thus, AP is either 2,4,6,8 or 8,6,4,2.