Question

Dived 56 into 4 parts which are in AP such that the ratio of product of extremes to the product of mean is 5:6

Answer 1

$<i>$Let the numbers be ( a - 3d ), ( a - d ),
( a + d ) and ( a + 3d ).

A/q

Sum of these numbers = 56

•°•

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

4a = 56

a = 56 / 4

a = 14

Also,

Product of extremes / Product of means = 5 / 6 { Given }

[ ( a - 3d ) ( a + 3d ) ] / [ ( a - d ) ( a + d ) ] = 5 / 6

{ Using identity -

( a - b ) ( a + b ) = a² - b² }

•°•

[ ( a )² - ( 3d )² ] / [ ( a )² - ( d )² ] = 5 / 6

Putting value of a

[ ( 14 )² - ( 3d )² ] / [ ( 14 )² - ( d )² ] = 5 / 6

[ 196 - 9d² ] / [ 196 - d² ] = 5 / 6

Cross multiplication

6 [ 196 - 9d² ] = 5 [ 196 - d² ]

1176 - 54d² = 980 - 5d²

1176 - 980 = - 5d² + 54d²

196 = 49d²

d² = 196 / 49

d² = ( 14 )² / ( 7 )²

d² = [ 14 / 7 ]²

d = 14 / 7

d = 2

Now,

First term = a - 3d = 14 - 3 ( 2 ) = 14 - 6 = 8

Second term = a - d = 14 - 2 = 12

Third term = a + d = 14 + 2 = 16

Fourth term = a + 3d = 14 + 3 ( 2 ) = 14 + 6 = 20