Question

the 4 numbers in A.P. whose sum is 24 and their product is 945 are

Answer 1

Answer:

3, 5, 7, 9 or 6 - 3✓39, 6 - ✓39, 6 + ✓39, 6 + 3✓39

Step-by-step explanation:

Let the numbers be a - 3d, a - d, a + d, a + 3d.

Hence, as given,

( a - 3d ) + ( a - d ) + ( a + d ) + ( a + 3d ) = 24

4a = 24

a = 6

Also,

( a - 3d ) × ( a - d ) × ( a + d ) × ( a + 3d ) = 945

( a - 3d) ( a + 3d) × ( a - d ) ( a + d) = 945

Applying ( a - b ) ( a + b ) = a^2 - b^2,

( a^2 - 9d^2 ) ( a^2 - d^2) = 945

Substituting a = 6,

( 36 - 9d^2 ) ( 36 - d^2 ) = 945

1296 - 360d^2 + 9d^4 = 945

144 - 8d^2 + d^4 = 105

d^4 - 40d^2 + 39 = 0

d^4 - 39d^2 - d^2 + 39 = 0

d^2 ( d^2 - 39 ) - 1 ( d^2 - 39 ) = 0

( d^2 - 1 ) ( d^2 - 39 ) = 0

d = ±1 or ± ✓39

Hence, AP = 6 - 3 ( 1 ), 6 - ( 1 ), 6 + ( 1 ), 6 + 3 ( 1 ) = 3, 5, 7, 9

Also, AP = 6 - 3 ( ✓39 ), 6 - ( ✓39 ), 6 + ( ✓39 ), 6 + 3 ( ✓39 )

= 6 - 3✓39, 6 - ✓39, 6 + ✓39, 6 + 3✓39

P. S. Taking positive or negative values of d, both give the same terms.