Question

The sum of four numbers in an A.P. is 24. If the difference between the product of
means and the product of extremes is 8, find the numbers.​

Answer 1

Heya friend,

Here we go :

Before we move on to the question,

A tip :-

when the sum of any number (say 2 or 4) in an A.P. is given we will take them in such a manner that the common difference can be eliminated out when we add them and we can directly get the value of the first term.

Here,

we'll the 4 terms as :-

a-3d , a-d , a+d and a+3d

According to the question :-

a - 3d + a - d + a + d + a + 3d = 24

4a = 24

a = 6

and,

(a+d)(a-d) - (a-3d)(a+3d) = 8

Putting a = 6, we get,

(6+d)(6-d) - (6-3d)(6+3d) = 8

36 - $d^{2}$ - 36 + 9$d^{2}$ = 8

8 $d^{2}$ = 8

$d^{2}$ = 1

d = ±1

Now,

When a = 6 and d = 1

Numbers are :-

a - 3d = 6 - 3(1)

         = 3

a - d = 6 - (1)

       = 5

a + d = 6 + (1)

        = 7

a + 3d = 6 + 3(1)

          = 9

Now,

When a =  6 and d = -1

Numbers are :-

a - 3d = 6 - 3(-1)

         = 9

a - d = 6 - (-1)

       = 7

a + d = 6 + (-1)

        = 5

a + 3d = 6 + 3(-1)

          = 3

So,

either way, The numbers are :-

3, 5, 7 and 9

[Verification :- 3 + 5 + 7 + 9 = 24 and (5×7) - (3×9) = 8]

Thanks !

#BAL #answerwithquality