1.

Three positive integers x , y , z , are in AP such that sum of x, y and z is 33 and product of x, y and z is 1155 then the value of x is

Answer»

In the above Question, the following information is given -

Three POSITIVE integers x , y , Z , are in AP such that sum of x, y and z is 33 and PRODUCT of x, y and z is 1155 .

To find -

The value of x is -

Solution -

Here ,

Three positive integers x , y , z , are in AP such that sum of x, y and z is 33 and product of x, y and z is 1155 .

Let y = y

x = ( y - d )

x = ( y + d )

Now ,

Three positive integers x , y , z , are in AP such that sum of x, y and z is 33

So ,

( y - d ) + y + ( y + d ) = 33

=> 3y = 33

=> y = 11

Product -

=> ( y - d )( y + d ) y

=> ( 11 - d )( 11 + d ) × 11

=> ( 121 - d² ) × 11

=> 1331 - 11d²

But , this is equal to 1155 .

So ,

1331 - 11d² = 1155

=> 11 d² = 176

=> d² = 16

=> d = \neq 4

Thus , x can be 15 as well as 7 .

This is the required answer .

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