the sum of three numbers of AP is 27 and their product is 405 find the numbers
Answers
Solution:-
The three consecutive term of AP is ( a - d ) ( a ) ( a + d )
Now ,
The sum of three numbers of AP = 27
=> a - d + a + a + d = 27
=> 3a = 27
=> a = 27/ 3
=> a = 9
Now
The product of three number is 405
( a - d ) × ( a ) × ( a + d ) = 405
Using this identity
(a - b)(a + b ) = ( a² - b² )
We get
( a² - d² )a = 405
We have already find value of a = 9
{(9)² - d²} 9 = 405
(81 - d² ) 9 = 405
729 - 9d² = 405
-9d² = 405 - 729
-9d² = - 324
d² = 36
d = ± 6
Now
Case = 1
a = 9 and d = +6
put the value in this
( a - d ) ( a ) ( a + d )
(9 - 6) , 9 ,( 9 + 6 )
=> 3 , 9 , 15
Case = 2
a = 9 and d = - 6
put the value in this
( a - d ) ( a ) ( a + d )
( 9 - ( - 6), ( 9) , ( 9 - 6 )
=> 15 , 9 , 3
- the sum of three numbers of AP is 27 and their product is 405 find the numbers
_______________________
- Sum of 3 no. in AP = 27
- Product of 3no. in AP = 405
_______________________
- Find the numbers = ??
______________________
- let the no. be x -
let the 3no. be ( x - a ) , x , ( x + a )
- sum of 3no. of AP = 27
- product of 3 no. of AP is 405
- finding the no.
First no. = x - a = 9 - 6 = 3
Second no. = x = 9
Third no. = x + a = 9 + 6 = 15
___________________________
- Therefore the no. are 3 , 9 , 15