The sum of three numbers of an ap is 27 and product is 405. find the numbers
Answers
Required three numbers in given A.P are (3,9,15) Or (15,9,3)
Explanation:
Let (a-d),a,(a+d) are three consecutive terms in an A.P
According to the problem given,
i) Sum of three numbers = 27
=> a-d+a+a+d=27
=> 3a = 27
Divide each term by 3 , we get
=> a = 9 ----(1)
ii) product of the numbers= 405
=> (a-d)a(a+d)=405
=> (a²-d²)a=405
/* By algebraic identity:
(x-y)(x+y) = x²-y² */
=> (9²-d²)9 = 405
/* substitute a value */
Divide both sides of the equation by 9 , we get
=> 9²-d²=45
=> 81 - d² = 45
=> -d² = 45 - 81
=> -d² = -36
=> d² = 36
=> d =±√6²
=> d = ±6 ---(2)
Therefore,
case1:
If a = 9, d = 6
Required 3 terms are,
(a-d) = 9-6 = 3
a = 9
(a+d) = 9+6 = 15
case 2:
If a = 9, d = -6
Required 3 terms are
(a-d) = 9-(-6) = 9+6 = 15
a = 9
a+d = 9-6 = 3
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Required three numbers in given A.P are (3,9,15) Or (15,9,3)
Explanation:
Let (a-d),a,(a+d) are three consecutive terms in an A.P
According to the problem given,
i) Sum of three numbers = 27
=> a-d+a+a+d=27
=> 3a = 27
Divide each term by 3 , we get
=> a = 9 ----(1)
ii) product of the numbers= 405
=> (a-d)a(a+d)=405
=> (a²-d²)a=405
/* By algebraic identity:
(x-y)(x+y) = x²-y² */
=> (9²-d²)9 = 405
/* substitute a value */
Divide both sides of the equation by 9 , we get
=> 9²-d²=45
=> 81 - d² = 45
=> -d² = 45 - 81
=> -d² = -36
=> d² = 36
=> d =±√6²
=> d = ±6 ---(2)
Therefore,
case1:
If a = 9, d = 6
Required 3 terms are,
(a-d) = 9-6 = 3
a = 9
(a+d) = 9+6 = 15
case 2:
If a = 9, d = -6
Required 3 terms are
(a-d) = 9-(-6) = 9+6 = 15
a = 9
a+d = 9-6 = 3
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Hope it helps.