the sum of three integers in AP is 21 and their product is 280,find the three integers
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Let the three numbers be (a-d), a, (a+d)
Given that their sum = 21
=> a + d + a + a - d = 21
d will cancel out
=> 3a = 21
=> a = 7
Given that their product is 280
=> (7-d) × 7 × (7+d) = 280
=> (7-d) (7+d) = 280/7
=> (7-d) (7+d) = 40
=> 7^2 - d^2 = 40
=> 49 - d^2 = 40
=> d^2 = 9
=> d = +3 or -3
HENCE, THE NUMBERS ARE
》by using d = +3
7-3 , 7 , 7+3
= 4, 7, 10
》by using d=-3
7-(-3), 7, 7-3
= 10, 7, 4
___________________________
HOPE IT HELPS ^_^
___________________________
Let the three numbers be (a-d), a, (a+d)
Given that their sum = 21
=> a + d + a + a - d = 21
d will cancel out
=> 3a = 21
=> a = 7
Given that their product is 280
=> (7-d) × 7 × (7+d) = 280
=> (7-d) (7+d) = 280/7
=> (7-d) (7+d) = 40
=> 7^2 - d^2 = 40
=> 49 - d^2 = 40
=> d^2 = 9
=> d = +3 or -3
HENCE, THE NUMBERS ARE
》by using d = +3
7-3 , 7 , 7+3
= 4, 7, 10
》by using d=-3
7-(-3), 7, 7-3
= 10, 7, 4
___________________________
HOPE IT HELPS ^_^
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