The sum of three integers in AP is 15 and their product is 80 the integers are
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Consider the 3 integral AP as (a-d), a, (a+d).
where a refers to the 2nd term of this AP, & d is the common difference.
Now, Sum of these 3 integers = 15
(a-d) + a + (a+d) = 15
3a = 15
a = 5
Now, Their product = 80
(a-d) * (a) * (a+d) = a * (a² - d²) = 80
a * (a² - d²) = 80
put a = 5
5 * (5² - d²) = 80
(25 - d²) = 80/5 = 16
d² = 25 - 16 = 9
So, d = ±3.
Now, d = 3, and also d = -3.
So there exist 2 set of integers in AP, whose sum is 15 and their product is 80.
[(5 - (±3)), (5), (5 + (±3))]
→ [2, 5, 8] and [8, 5, 3]
Thankyou!!!
where a refers to the 2nd term of this AP, & d is the common difference.
Now, Sum of these 3 integers = 15
(a-d) + a + (a+d) = 15
3a = 15
a = 5
Now, Their product = 80
(a-d) * (a) * (a+d) = a * (a² - d²) = 80
a * (a² - d²) = 80
put a = 5
5 * (5² - d²) = 80
(25 - d²) = 80/5 = 16
d² = 25 - 16 = 9
So, d = ±3.
Now, d = 3, and also d = -3.
So there exist 2 set of integers in AP, whose sum is 15 and their product is 80.
[(5 - (±3)), (5), (5 + (±3))]
→ [2, 5, 8] and [8, 5, 3]
Thankyou!!!
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