find three consecutive terms in A•P• whose sum is 15 and product is 80•
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Let the the three consecutive terms of the AP be a-d, a, a+d.
There sum of these terms is 15.
(a-d) + a + (a+d) = 15
3a = 15
a = 5
The product of the terms is 80.
(a-d)(a)(a+d) = 80
a(a-d)(a+d) = 80
a(a^2 - d^2) = 80. [(a+b)(a-b) = a^2 - b^2]
5[(5)^2 - d^2] = 80
25 - d^2 = 16
d^2 = 25 - 16
d^2 = 9
d = 3. [Taking square root]
The three terms are
a-d = 5 - 3 = 2
a = 5
a+d = 5 + 3 = 8
The three terms of the AP are 2, 5 and 8.
There sum of these terms is 15.
(a-d) + a + (a+d) = 15
3a = 15
a = 5
The product of the terms is 80.
(a-d)(a)(a+d) = 80
a(a-d)(a+d) = 80
a(a^2 - d^2) = 80. [(a+b)(a-b) = a^2 - b^2]
5[(5)^2 - d^2] = 80
25 - d^2 = 16
d^2 = 25 - 16
d^2 = 9
d = 3. [Taking square root]
The three terms are
a-d = 5 - 3 = 2
a = 5
a+d = 5 + 3 = 8
The three terms of the AP are 2, 5 and 8.
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