Question

Find three positive integers in AP such that their sum is 30 and twice the square of the smaller☆ ♡Plss answer ♡

Answer 1

Answer:

Let the three terms of the AP be

a-d, a and a+d, so that

a-d+ a + a+d = 24, or

a = 24/3 = 8 …(1)

Their product = (a-d)(a)(a+d) = 480, or

a(a^2-d^2)= 480, or

(a^2-d^2) = 480/8 = 60

8^2-d^2=60, or

64–60 = 4 = d^2, or d = +2 or -2.

Hence the three terms of the AP are 6, 8 and 10, or 10, 8 and 6.

Answer 2

Step-by-step explanation:

Let the three terms of the AP be

a-d, a and a+d, so that

a-d+ a + a+d = 24, or

a = 24/3 = 8 …(1)

Their product = (a-d)(a)(a+d) = 480, or

a(a^2-d^2)= 480, or

(a^2-d^2) = 480/8 = 60

8^2-d^2=60, or

64–60 = 4 = d^2, or d = +2 or -2.

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