If the sum of three numbers in an A.P. is 24 | vedclass

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(A) Let the three numbers in $A.P.$ be $(a-d), a,$ and $(a+d).$According to the given information:$(a-d) + a + (a+d) = 24$$3a = 24$$a = 8$Also,the pro

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$5, 8, 11$

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(A) Let the three numbers in $A.P.$ be $(a-d), a,$ and $(a+d).$According to the given information:$(a-d) + a + (a+d) = 24$$3a = 24$$a = 8$Also,the product is given as:$(a-d) 	imes a 	imes (a+d) = 440$$(8-d) 	imes 8 	imes (8+d) = 440$$(8-d)(8+d) = \frac{440}{8}$$64 - d^2 = 55$$d^2 = 64 - 55 = 9$$d = \pm 3$If $d = 3,$ the numbers are $(8-3), 8, (8+3),$ which are $5, 8, 11.$If $d = -3,$ the numbers are $(8-(-3)), 8, (8+(-3)),$ which are $11, 8, 5.$Thus,the numbers are $5, 8, 11.$

If the sum of three numbers in an $A.P.$ is $24$ and their product is $440,$ find the numbers.

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