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Find three numbers in A.P. whose sum is 21 and their product is 231.
Let, three numbers are (a-d), a, (a+d) So, (a-d)+a+(a+d)=21 => a=7 Again, (7-d)×7×(7+d)=231 => (7-d)×(7+d)=231÷7 => 7^2 - d^2 = 33=> d^2 = 49 - 33=> d^2 = 16=> d = 4 So, the numbers are 7-4= 3, 7 and 7+4=11
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