Find three numbers in A.P. whose sum is 21 and their product is 231.

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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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Three numbers are a_d, a, a+d a_d+a+a+d=21 3a=21 a=21/3 a=7 Product =231 a_d*a*a+d=231 7_d*7*7+d=231 7_d*7+d=231/7 (7-d)(7+d)=33 49-d square =33 d Square =33-49 d^2=16 d=4 So number are 7_4,7,,7+4 3,7,11
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Let the numbers be (a-d), a, (a+d). According to first given condition, (a-d)+a+(a+d)=21 =>> a=7 according to second given condition, (a-d) x a x (a+d)=231 --->(7-d) x 7x (7+d)=231 ---> 7^2 - d^2...
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Let the numbers be (a-d), a, (a+d). According to first given condition, (a-d)+a+(a+d)=21 =>> a=7 according to second given condition, (a-d) x a x (a+d)=231 --->(7-d) x 7x (7+d)=231 ---> 7^2 - d^2 = (231/7)=33 --->49 - 33=d^2 --->d^2 =16 Hence d=4 Therefore the required numbers are, (7-4),7,(7+4) i.e 3,7,11 read less
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Let three no.in A.p.be a-d,a,a+d a-d+a+a+d=21 3a=21 a=7 (a-d)a(a+d)=231 Put a=7 (7-d)7(7+d)=231 (49_d2)=231 (49-d2)=231/7 49-d2=33 d2=16 d=4 or d=-4 Three no. are 7-4,7.7+4 3, 7, 11 or 11, 7, 3.
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Let, Given, Three Numbers A.P SUM = 21 ----- eq(1) Product = 31 ------- eq(2)Let,Three numbers Are Given below (a-d), a, (a+d) -------- (3)Put the Vallue of eq 1 (a-d), a, (a+d) = 21Solve The equation 3a = 21 a = 7 -------- Eq(4)Put The vallue Of eq No (4) to Eq No 3...
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Let, Given, Three Numbers A.P SUM = 21 ----- eq(1) Product = 31 ------- eq(2)Let,Three numbers Are Given below(a-d), a, (a+d) -------- (3)Put the Vallue of eq 1(a-d), a, (a+d) = 21Solve The equation3a = 21a = 7 -------- Eq(4)Put The vallue Of eq No (4) to Eq No 3Then, (7-d)×7×(7+d) = 231then Solve eq (7-d)×(7+d) = 231÷7 (7-d)×(7+d) = 33Use The Formulla (a-b)×(a+b)=a^2-b^2 7^2 - d^2 = 33 Square Of The given Equation 49 -d^2 = 33 49 - 33 =d^2 16 =d^2 . . . d = 4 ------- (5)put The vallu Of a & b in Eq 3 (7-4),7,(7+4)Then Three Numbers Are 3, 7,11 read less
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