Let the numbers are a-d,,a, a+dso a-d+a +a+d=243a=24 , a=8and (8-d)x 8x(8+d)=4408(64-d2)=440512-8d2=4408d2=72 do d= 3\xa0hence the no are 5,8 and\xa011\xa0
Take, (a-d),a, and , (a+d) now, take their sum as, 24, yu will find, a and then their product = 440, yu ill, find d.
Hey can u please explain who do u get 8 5 and 11
5,8,11…..

Solution:-Let the required terms of the given AP be a-d, a and a+d *Where the first term is a-d*The common difference = d*Given : The sum of the three parts = 24*∴ (a-d)+(a)+(a+d) = 243a = 24 #a = 8#Given : The product of these three terms = 440∴ (a-d) (a) (a+d) = 440#(8-d) (8) (8+d) = 440#- 8d² + 512 = 440#- 8d² = 440 – 512#- 8d² = – 72#d² = 72/8#d² = 9#d = √9#d= 3#So the three required terms of AP is 8 – 3 = 5 ; 8 and 8 + 3 = 11Three terms are 5, 8, 11