0

Find the condition for 3 given numbers x1, x2 and x3 to be terms of an arithmetic progression (obviously the same).

Well, I don't know what exactly the problem is asking: Assuming, wlg that x3>x2>x1: It must be x2-x1= mr (where m is an integer and r is the common difference. Also x3-x1=nr and therefore x3-x2=(n-m)*r. I don't know what else!!

Harry Gartner
  • 453
  • For 3 numbers in AP, the second number is always the arithmetic mean of the first and the second. Similarly is the case of GP you take the geometric mean of the first and third number to obtain the second term. – S.S Nov 16 '17 at 12:43
  • @ShatabdiSinha: not necessarily 3 consecutive terms – Harry Gartner Nov 16 '17 at 12:55
  • yup i mean only for 3 consecutive terms in progression. – S.S Nov 16 '17 at 12:57

1 Answers1

0

we have $$x_1=x_1$$,$$x_2=x_1+d$$,$$x_3=x_1+2d$$ and we get with $$d=x_2-x_1$$ $$x_3=x_1+2x_2-2x_1$$ or $$x_2=\frac{x_1+x_3}{2}$$

Dr. Sonnhard Graubner
  • 95,283