Given the sequence: 2x-3, 7x, 11x; find x if the sequence is arithmetic and then geometric.
I already have the answers which are -3 for arithmetic and -11/9 for geometric, but I don't understand how to do this. If anyone can help me out without being rude, it would be greatly appreciated.
For an arithmetic sequence, the difference between successive terms is a constant, usually referred to as the 'common difference' which I denote by the letter 'd'. From the sequence you've stated; 7x - (2x-3) = d and also 11x - 7x = d therefore 7x - (2x-3) = 11x - 7x 7x - 2x + 3 = 4x 5x - 4x = -3 x = -3 For a geometric sequence, the ratio between successive terms is a constant referred to as the 'common ratio' which I denote by the letter 'r'. From your given sequence; 7x/(2x-3) = r and 11x/7x = r therefore 7x/(2x-3) = 11x/7x (7x) x (7x) = (11x) x (2x-3) 49x^2 = 22x^2 -33x 49x^2 - 22x^2 = -33x 27x^2 = -33x x^2/x = -33/27 x = -11/9
Arithmetic: there is constant number added from one term to the next so (7x)-(2x-3)=(11x)-(7x) 5x+3=4x x=-3
Geometric: there is a constant number multiplied from one term to the next so (7x)/(2x-3)=(11x)/(7x) (7x)(7x)=(2x-3)(11x) 49x^2=22x^2-33x 27x^2+33x=0 (3x)(9x+11)=0 3x=0 or 9x+11=0 x=0 or x=-11/9 but it not work for the x=0 so it is only x=-11/9