Help on arithmetic and geometric sequences?

Arithmetic series

=> 7x - (2x - 3) = 11x - 7x

so, 5x + 3 = 4x

=> x = -3

Geometric series

=> 7x/(2x - 3) = 11x/7x

so, 7x/(2x - 3) = 11/7

=> 7(7x) = 11(2x - 3)

i.e. 49x = 22x - 33

=> 27x = -33

i.e. x = -33/27 = -11/9

Not rude but to the point!

:)>

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

I can help you out if it is arithmetic

what is the difference between the 2nd and 3nd term?

11x-7x = 4x and the difference between 2nd and 1st 7x-(2x-3) = 5x+3 the differences should be the same so 4x=5x+3

and -x=3 and x=-3

geometric progression is

y(2x-3) = 7x and 7xy = 11x then set those two equal and solve

y= 7x / (2x-3 and y = 11x / 7x or y = 11/7 substitute into first equation

11/7 = 7x / (2x-3) cross multiply

11(2x-3) = 49x

22x - 33 = 49x

-33 = 27x

x= -33/27 or - 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