geometric sequence

[waski]

I can't get started on this problem about geometric sequences

the sum of the first 2 terms is 16, and the sum of the reciprocals of the first 2 terms is 1/3.

find the fist 2 terms

thanks

 

[galactus]

Couldn't you just solve:

\(\displaystyle \L\\x+y=16\)

\(\displaystyle \L\\\frac{1}{x}+\frac{1}{y}=\frac{1}{3}\)?.

 

[waski]

thanks for the reply

the algebra gets to messy for me to handle :?

 

[galactus]

Solve the first one for y and sub into the second:

\(\displaystyle \L\\y=16-x\)

This gives:

\(\displaystyle \L\\\frac{1}{x}+\frac{1}{16-x}=\frac{1}{3}\)

Now, it'll be good practice to solve this yourself.

Your LCD is 3x(16-x)

\(\displaystyle \L\\3\not{x}(16-x)\frac{1}{\not{x}}+3x\sout{(16-x)}\frac{1}{\sout{16-x}}=\not{3}x(16-x)\frac{1}{\not{3}}\)

Now finish. You'll have a quadratic that is easily factorable or solvable by the quadratic formula.