document.write( "Question 39248: did I do this right: Use the geometric sequence of numbers 1, 3, 9, 27, … to find the following:
\n" ); document.write( "a) What is r, the ratio between 2 consecutive terms?
\n" ); document.write( "Answer: r=3
\n" ); document.write( "Show work in this space.
\n" ); document.write( "each are multiples of 3
\n" ); document.write( "1*3=3, 3*3=9, 9*3=27 and so on\r
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\n" ); document.write( "\n" ); document.write( "b) Using the formula for the nth term of a geometric sequence, what is the 10th term?
\n" ); document.write( "Answer:n=19683
\n" ); document.write( "Show work in this space.
\n" ); document.write( "a(n)=a(1)(r^n-1)
\n" ); document.write( "a(n)=1(1)*(3^10-1
\n" ); document.write( "a(n)=3^9=19683\r
\n" ); document.write( "\n" ); document.write( "c) Using the formula for the sum of a geometric series, what is the sum of the first 10 terms?
\n" ); document.write( "Answer: s=29524
\n" ); document.write( "Show work in this space.
\n" ); document.write( "s(n)= a(1)(1-r^n)/1-r
\n" ); document.write( "s(n)= 1(1)*(1-3^10)/(1-3)=29524 \n" ); document.write( "

Algebra.Com's Answer #24669 by fractalier(6550) $\"\"$ $\"About$

You can put this solution on YOUR website!
It all looks good. The only thing I would mention is that the way to find the common ratio of a geometric sequence or series is to merely divide any a-sub-(n+1) term by a-sub-n... \n" ); document.write( "

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