document.write( "Question 202544: Can ANYONE help me with these two problems? Please!\r
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\n" ); document.write( "\n" ); document.write( "Type the general term for the following sequences:\r
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\n" ); document.write( "\n" ); document.write( "#1) 3, 9, 27, 81, 243, . . .
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\n" ); document.write( "\n" ); document.write( "#2) 3/4, 6/5, 9/6, 12/7, 15/8...
\n" ); document.write( "__________\r
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\n" ); document.write( "\n" ); document.write( "Thanks,
\n" ); document.write( "Sarah
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Algebra.Com's Answer #152745 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
I'll do the first one to get you going\r
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\n" ); document.write( "\n" ); document.write( "# 1\r
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\n" ); document.write( "\n" ); document.write( "Notice that each term is increasing exponentially. So this sequence might be a geometric sequence. To find out, let's simply divide the terms.\r
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\n" ); document.write( "\n" ); document.write( "First divide the 2nd term 9 by the 1st term 3 to get \r
\n" ); document.write( "\n" ); document.write( "\"9%2F3=3\" \r
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\n" ); document.write( "Now divide the 3rd term 27 by the 2nd term 9 to get \r
\n" ); document.write( "\n" ); document.write( "\"27%2F9=3\" \r
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\n" ); document.write( "Now divide the 4th term 81 by the 3rd term 27 to get \r
\n" ); document.write( "\n" ); document.write( "\"81%2F27=3\" \r
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\n" ); document.write( "Now divide the 5th term 243 by the 4th term 81 to get \r
\n" ); document.write( "\n" ); document.write( "\"243%2F81=3\" \r
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\n" ); document.write( "\n" ); document.write( "So if we pick ANY term and divide it by the previous term, we'll always get 3. \r
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\n" ); document.write( "\n" ); document.write( "This is the common ratio between the terms. So this means that \"r=3\".\r
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\n" ); document.write( "\n" ); document.write( "Now since we've started at 3, this means that \"a=3\"\r
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\n" ); document.write( "\n" ); document.write( "Since the general geometric sequence is \"a%5Bn%5D=ar%5En\", this means the sequence is\r
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\n" ); document.write( "\n" ); document.write( "\"a%5Bn%5D=3%2A3%5En\" where \"n\" starts at \"n=0\"\r
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\n" ); document.write( "\n" ); document.write( "Answer:\r
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\n" ); document.write( "\n" ); document.write( "So the sequence is \"a%5Bn%5D=3%2A3%5En\" where \"n\" starts at \"n=0\"\r
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\n" ); document.write( "\n" ); document.write( "Note: if you want to start at \"n=1\", then you need to replace \"n\" with \"n-1\" to get the new sequence \"a%5Bn%5D=3%2A3%5E%28n-1%29\"\r
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