document.write( "Question 495345: WHAT IS THE 200TH NUMBER IN THIS SEQUENCE
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Algebra.Com's Answer #336073 by Edwin McCravy(20060)\"\" \"About 
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document.write( "1,4,7,10,13,16\r\n" );
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document.write( "To get from the first term 1 to the 2nd term 4 we add 3.\r\n" );
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document.write( "To get from the 2nd term 4 to the 3rd term 7 we add 3.  Therefore\r\n" );
document.write( "To get from the 1st term 1 to the 3rd term 7 we must add 3 2 times.\r\n" );
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document.write( "To get from the 3rd term 7 to the 4th term 10 we add 3.  Therefore\r\n" );
document.write( "To get from the 1st term 1 to the 4th term 10 we must add 3 3 times.\r\n" );
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document.write( "To get from the 4th term 10 to the 5th term 13 we add 3.  Therefore\r\n" );
document.write( "To get from the 1st term 1 to the 5th term 13 we must add 3 4 times.\r\n" );
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document.write( "To get from the 5th term 13 to the 6th term 16 we add 3.  Therefore\r\n" );
document.write( "To get from the 1st term 1 to the 6th term 16 we must add 3 5 times.\r\n" );
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document.write( "We assume that this pattern continues forever:\r\n" );
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document.write( "To get from the (n-1)th term an-1 to the nth term an we add 3.  Therefore\r\n" );
document.write( "To get from the 1st term 1 to the nth term an we must add 3 n-1 times.\r\n" );
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document.write( "and we end up with \r\n" );
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document.write( "To get from the 199th term a199 to the 200th term a200 we add 3.  Therefore\r\n" );
document.write( "To get from the 1st term 1 to the 200th term a200 we must add 3 199 times.\r\n" );
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document.write( "Therefore \r\n" );
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document.write( "a200 = 1 + 199×3 = 1 + 597 = 598\r\n" );
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document.write( "Actually the formula for the nth term when there is a common\r\n" );
document.write( "difference d, such as the 3 that is added each time, is\r\n" );
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document.write( "an = a1 + (n-1)d \r\n" );
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document.write( "where a1 is the first term and d is the common difference\r\n" );
document.write( "and n is the number of terms.\r\n" );
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document.write( "This is called an \"arithmetic sequence\".\r\n" );
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document.write( "So we could have just substituted a1 = 1, d = 3 and n = 200\r\n" );
document.write( "into\r\n" );
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document.write( "an = a1 + (n-1)d\r\n" );
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document.write( "a200 = 1 + (200-1)3\r\n" );
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document.write( "a200 = 1 + (199)3\r\n" );
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document.write( "a200 = 1 + 597\r\n" );
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document.write( "a200 = 598\r\n" );
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document.write( "Edwin
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