document.write( "Question 83751: Use the arithmetic sequence of numbers 1, 3, 5, 7, 9,….to find the following:\r
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\n" ); document.write( "\n" ); document.write( "a) What is d, the difference between any two consecutive terms?\r
\n" ); document.write( "\n" ); document.write( "b) Using the formula for the nth term of an arithmetic sequence, what is 101st term?\r
\n" ); document.write( "\n" ); document.write( "c) Using the formula for the sum of an arithmetic sequence, what is the sum of the first 20 terms?\r
\n" ); document.write( "\n" ); document.write( "d) Using the formula for the sum of an arithmetic sequence, what is the sum of the first 30 terms?\r
\n" ); document.write( "\n" ); document.write( "e) What observation can you make about these sums of this sequence (HINT: It would be beneficial to find a few more sums like the sum of the first 2, then the first 3, etc.)? Express your observations as a general formula in “n.”\r
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Algebra.Com's Answer #60240 by jim_thompson5910(35256)\"\" \"About 
You can put this solution on YOUR website!
a)
\n" ); document.write( "The difference is the factor between each term. So going from 1 to 3, 3 to 5, 5 to 7, you see that its adding 2 each time. To verify, pick one term and subtract the previous term from it. So lets say I choose 7: I'm going to subtract 5 from it to get a difference of 2. If I pick 5, and subtract 3, I get a difference of 2.So the difference is: d=2\r
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\n" ); document.write( "\n" ); document.write( "b)
\n" ); document.write( "Using what we found earlier, I know that the sequence counts up by 2 each term. So if I'm at 1 (the 1st term) and I go to 3, this means I increase by 2 each term. If I let n=0 then the term is 1, and if I let n=1 then the term is 3. This basically tells me that the arithmetic sequence is 2n+1. To verify, simply plug in the 1st term (n=0) and you'll get 1. Plug in the 2nd term (n=1) you'll get 3, if I let n=2 I get 5, etc. If I wanted to know the 101st term, let n=100 (zero is the first term) and it comes to
\n" ); document.write( "\"2%2Ahighlight%28100%29%2B1=201\" So the 101st term is 201\r
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\n" ); document.write( "\n" ); document.write( "c)
\n" ); document.write( "Using the sum of arithmetic series formula:
\n" ); document.write( "\"s=%28n%2F2%29%2A%28a%5B1%5D%2Ba%5Bn%5D%29\" a[1]=first term, a[n]=nth term (ending term which is the 20th term), and n is the number of terms
\n" ); document.write( "\"s=%2820%2F2%29%2A%281%2B39%29\" Plug in values
\n" ); document.write( "\"s=%2810%29%2A%2840%29\"Simplify
\n" ); document.write( "\"s=400\" So the sum of the first 20 terms is 400.\r
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\n" ); document.write( "\n" ); document.write( "d)
\n" ); document.write( "Again using the same formula
\n" ); document.write( "\"s=%28n%2F2%29%2A%28a%5B1%5D%2Ba%5Bn%5D%29\" a[1]=first term, a[n]=nth term (ending term which is the 30th term), and n is the number of terms
\n" ); document.write( "\"s=%2830%2F2%29%2A%281%2B59%29\" Plug in values
\n" ); document.write( "\"s=%2815%29%2A%2860%29\"Simplify
\n" ); document.write( "\"s=900\"\r
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\n" ); document.write( "\n" ); document.write( "e)
\n" ); document.write( "Sum of the first 2 terms
\n" ); document.write( "1+3=4
\n" ); document.write( "Sum of the first 3 terms
\n" ); document.write( "1+3+5=9
\n" ); document.write( "Sum of the first 4 terms
\n" ); document.write( "1+3+5+7=16
\n" ); document.write( "Sum of the first 5 terms
\n" ); document.write( "1+3+5+7+9=25
\n" ); document.write( "Sum of the first 6 terms
\n" ); document.write( "1+3+5+7+9+11=36
\n" ); document.write( "Sum of the first 7 terms
\n" ); document.write( "1+3+5+7+9+11+13=49
\n" ); document.write( "Sum of the first 8 terms
\n" ); document.write( "1+3+5+7+9+11+13+15=64
\n" ); document.write( "Sum of the first 9 terms
\n" ); document.write( "1+3+5+7+9+11+13+15+17=81
\n" ); document.write( "Sum of the first 10 terms
\n" ); document.write( "1+3+5+7+9+11+13+15+17+19=100 \r
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\n" ); document.write( "\n" ); document.write( "Notice how the partial sums are all perfect squares. So the sums follow the sequence \"n%5E2\"
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