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If the sum of n terms is 2n+1 then the sum of n+1 terms, using the same formula, is 2*(n+1)+1 = 2n+2+1 = 2n+3

So the (n+1)th term is sum to n+1 minus sum to n = (2n+3) - (2n+1) = 2

So each term is 2. But if each term is 2, then the sum of n terms must be even. The sum is clearly odd - which leads to a contradiction.

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Q: What is the nth term of ap if sum of n terms of ap is 2n plus 1?
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What is the sum of the 3rd and 5th term in the sequence where the nth term is 3n-4?

16


What is the sum of the positive even integers less than one hundred?

This can be found out by using the formula for sum of n terms of an arithmetic progression. here, n is not known. a=2 d=2 (since it's even) nth term=98 (last even number less than 100) using formula for nth term, nth term=a+(n-1)d 98=2+(n-1)2 therefore, n=49 so, sum of n terms=n/2[2*a+(n-1)d] putting n=49, a=2, and d=2, sum=2450


The sum of the first 5 terms of an arithmetic series is 85. The sum of the first 6 terms is 123. Determine the first four terms of the series.?

Suppose the first term is a, the second is a+r and the nth is a+(n-1)r. Then the sum of the first five = 5a + 10r = 85 and the sum of the first six = 6a + 15r = 123 Solving these simultaneous equations, a = 3 and r = 7 So the first four terms are: 3, 10, 17 and 24


What is the nth term and sum of n terms of 6 13 24 39?

The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6


What is the sum of the arithmetic sequence?

The sum of an arithmetical sequence whose nth term is U(n) = a + (n-1)*d is S(n) = 1/2*n*[2a + (n-1)d] or 1/2*n(a + l) where l is the last term in the sequence.

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Give the simple formula for the nth term of the following arithmetic sequence What if your answer will be of the form an plus b?

If the formula for additional terms was the summation of the term before it, the nth term of the series would be the sum of all terms prior. In other words it would be the summation of a through n minus 1.


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What is the sum of the 3rd and 5th term in the sequence where the nth term is 3n-4?

16


What is the sum of the positive even integers less than one hundred?

This can be found out by using the formula for sum of n terms of an arithmetic progression. here, n is not known. a=2 d=2 (since it's even) nth term=98 (last even number less than 100) using formula for nth term, nth term=a+(n-1)d 98=2+(n-1)2 therefore, n=49 so, sum of n terms=n/2[2*a+(n-1)d] putting n=49, a=2, and d=2, sum=2450


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What is the nth term for 0 1 1 2 3 5 8 13?

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The sum of the first 5 terms of an arithmetic series is 85. The sum of the first 6 terms is 123. Determine the first four terms of the series.?

Suppose the first term is a, the second is a+r and the nth is a+(n-1)r. Then the sum of the first five = 5a + 10r = 85 and the sum of the first six = 6a + 15r = 123 Solving these simultaneous equations, a = 3 and r = 7 So the first four terms are: 3, 10, 17 and 24


What is the nth term and sum of n terms of 6 13 24 39?

The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6The simplest rule that will generate the 4 terms, and the nth term isUn = 2n2 + n + 3Then Sn = Sum for k = 1 to n of (2k2 + k + 3)= 2*sum(k2) + sum(k) + sum(3)= 2*n*(n+1)*(2n+1)/6 + n*(n+1)/2 + 3*n= (2n3 + 3n2 + n)/3 + (n2 + n)/2 + 3n= (4n3 + 9n2 + 23n)/6


The 'nth term of an Arithmetic Progression is 3n-2.Find the sum of first n terms.What is the sum of first 10 terms?

The sum of the 1st n terms is : N(3N-1)/2 Explanation : The sum from 1 to N of (3m-2) = 3 * sumFrom1toN(m) - sumFrom1toN(2) = 3 * (N*(N+1)/2) -2*N = N(3N-1)/2 For N=10 => 145