Odd- and Even-numbered Terms

Algebra Level 3

Let $a_1, a_2, a_3, \ldots a_{2n}, a_{2n+1}$ be an arithmetic progression. If the sum of all the odd-numbered terms is $130$ and the sum of all the even-numbered terms is $117,$ what is $2n+1?$

21

23

25

19

1 solution

Giorgio de Fornasari
Oct 1, 2015

The progression ends with an odd term $a_{2n+1}$ so the odd terms are one more than the even terms.

Calling "d" the common difference of the progression, the difference between the sums of odd and even terms equals to the difference between the last two terms:

$a_{2n+1} - a_{2n} = [a_{1}+(2n \times d)] - [a_{1}+((2n-1) \times d)] = d = 130 -117 = 13$

The total amount of the progression is $130+117 = 247$

Being 247 a multiple of 13 $(\frac {247}{13} = 19)$ , we can easily say $a_{1} = 0$ and $2n+1 = 19$

Hi there. Your solution is not correct. The difference in the two sums is $130-117=(a_{2n+1}-a_{2n})+(a_{2n-1}-a_{2n-2})+...+(a_3-a_2)+a_1=2nd+a_1$ . And at the end, you assumed $247$ is the last term of the sequence instead of the sum of the sequence.

James Wilson - 3 years, 5 months ago

If you start by assuming $a_1=0$ , then the number of terms is $19$ , but the common difference is $\frac{13}{9}$ .

James Wilson - 3 years, 5 months ago

Odd- and Even-numbered Terms

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