Given an arithmetic sequence $\{a_n\}$, we know that $a_2 + a_8 = 1$. The sum of the first 3 terms is 20, the sum of the last 3 terms is 130, and the sum of all terms is 200. We want to find the number of terms $n$.

1. Problem Description

Given an arithmetic sequence $\{a_n\}$, we know that $a_2 + a_8 = 1$. The sum of the first 3 terms is 20, the sum of the last 3 terms is 130, and the sum of all terms is

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0. We want to find the number of terms $n$.

2. Solution Steps

Let the first term be $a_1$ and the common difference be $d$. Then $a_n = a_1 + (n-1)d$.

We are given that $a_2 + a_8 = 1$. In terms of $a_1$ and $d$, this means

$a_1 + d + a_1 + 7d = 1$, so $2a_1 + 8d = 1$.

The sum of the first 3 terms is $a_1 + a_2 + a_3 = a_1 + (a_1 + d) + (a_1 + 2d) = 3a_1 + 3d = 20$, which implies $a_1 + d = \frac{20}{3}$.

The sum of the last 3 terms is $a_{n-2} + a_{n-1} + a_n = (a_1 + (n-3)d) + (a_1 + (n-2)d) + (a_1 + (n-1)d) = 3a_1 + (3n - 6)d = 3a_1 + 3(n-2)d = 130$. This gives $a_1 + (n-2)d = \frac{130}{3}$.

The sum of all terms is $S_n = \frac{n(a_1 + a_n)}{2} = \frac{n(a_1 + a_1 + (n-1)d)}{2} = \frac{n(2a_1 + (n-1)d)}{2} = 200$. This gives $n(2a_1 + (n-1)d) = 400$.

From $a_1 + d = \frac{20}{3}$, we have $a_1 = \frac{20}{3} - d$. Substituting into $2a_1 + 8d = 1$, we get

$2(\frac{20}{3} - d) + 8d = 1$, so $\frac{40}{3} - 2d + 8d = 1$, which means $6d = 1 - \frac{40}{3} = \frac{-37}{3}$. Thus $d = \frac{-37}{18}$.

Then $a_1 = \frac{20}{3} - (\frac{-37}{18}) = \frac{120 + 37}{18} = \frac{157}{18}$.

We have $a_1 + (n-2)d = \frac{130}{3}$. Substituting the values of $a_1$ and $d$, we have

$\frac{157}{18} + (n-2)(\frac{-37}{18}) = \frac{130}{3}$. Multiply by 18 to get $157 - 37(n-2) = 6 \cdot 130 = 780$.

$157 - 37n + 74 = 780$, so $231 - 37n = 780$. This gives $-37n = 549$, which is impossible because $n$ must be an integer.

We have $a_1+d = 20/3$ and $a_1+(n-2)d = 130/3$. Subtract the first equation from the second to get $(n-3)d=110/3$.

Also $S_n = n/2 (a_1+a_n)=n/2 (a_1+a_1+(n-1)d)=n/2 (2a_1+(n-1)d)=200$, so $2a_1+(n-1)d=400/n$.

And $2a_1+8d=1$.

From $a_1+d = 20/3$ and $a_1+(n-2)d = 130/3$, we sum them to get $2a_1+(n-1)d = 150/3=50$.

Hence $400/n=50$, so $n=8$.

3. Final Answer

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