1. Let $\left(a_{n}\right)$ be a non-constant arithmetic sequence with the first term $a_{1}=1$. The terms $a_{2}, a_{5}, a_{11}$ form a geometric sequence. Calculate the sum of the first 2009 terms of the sequence $\left(a_{n}\right)$.
IV/1. Let $d$ be the difference of the sequence $\left(a_{n}\right)$. Then $a_{2}=1+d, a_{5}=1+4 d$ and $a_{11}=1+10 d$. Since $a_{2}, a_{5}$ and $a_{11}$ form a geometric sequence, we have $(1+4 d)^{2}=(1+d)(1+10 d)$, or $6 d^{2}=3 d$. Since the sequence is non-constant, $d=\frac{1}{2}$, and the sum of the first 2009 terms is $2009+\frac{2009 \cdot 2008}{2} \cdot \frac{1}{2}=1010527$.

Note. The condition that $a_{2}, a_{5}$ and $a_{11}$ form a geometric sequence can also be written in the form $(d+1) q^{2}=(4 d+1) q=10 d+1$, where $q$ denotes the quotient of two consecutive terms of this sequence.

The notation $a_{2}=1+d, a_{5}=1+4 d$ and $a_{11}=1+10 d \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ point

The equation $(1+4 d)^{2}=(1+d)(1+10 d)$ or $(d+11) q^{2}=(4 d+1) q=10 d+1 \ldots \ldots \ldots$.

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![](https://cdn.mathpix.com/cropped/2024_06_07_2dd3bccc19bc96d6d7dfg-16.jpg?height=60&width=1633&top_left_y=1923&top_left_x=220)

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