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# A sequence is defined by$u _ { n } = \frac { 71 - 7 n } { 2 }$. a. Show that the sequence is arithmetic. b. Find$u_1$and d. c. Find$u_{75}$. d. For what values of n are the terms of the sequence less than -200?

Solution

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#### (a)

We know that $u_n=\dfrac{71-7n}{2}$. Therefore, $u_{n+1}=\dfrac{71-7(n+1)}{2}$.

Now, we have:

\begin{align*} u_{n+1}-u_n&=\dfrac{71-7(n+1)}{2} - \dfrac{71-7n}{2}\\ &=\dfrac{71-7n-7-71+7n}{2}\\ &=\dfrac{-7}{2} \end{align*}

This is arithmetic sequence because consecutive terms differ by $\dfrac{-7}{2}$ for each $n$.

#### (b)

$d=\dfrac{-7}{2}$

$u_1=\dfrac{71-7}{2}=32$

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