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Question

A sequence is defined by

un=717n2u _ { n } = \frac { 71 - 7 n } { 2 }

. a. Show that the sequence is arithmetic. b. Find

u1u_1

and d. c. Find

u75u_{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 un=717n2u_n=\dfrac{71-7n}{2}. Therefore, un+1=717(n+1)2u_{n+1}=\dfrac{71-7(n+1)}{2}.

Now, we have:

un+1un=717(n+1)2717n2=717n771+7n2=72\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 72\dfrac{-7}{2} for each nn.

(b)

d=72d=\dfrac{-7}{2}

u1=7172=32u_1=\dfrac{71-7}{2}=32

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