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Chapter 7: Problem 58

Show that the sum of a finite arithmetic sequence is 0 if and only if the last term equals the negative of the first term.

Short Answer

Expert verified
The sum of a finite arithmetic sequence is given by the formula \(S_n = \frac{n}{2} (a_1 + a_n)\), where \(S_n\) is the sum, \(n\) is the number of terms, \(a_1\) is the first term, and \(a_n\) is the last term. If \(a_n = -a_1\), then the sum is 0: \(S_n = \frac{n}{2} (a_1 - a_1) = 0\). Conversely, if \(S_n = 0\), then the equation \(a_1 + a_n = 0\) implies that \(a_n = -a_1\). Hence, the sum of a finite arithmetic sequence is 0 if and only if the last term equals the negative of the first term.

Step by step solution

01

Recall the formula for the sum of a finite arithmetic sequence

The sum of a finite arithmetic sequence can be calculated using the following formula: \[S_n = \frac{n}{2} (a_1 + a_n)\] where \(S_n\) is the sum of the sequence, \(n\) is the number of terms in the sequence, \(a_1\) is the first term, and \(a_n\) is the last term.
02

Show that the sum is 0 when the last term equals the negative of the first term

We must show that \[S_n = \frac{n}{2} (a_1 + a_n) = 0\] if and only if \(a_n = -a_1\). First, assume that \(a_n = -a_1\). Then, \[S_n = \frac{n}{2} (a_1 + (-a_1)) = 0\] Since the sum is indeed 0 when the last term equals the negative of the first term, the first part of the proof is complete.
03

Show that the sum is 0 only if the last term equals the negative of the first term

Now let's consider that \(S_n = 0\), and we must show that this implies \(a_n = -a_1\). From the equation \(S_n = \frac{n}{2} (a_1 + a_n) = 0\), we can divide by \(\frac{n}{2}\) to get: \(a_1 + a_n = 0\) From this equation, we can conclude that: \[a_n = -a_1\] Thus, the sum of the finite arithmetic sequence is 0 if and only if the last term equals the negative of the first term. This completes the proof.

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Most popular questions from this chapter

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For Example 2 , the author wanted to find a polynomial \(p\) such that $$ p(1)=1, p(2)=4, p(3)=9, p(4)=16, p(5)=31 $$ Carry out the following steps to see how that polynomial was found. (a) Note that the polynomial $$ (x-2)(x-3)(x-4)(x-5) $$ is 0 for \(x=2,3,4,5\) but is not zero for \(x=1\). By dividing the polynomial above by a suitable number, find a polynomial \(p_{1}\) such that \(p_{1}(1)=1\) and $$ p_{1}(2)=p_{1}(3)=p_{1}(4)=p_{1}(5)=0 $$. (b) Similarly, find a polynomial \(p_{2}\) of degree 4 such that \(p_{2}(2)=1\) and $$ p_{2}(1)=p_{2}(3)=p_{2}(4)=p_{2}(5)=0 $$ (c) Similarly, find polynomials \(p_{j},\) for \(j=3,4,5,\) such that each \(p_{j}\) satisfies \(p_{j}(j)=1\) and \(p_{j}(k)=0\) for values of \(k\) in \\{1,2,3,4,5\\} other than \(j\). (d) Explain why the polynomial \(p\) defined by $$ p=p_{1}+4 p_{2}+9 p_{3}+16 p_{4}+31 p_{5} $$. satisfies $$ p(1)=1, p(2)=4, p(3)=9, p(4)=16, p(5)=31 $$.
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