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Chapter 7: Problem 26

Find the \(200^{\text {th }}\) term of an arithmetic sequence whose fifth term is 23 and whose sixth term is \(25 .\)

Short Answer

Expert verified
The 200th term of the arithmetic sequence is \(a_{200} = 413\).

Step by step solution

01

STEP 1: Identify the common difference

The common difference (d) is the difference between any two terms in an arithmetic sequence. Here it can be found by calculating the difference between the sixth term (25) and the fifth term (23). Therefore, \( d = a_6 - a_5 = 25 - 23 = 2 \).
02

STEP 2: Calculate the first term

In order to find the 200th term, we need to know the first term of the sequence. We can find it using the formula \( a = a_n - (n - 1) * d \). We know the fifth term (23) and the common difference (2), so by substituting those values, we find the first term: \( a = a_5 - (5 - 1) * d = 23 - 4 * 2 = 15 \).
03

STEP 3: Find the 200th term

We can now find the 200th term (a_200) using the arithmetic sequence formula a_n = a + (n - 1) * d. Substitute a = 15, n = 200, and d = 2 into the formula to get: \( a_{200} = 15 + (200 - 1) * 2 = 15 + 398 = 413 \). Therefore, the 200th term in the arithmetic sequence is 413.

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Most popular questions from this chapter

Suppose \(x\) is a positive number. (a) Explain why \(x^{1 / n}=e^{(\ln x) / n}\) for every nonzero number \(n\). (b) Explain why $$ n\left(x^{1 / n}-1\right) \approx \ln x $$ if \(n\) is very large. (c) Explain why $$ \ln x=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right) $$ [A few books use the last equation above as the definition of the natural logarithm.]

Evaluate \(\lim _{n \rightarrow \infty} n\left(\ln \left(7+\frac{1}{n}\right)-\ln 7\right)\)

Find the eighth term of a geometric sequence whose fourth term is 7 and whose fifth term is 4.

Evaluate \(\lim _{n \rightarrow \infty} \frac{3 n+5}{2 n-7}\).

Find the smallest integer \(n\) such that \(0.8^{n}<10^{-100}\).
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