Find the sum of the first 40 terms of the arithmetic sequence if the first term is 16 and tenth term is 70
The answer is: 5320
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Answer (1)
Answer:What you're solving for You are finding the sum of the first \(\text{40}\) terms of an arithmetic sequence. What's given in the problem The first term of the arithmetic sequence is \(a_{1}=16\). The tenth term of the arithmetic sequence is \(a_{10}=70\). Helpful information The formula for the \(n\)-th term of an arithmetic sequence is \(a_{n}=a_{1}+(n-1)d\).The formula for the sum of the first \(n\) terms of an arithmetic sequence is \(S_{n}=\frac{n}{2}(a_{1}+a_{n})\) or \(S_{n}=\frac{n}{2}(2a_{1}+(n-1)d)\).Step-by-step explanation:How to solve First, find the common difference, then calculate the \(\text{40th}\) term or use the sum formula with the common difference. Step 1 . Find the common difference \(d\). Use the formula for the \(n\)-th term: \(a_{n}=a_{1}+(n-1)d\). Substitute the given values: \(a_{10}=16+(10-1)d\). So, \(70=16+9d\). Solve for \(d\): \(9d=70-16=54\). \(d=\frac{54}{9}=6\). Step 2 . Calculate the sum of the first \(\text{40}\) terms. Use the sum formula: \(S_{n}=\frac{n}{2}(2a_{1}+(n-1)d)\). Substitute \(n=40\), \(a_{1}=16\), and \(d=6\): \(S_{40}=\frac{40}{2}(2(16)+(40-1)6)\). \(S_{40}=20(32+39\times 6)\). \(S_{40}=20(32+234)\). \(S_{40}=20(266)\). \(S_{40}=5320\). Solution The sum of the first \(\text{40}\) terms of the arithmetic sequence is \(\text{5320}\).
