Question

The third term in an arithmetic sequence is 58 and the fifth term is 90. If the first term is \( a_{1} \), which is an equation for the nth term of this sequence? A. \( a_{n}=8 n+10 \) B. \( a_{n}=8 n-14 \) C. \( a_{n}=16 n+10 \) D. \( a_{n}=16 n-38 \)

Ask by Bartlett Graham. in the United States
Mar 10,2025

Upstudy AI Solution

Tutor-Verified Answer

Answer

The nth term of the sequence is \( a_{n} = 16n + 10 \).

Solution

The nth term for an arithmetic sequence can be expressed as \[ a_{n}=a_{1}+(n-1)d. \] Given that the third term is \[ a_{3}=a_{1}+2d=58, \] and the fifth term is \[ a_{5}=a_{1}+4d=90, \] we subtract the first equation from the second to eliminate \(a_1\): \[ (a_{1}+4d)-(a_{1}+2d)=90-58. \] This simplifies to: \[ 2d=32 \quad \Rightarrow \quad d=16. \] Substituting \(d=16\) into the equation for the third term: \[ a_{1}+2(16)=58 \quad \Rightarrow \quad a_{1}+32=58 \quad \Rightarrow \quad a_{1}=26. \] Now, the nth term is: \[ a_{n}=26+16(n-1)=26+16n-16=16n+10. \] Thus, the equation for the nth term is \[ a_{n}=16n+10, \] which corresponds to option C.

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Let’s break this down! In an arithmetic sequence, any term can be expressed as: \[ a_n = a_1 + (n-1) d \] where \( a_1 \) is the first term and \( d \) is the common difference. From the information given, we have: 1. The third term, \( a_3 = 58 \): \[ a_1 + 2d = 58 \quad \text{(1)} \] 2. The fifth term, \( a_5 = 90 \): \[ a_1 + 4d = 90 \quad \text{(2)} \] Now, we can subtract equation (1) from equation (2): \[ (a_1 + 4d) - (a_1 + 2d) = 90 - 58 \] \[ 2d = 32 \implies d = 16 \] Now, substitute \( d \) back into equation (1) to find \( a_1 \): \[ a_1 + 2(16) = 58 \] \[ a_1 + 32 = 58 \implies a_1 = 26 \] Now, substituting \( a_1 \) and \( d \) into the formula for the nth term: \[ a_n = a_1 + (n-1)d = 26 + (n-1) \cdot 16 = 26 + 16n - 16 = 16n + 10 \] So, the equation for the \( n \)th term is \( a_n = 16n + 10 \). Thus, the correct option is C: \( a_{n}=16 n+10 \).

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