how to solve Find the 1st term, d, and the 5th term of an arithmetic sequence whose 3rd term is 35 and whose 10th term is 77

Here's how to solve this problem step-by-step:1. Understand the Arithmetic Sequence Formula:An arithmetic sequence follows the pattern:  [tex]$\sf{a_n=a_1+(n-1)d}$[/tex]Where:[tex]$\sf{a_n}$[/tex] is the nth term[tex]$\sf{a_1}$[/tex] is the first term[tex]$\sf{n}$[/tex] is the term number[tex]$\sf{d}$[/tex] is the common difference2. Set up Equations:We're given:[tex]$\sf{a_3 =35}$[/tex]  (3rd term is 35)[tex]$\sf{a_{10}=77}$[/tex] (10th term is 77)Using the formula, we can create two equations:Equation 1: [tex]$\sf{35 = a_1 + (3-1)d \implies 35 = a_1 + 2d}$[/tex]Equation 2: [tex]$\sf{77 = a_1 + (10-1)d \implies 77 = a_1 + 9d}$[/tex]3. Solve the System of Equations:We now have a system of two linear equations with two variables ([tex]$\sf{a_1}$[/tex] and [tex]$\sf{d}$[/tex]). We can solve this using substitution or elimination.  Let's use elimination:Subtract Equation 1 from Equation 2:[tex]$\sf{(77 = a_1 + 9d) - (35 = a_1 + 2d)}$[/tex]This simplifies to:[tex]$\sf{42 = 7d}$[/tex]Solving for d:[tex]$\sf{d = \frac{42}{7} = 6}$[/tex]4. Find the First Term ([tex]$\Large\sf{a_1}$[/tex]):Substitute the value of d (6) back into either Equation 1 or Equation 2. Let's use Equation 1:[tex]$\sf{35 = a_1 + 2(6)}$[/tex][tex]$\sf{35 = a_1 + 12}$[/tex][tex]$\sf{a_1 = 35 - 12 = 23}$[/tex]5. Find the 5th Term ([tex]$\Large\sf{a_5}$[/tex]):Use the formula [tex]$\sf{a_n=a_1+(n-1)d}$[/tex] with n = 5:[tex]$\sf{a_5 = 23 + (5-1)(6) = 23 + 24 = 47}$[/tex]Final answers:1st term ([tex]$\sf{a_1}$[/tex]) = 23Common difference (d) = 65th term ([tex]$\sf{a_5}$[/tex]) = 47

Answered by PrincessUmbriel | 2025-07-23