Answer:
[tex]a_{14}=\dfrac{52}{3}[/tex]
Step-by-step explanation:
General form of an arithmetic sequence:
[tex]\boxed{a_n=a+(n-1)d}[/tex]
where:
Given terms of an arithmetic sequence:
Substitute the values into the general formula to create two equations:
[tex]\sf\underline{Equation \:1}\\\begin{aligned}a_4 & = 14\\\implies a+(4-1)d & = 14\\a + 3d & = 14\end{aligned}[/tex]
[tex]\sf\underline{Equation \:2}\\\begin{aligned}a_{10} & = 16\\\implies a+(10-1)d & = 16\\a + 9d & = 16\end{aligned}[/tex]
Subtract Equation 1 from Equation 2 to eliminate a:
[tex]\begin{array}{l r l}& a+9d & = 16\\- & a+3d & = 14\\\cline{2-3}&6d & = \:\:2\end{array}[/tex]
Solve for d:
[tex]\begin{aligned}\implies 6d & = 2\\\dfrac{6d}{6} & = \dfrac{2}{6}\\d & = \dfrac{1}{3} \end{aligned}[/tex]
Substitute the found value of d into one of the equations and solve for a:
[tex]\begin{aligned}a + 3d & = 14\\ \implies a+3 \left(\dfrac{1}{3}\right)&=14\\a+1&=14\\a+1-1&=14-1\\a&=13\end{aligned}[/tex]
Substitute the found values of a and d into the general formula to create an equation for the nth term of the arithmetic sequence.
[tex]\implies a_n=13+(n-1)\dfrac{1}{3}[/tex]
[tex]\implies a_n=13+\dfrac{1}{3}(n-1)[/tex]
To find the 14th term, simply substitute n = 14 into the equation:
[tex]\begin{aligned}a_n & = 13+\dfrac{1}{3}(n-1)\\\implies a_{14}& = 13+\dfrac{1}{3}(14-1)\\&=13+\dfrac{1}{3}(13)\\ & = \dfrac{39}{3}+\dfrac{13}{3}\\&=\dfrac{52}{3}\end{aligned}[/tex]
Learn more about arithmetic sequences here:
https://brainly.com/question/27953040