Respuesta :
Let [tex]a_n[/tex] be the [tex]n[/tex]-th term in the sequence. Consecutive terms have a fixed difference [tex]d[/tex] between them, i.e.
[tex]a_n - a_{n-1} = d[/tex]
for [tex]n>1[/tex]. By substitution, we have
[tex]a_{11} = a_{10} + d[/tex]
[tex]a_{12} = a_{11} + d = a_{10} + 2d[/tex]
[tex]a_{13} = a_{12} + d = a_{10} + 3d[/tex]
and so on. Notice how the subscript and coefficient of [tex]d[/tex] on the right side add up to the subscript on the left side. This means
[tex]a_{30} = a_{10} + 20d \implies 83 = 26 + 20d \implies d = \dfrac{57}{20}[/tex]
and
[tex]a_{50} = a_{10} + 40d = 26 + 40\cdot\dfrac{57}{20} = \boxed{140}[/tex]
Answer: a₅₀=140
Step-by-step explanation:
[tex]a_{10}=26\ \ \ \ a_{30}=83\ \ \ \ \ a_{50}=?\\a_{10}=a_1+9d=26\ \ \ \ \ (1)\\a_{30}=a_1+29d=83\\[/tex]
Multiply both parts of the equation by 2:
[tex](2)(a_1)+(2)(29d)=(83)(2)\\2a_1+58d=166\ \ \ \ \ (2)[/tex]
Subtract equation (2) from equation (1):
[tex]a_1+49d=140\\a_{50}=140[/tex]