Respuesta :
Answer:
50th term is 248
Step-by-step explanation:
the formula for nth term of arithmetic sequence is
[tex]a_n = a+(n-1) d[/tex]
'a' is the first term and d is the constant difference
n is the nth term
3rd term of arithmetic sequence is 13
Now we plug in 3 for n in the formula
[tex]a_3 = a+(3-1) d[/tex]
[tex]13 = a+2d[/tex] --------> equation 1
Given 4th term is 18
The difference between 3rd and 4th term is common difference 'd'
18 - 13 = 5
so d= 5
Now we plug in 5 for 'd' in [tex]13 = a+2d[/tex]
[tex]13 = a+2(5)[/tex]
13= a+ 10
subtract 10 from both sides
a= 3
We know a=3, d= 5 Lets find 50th term
[tex]a_n = a+(n-1) d[/tex]
[tex]a_{50} = 3+(50-1)5[/tex]
[tex]a_{50} = 248[/tex]
The [tex]50th[/tex] term of the series whose [tex]3rd[/tex] and [tex]4th[/tex] terms are [tex]13[/tex] and [tex]18[/tex] is [tex]248[/tex].
The general equation of the [tex]nth[/tex] term of an arithmatic progression is [tex]t_n=a+(n-1)d[/tex] where,
[tex]a[/tex] is the first term of the series
[tex]d[/tex] is the common difference of the series
[tex]n[/tex] is the total number of terms in the series
According tot he question,
[tex]t_3=13\\a+(3-1)d=13\\a+2d=13----(i)[/tex]
The common difference is calculated as-
[tex]d=t_4-t_3\\d=18-13\\d=5[/tex]
Substitute the value of [tex]d[/tex] in the equation (i)-
[tex]a+2(5)=13\\a=13-10\\a=3[/tex]
Now, to evaluate the [tex]50th[/tex] term of the series-
[tex]t_{50}=a+(n-1)d\\t_{50}=3+(50-1)5\\t_{50}=3+49\times 5\\t_{50}=248[/tex]
Hence, the [tex]50th[/tex] term of the series whose [tex]3rd[/tex] and [tex]4th[/tex] terms are [tex]13[/tex] and [tex]18[/tex] is [tex]248[/tex].
Learn more about arithmetic sequence here:
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