Respuesta :
Answer and explanation:
Given : An arithmetic sequence is given as 5, 11, 17, ………, 599.
An arithmetic series is in the form [tex]a,a+d,a+2d,a+3d,......a+(n-1)d[/tex]
Where, a is the first term and d is the common term.
(i) State the values of ‘a’, the first term of the sequence
The first term in the given sequence is a=5
(ii) Find the value of ‘d’, the common difference of the sequence
The common difference of the given sequence is d=11-5=6
(iii) Find T15, the 15th term of the sequence
The nth term of the sequence is given by, [tex]a_n=a+(n-1)d[/tex]
[tex]a_{15}=5+(15-1)6[/tex]
[tex]a_{15}=5+(14)6[/tex]
[tex]a_{15}=5+84[/tex]
[tex]a_{15}=89[/tex]
(iv) Find the total number of terms, n, in the sequence, where 599 is the last term
Last term l=599 , a=5 , d=6
Applying last term formula, [tex]l=a+(n-1)d[/tex]
[tex]599=5+(n-1)6[/tex]
[tex]594=(n-1)6[/tex]
[tex]n-1=\frac{594}{6}[/tex]
[tex]n-1=99[/tex]
[tex]n=99+1[/tex]
[tex]n=100[/tex]
(v) Find the sum of all the terms of the sequence
a=5 , d=11, l=599 , n=100
The sum of the sequence is given by, [tex]S_n=\frac{n}{2}(a+l)[/tex]
[tex]S_{100}=\frac{100}{2}(5+599)[/tex]
[tex]S_{100}=50(604)[/tex]
[tex]S_{100}=30200[/tex]
