Answer:
The first three terms of the series are 8, 10 and 12. The number of terms is 12 to make the sum 228.
Step-by-step explanation:
The series is defined as
[tex]t_n=2(n+3)[/tex]
Put n=1.
[tex]t_1=2(1+3)=2\times 4=8[/tex]
Put n=2.
[tex]t_2=2(2+3)=2\times 5=10[/tex]
Put n=3.
[tex]t_3=2(3+3)=2\times 6=12[/tex]
The first three terms of the series are 8, 10 and 12.
It is an arithmetic series. The first terms is 8 and the common difference is
[tex]d=a_2-a_1=10-8=2[/tex]
The sum of n terms of an arithmetic series is
[tex]S_n=\frac{n}{2}[2a+(n-1)d][/tex]
[tex]288=\frac{n}{2}[2(8)+(n-1)2][/tex]
[tex]288=\frac{2n}{2}[8+n-1][/tex]
[tex]288=n[n+7][/tex]
[tex]0=n^2+7n-288[/tex]
[tex]0=n^2+19n-12n-288[/tex]
[tex]0=n(n+19)-12(n+19)[/tex]
[tex]0=(n+19)(n-12)[/tex]
Equate each factor equal to zero.
[tex]n=-19,12[/tex]
The number of terms can not be negative, therefore the value of n must be 12.
