Question

Write the first three terms of the series for which tn = 2(n+3). Find the number of terms of the series required to make the sum 228.

Answer 1

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

$t_n=2(n+3)$

Put n=1.

$t_1=2(1+3)=2\times 4=8$

Put n=2.

$t_2=2(2+3)=2\times 5=10$

Put n=3.

$t_3=2(3+3)=2\times 6=12$

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

$d=a_2-a_1=10-8=2$

The sum of n terms of an arithmetic series is

$S_n=\frac{n}{2}[2a+(n-1)d]$

$288=\frac{n}{2}[2(8)+(n-1)2]$

$288=\frac{2n}{2}[8+n-1]$

$288=n[n+7]$

$0=n^2+7n-288$

$0=n^2+19n-12n-288$

$0=n(n+19)-12(n+19)$

$0=(n+19)(n-12)$

Equate each factor equal to zero.

$n=-19,12$

The number of terms can not be negative, therefore the value of n must be 12.