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

The first three terms of the series are 8, 10 and 12. The number of terms is 12 to make the sum 228.

Explanation:

The series is defined as

`t_n=2(n+3)`

Put n=1.

`t_1=2(1+3)=2* 4=8`

Put n=2.

`t_2=2(2+3)=2* 5=10`

Put n=3.

`t_3=2(3+3)=2* 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=(n)/(2)[2a+(n-1)d]`

`288=(n)/(2)[2(8)+(n-1)2]`

`288=(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.