Question

An arithmetic sequence is given as 5, 11, 17, ………, 599.

(i) State the values of ‘a’, the first term of the sequence

[2 marks]

(ii) Find the value of ‘d’, the common difference of the sequence

[2 marks]

Find T15, the 15th term of the sequence
[3 marks]

Find the total number of terms, n, in the sequence, where 599 is the last term
[4 marks]

Find the sum of all the terms of the sequence

Answer 1

Answer and explanation:

Given : An arithmetic sequence is given as 5, 11, 17, ………, 599.

An arithmetic series is in the form
`a,a+d,a+2d,a+3d,......a+(n-1)d`

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,
`a_n=a+(n-1)d`

`a_(15)=5+(15-1)6`

`a_(15)=5+(14)6`

`a_(15)=5+84`

`a_(15)=89`

(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,
`l=a+(n-1)d`

`599=5+(n-1)6`

`594=(n-1)6`

`n-1=(594)/(6)`

`n-1=99`

`n=99+1`

`n=100`

(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,
`S_n=(n)/(2)(a+l)`

`S_(100)=(100)/(2)(5+599)`

`S_(100)=50(604)`

`S_(100)=30200`