Question

How do I find the nth term of the arithmetic sequence whose initial term and common difference d are given below? (A) `a_1=4, d=3` (B) `a_1= -8, d=5`

Answer 1

(A)`" "n ^(th)=3n+1`

(B)`" "n^(th)=5n-13`

Explanation:

the `n^(th)`term of an Ap is given by

`n^(th)=a+(n-1)d`

`(A)`

`n^(th)=4+(n-1)3`

`n^(th)=4+3n-3`

`:.n ^(th)=3n+1`

#(B)

`n^(th)=-8+(n-1)5`

`n^(th)=-8+5n-5`

`:.n^(th)=5n-13`

Answer 2

a) `color(white)("d")a_n=+4+3(n-1)`

b) `color(white)("d")a_n=-8+5(n-1)`

Explanation:

`color(blue)("Solution to part A ( in detail )")`

Given: `a_1=4 and d=3`

I am doing it this way to get you used to the way sequences are written. I will be building up the the `n^("th")` term

Let the position count in the sequence be `i`

`i=1->a_i->color(white)("dd")a_1=4+ 0d larr" First term"`
`i=2->a_i->color(white)("dd")a_2=4+ 1d`
`i=3->a_i->color(white)("dd")a_3=4+ 2d`
`i=4->a_i->color(white)("dd")a_4=4+ 3d`
`i=5->a_i->color(white)("dd")a_5=4+ 4d`

Looking at the behaviour we have:

Set `d=3 and i=n` giving:

`a_i=a_1+(i-1)d`

`a_n=4+(n-1)3`

Writing this as required by convention:

`a_n=4+3(n-1)`
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("Solution to part B - using the above")`

Set `a_1=-8, d=5 and i=n` giving:

`a_i=a_1+(i-1)d`

`a_n=-8+(n-1)5`

`a_n=-8+5(n-1)`