Question

4. The 6th term of an arithmetic series is 13, the sum of the first 40 terms is 3420. Find

the sum of the first 30 terms.

Answer 1

2145

Explanation:

Here is the formula for the sum of the first n terms.

`S_n=(n)/(2)(2a_1+(n-1)d)`

Here is the formula for the nth term.

`a_n=a_1+(n-1)d`

We are given

`a_6=13`

`S_(40)=3420`

`3420=(40)/(2)(2a_1+(40-1)d)`

`13=a_1+(6-1)d`

We don't have a lot to work with but we can get it done.

Lets solve for
`a_1` in
`S_n=(n)/(2)(2a_1+(n-1)d)`.

Multiply each term by
`(2)/(n)`.

`(n)/(2)(2a_1+(n-1)d)(2)/(n) =S_n(2)/(n)`

Simplify.

`(n)/(2)*(2a_1+nd-d)(2)/(n) =S_n(2)/(n)`

Apply the distributive property.

`[(n)/(2) (2a_1)+(n)/(2) (nd)+(n)/(2) (-d)](2)/(n) =S_n(2)/(n)`

Simplify.

`(2(n)/(2)a_1+(n^2d)/(2) -(n)/(2)d)(2)/(n) =S_n(2)/(n)`

`(na_1+(n^2d)/(2) -(n)/(2)d)(2)/(n) =S_n(2)/(n)`

`(na_1+(n^2d)/(2) -(dn)/(2))(2)/(n) =S_n(2)/(n)`

`((na_1+(n^2d)/(2) -(dn)/(2))*2 )/(n)=S_n(2)/(n)`

`(n(2a_1+nd-d))/(n) =S_n(2)/(n)`

`2a_1+nd-d=S_n(2)/(n)`

`2a_1+nd-d=(2S_n)/(n)`

`2a_1=(2S_n)/(n)-nd+n`

Divide each term by 2.

`(2a_1)/(2)=((2S_n)/(n) )/(2) +(-nd)/(2) +(d)/(2)`

Simplify.

`a_1=(S_n)/(n) -(nd)/(2) +(d)/(2)`

Insert our answer for
`a_1` into
`a_n=a_1+(n-1)d`

`a_n=(S_n)/(n_(40) ) -(n_(40) d)/(2) +(d)/(2)+(n_6-1)d\\`

I continued my work on paper. I will attach it below.