The formula for the sum of arithmetic series is:

`S_n = (a_1+a_n)n/2`

where `S_n` - sum of n terms

`a_1` - first term of the arithmetic series

`a_n` - nth term or last term and

`n` - number of terms

Substitute `S_n=189` and `a_1=49` to the formula.

`189=(49+a_n)n/2`

`378=49n+a_n n` (Let this be EQ1.)

To determine the nth term, use the formula:

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

where d - is the common difference between terms.

Base on the given series, the common difference is

`d= 42-49= 35-42 = -7`

Then, substitute values of *a1* and *d* to the formula of nth term.

`a_n=49+(n-1)(-7)`

`a_n=49-7n+7`

`a_n= 56-7n` (Let this be EQ2.)

To solve for n, use substitution method of system of equations. Substitute EQ2 to EQ1.

`378=49n+a_n n`

`378=49n+(56-7n)n`

`378=49n+56n-7n^2`

`378=105n-7n^2`

Express the equation in a quadratic form `ax^2+bx+c=0` .

`7n^2-105n+378=0`

To simplify, divide both sides by 7.

`n^2-15n+54=0`

Factor left side.

`(n-6)(n-9)=0`

Set each factor to zero and solve for n.

`n-6=0 ` and `n-9=0`

`n=6` `n=9`

To check, substitute the values of n to EQ2 and to the formula of the sum of arithmetic series.

`n=6` , `a_n=56-7(6)=14`

`S_n=(49+14)6/2`

`189= 189` (True)

`n=9` , `a_n=56-7(9)=-7`

`S_n=(49-7)9/2`

`189=189` (True)

-------------------------------------------------------------------------------------------- **The arithmetic series 49,42,35,.... will have a sum of 189 if there are 6 or 9 terms in the series.**

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